Transcript for:
Algebra 1 Regents Exam Review

[Music] hello and welcome to the e-math instruction algebra 1 regent review my name is Kirk Wiler and tonight we're going to be going through the January 2025 Algebra 1 exam now some of you may have been expecting me to go through the June 2024 algebra 1 exam i apologize we put out a social media post that said that I think um about a week ago or so and that was that was our mistake um we're going to be going through the January 2025 exam and the reason I wanted to go through this exam was because the June 2024 was the first exam for algebra 1 under the new next generation learning standards so I wanted to have one that was a little bit you know more tested let's say and I think that this is an excellent exam we're going to be going through it we're going to be using the TI Inspire calculator as we go um so make sure to have your calculator ready i would highly suggest that you have a copy of the exam sitting in front of you so that you can work on problems as I go through them all right keep in mind that even though we are live right now on Monday night you can also watch this at another time you can pause it if you want you can watch it on Tuesday morning especially or on Wednesday morning if you've let's say got a bio exam or something like that tomorrow but why don't we get going right away and before we kind of launch into the exam itself I'd like to go almost to the absolute end of the exam to remind you about the formula sheet so let's take a look at the formula sheet first all right now the formula sheet on the algebra 1 exam is specific to the exam itself and I am going to get there just things are kind of getting in my way a little bit here we go almost there there we go right now this reference sheet has a lot on it okay and a lot that's useful okay and some things that aren't so useful like telling you that this is a quadratic equation rather that they tell you that it's a quadratic function but whatever we won't quibble now one of the things you're almost guaranteed to use on the exam on Wednesday is the quadratic formula so this is a great one to have also the axis of symmetry equation for a parabola this is fantastic especially if you're looking for the turning point of a parabola they also give you the slope formula for a line they also give you two forms of a line they show you the general form of an exponential equation the form of an of um sort of compound interest equation an arithmetic sequence and a geometric geometric sequence formula how to calculate the inter quartile range and as well how to determine if you've got outliers in a particular data set all right now in our exam to be honest the two formulas that I know that we're going to use actually more than two formulas are the quadratic formula the slope formula and we'll be using the arithmetic sequence formula on this particular exam the rest of these I don't know maybe a little bit on the equation of a line but for the most part the quadratic formula we're going to use the arithmetic sequence formula and the slope formula so let's get into it and let's begin this exam let me head all the way back up to the top and let's jump right in with question number one when factored the expression x cubed minus 36x is equivalent to which of the following all right so factoring right it's a major concentration in algebra 1 it means to break something up so that it is an equivalent product now this particular expression x cubed - 36x the first thing I always want you to look for when you factor is a greatest common factor and I notice in this binomial both terms contain an x to the 1st x the 3r so my greatest common factor here will be an x to the 1st when I factor an x to the 1st out and don't lose that thing I will get my red pen showing up but I'll get x * x^2 - 36 very often in these situations then what is left within parentheses can be factored again now there's only two other types of factoring there is tromial factoring when you have a three-term polomial or there's the difference of perfect squares which is what we have going on here because we have x^2 minus 6^2 right 6 * 6 is 36 so this expression can be factored into x * x + 6 * x - 6 and we cannot factor it any farther than that and that tells us that the correct choice is choice 3 all right let's take a look at question number two which equation represents the line that passes through the points -1 comma 8 and 4a -2 great now there's a lot of different ways that you could do this problem not the least of which is start by just checking to see if these points lie on each individual line by substituting them in that's maybe not the most efficient way of doing it so what I'm going to do is I'm first going to calculate the slope of this line all right so let's do that all right the slope formula which is on that formula sheet is y2 - y1 all over x2 - x1 this is the change in y over the change in x also known as the rise over the run so I'm going to have -2 - 8 in my numerator and I'm going to have 4 - -1 in my denominator that's going to leave me with -10 in the numerator 5 in the denominator and that's going to reduce to a slope of -2 all right immediately that means that I can eliminate choices three and four because they have slopes of - one/2 and of course those are there to try to catch the person who flip-flops this slope calculation now what I don't know yet is whether it's choice one or choice two because I don't yet know what the y intercept is so let's go through finding that just in case something like this comes up in a free response problem and you have to come up with the full equation of the line right now what I know is that the line's equation looks something like this y = -2x + b now to find that y intercept what I'm going to do is take either one of my two points i'll just go with the first one and I'm going to substitute it into the equation so I'm going to put 8 in for y and I'm going to put -1 in for x and what that leaves me is only my b left so that's going to be 8 = 2 + b i can now solve for b by subtracting two from each side and I find b is equal to 6 and that tells me that the correct choice is choice one all right awesome finding the equation of a line if you know two points that are on the line that is an absolute bedrock skill for algebra 1 let's take a look at number three a geometric sequence is shown below 1/2 comma 2 comma 8a 32 comma dot dot dot what is the common ratio all right so a geometric sequence you studied two real types of sequences in algebra 1 an arithmetic and a geometric and we're going to see them both in the multiple choice section of this test now a geometric sequence you get by basically taking a term and multiplying it by this common ratio to get your next term and then you multiply that one by the same common ratio to get to the next term and the next term and the next term so in other words I go from 1/2 to 2 by multiplying by r 2 to 8 by multiplying by the same r 8 to 32 by multiplying by the same r etc so many of you can probably just look at this and say well what do I have to multiply 2 by to get 8 well that's four what do I have to multiply 8 by to get 32 well that's four so the common ratio is equal to four but if you needed to you could always do something like this you could say well if I take two and I multiply it by r I would have to get my next term which is 8 so I can just divide by two on both sides and I get my common ratio is equal to four all right that's it let's take a look at a pure terminology question number four what is the constant term of the polomial 2x cubed - x + 5 + 4x^2 all right so this really gets into some terminology specifically the constant term of a polomial now constant means not changing right and that is in direct opposition to a term that would contain a variable so the constant term is the term that doesn't contain a variable and there's only one term in this polomial that doesn't contain a variable and that's five all right as well another way to think about this is if I wrote this polomial in standard form and that means I write it in terms of decreasing exponents it would look like this 2x cubed + 4x^2 - x + 5 there's my constant term now watch out right if for some reason that five wasn't there my constant term wouldn't begx because x changes as x changes right so in that case the constant term would just be zero all right but typically it is the number that is sort of added or subtracted from the end all right let's move on to page two actually I guess it's page three but whatever let's take a look at number five a landscaping company charges a set fee for spring cleanup plus an hourly labor rate the total cost is modeled by the function C of X= 55X plus 80 in this function what does the 55 represent choice one the set fee for the cleanup choice two the hourly labor rate for the cleanup choice three the profit earned by the company for one cleanup and choice four the number of hours of labor required for one cleanup well this is really cool right we we know now I wish they had kind of like really laid this out a little bit better and what I mean by that is it would have been really nice if they had defined what X was and what we have to assume in here is that X is the number of hours that the person or that the company has been working all right in which case 55 which is the slope of this line right slopes are always rates always always rates right and therefore the 55 is the hourly labor rate for a cleanup side note the 80 the plus 80 right that is the set fee right so in other words they're going to charge you $80 when they come out to do this cleanup no matter how many hours they work additionally they're going to charge you $55 an hour for every hour that you work all right let's take a look at number six which expression is equivalent to 5x^2 - 2x + 4 minus the quantity 3x^2 + 3x -1 well I'm sure you've seen problems like this before right now these this first set of parentheses is doing absolutely nothing why because there's no number outside of it that is multiplying it right so really I don't need them there i can just rewrite that first set as 5x^2 - 2x + 4 but this second set of parentheses I really want to look at this as being multiplied by -1 and the upshot of that is that the sign of every single thing is going to change inside of that parenthesis making it - 3x^2 - 3x and here's the kicker + 1 really really critical now when I simplify this I simply need to combine like terms right 5x^2 - 3x^2 is going to be 2x^2 -2x and -3x is going to be -5x and pos4 and pos1 is going to be pos5 so 2x^2 - 5x + 5 okay let's take a look at number seven a system of inequalities is graphed on the set of axes below which point is a solution to this system all right well let's plot the four choices and then talk about how we can identify which point is a solution to the system so I'm just going to plot the four points the first point is 1 comma 1 that's hanging out right there 2 -2 that's hanging out right there one comma eight one two three have to do my counting well otherwise I'm gonna catch flack for some eagle-eyed students and then four comma 2 okay so right the key is a point will be a solution to a system of inequalities if it lies where both parts of the shading overlap each other Okay now two of these points are a little bit tricky here and here they almost appear like they could be in the portion of the graph where there is double shading if you will or where the shading overlaps but they are on the dashed line and points along the dashed line are not in the solution set points along the solid line are but points along the dashed line are not and both of these fall along that dashed line so they are not in the solution set this particular point is only in the shading that's up here this point on the other hand is in that section where the shading overlaps each other so the correct answer here is 4 2 and that's the kind of system of inequalities problem I like on the algebra 1 test i hate the ones where you actually have to graph the system so this one is quite nice right still tests you on some good information but pretty easy problem all right let's take a look at problem number eight in an arithmetic sequence the first term is 25 and the third term is 15 what is the 10th term in this sequence all right so I promised you we would have both a geometric sequence problem and an arithmetic sequence problem and here's our arithmetic sequence now let's kind of draw this out a little bit right what do we know well what we know is we know that the first term of this sequence is 25 and the third term is 15 and what we want to know is what the 10th term is but let's not worry about that just for the moment right when you have an arithmetic sequence what you do to go from one term to the next term is to either add or subtract the same number again and again and again now obviously to go from 25 down to 15 we must be adding a negative number right and we do it twice now it's probably pretty obvious that what I do is I have to subtract five twice in order to go from 25 to 15 if that's not obvious you could set up an equation to figure it out but hopefully it's pretty obvious that I'm going to subtract five or add -5 if you will right to do that now I I have to do that a bunch more times to get to the 10th term and you could sit there on your calculator and just do okay -5 - 5 - 5 - 5 you just have to be really careful right or you could use the formula that says that the nth term of an arithmetic sequence will be the first term plus the common difference which is -5 times one less than the number of terms that you're trying to figure out in other words to get to the 10th term I need to subtract five nine times not 10 times nine times think about it this way to get to the second term I only subtract five once to get to the third term I subtract five twice so I can get to the 10th term wow that is a funny looking a i can get to the 10th term by taking my first term 25 and adding to it 9 -5s or subtracting -5 nine times you can do that obviously on your calculator pretty quickly but I'm going to do it in my head i'm going to have 25 minus 45 which is going to be -20 you really have to watch out because the number one mistake that students will make here of course is that they'll see ah the 10th term so they'll say I'm going to take 25 and I'm going to subtract five 10 times to get to the 10th term which would then leave you with - 255 problem right all right let's keep going problem number nine when the formula p = 2 L + 2 W is solved for W the result is what all right well as much as possible I will always encourage you when you're solving a linear equation to literally undo what has been done to the variable that you're trying to solve for so I'm trying to solve for W well what do I see on this side of the equation with W well I see that W has been multiplied by two and then I see that 2 L has been added on right so in order to get W by itself the first thing I'm going to do is subtract the 2L from both sides and then I'm going to divide by two right so let me put that down here so that we have it a little bit better i've got P= 2 L + 2 W i'm gonna subtract 2 L from both sides that's going to leave me with P minus 2 L is equal to 2 W and now I'm just going to divide by two on both sides and the first thing I want to do is just see if this answer is sitting there is P - 2 L / 2 sitting there you bet it is that's choice two if it's not sitting there then there probably means that I have to do some kind of simplification but here my answer is right there so I move on to problem number 10 market Street Pizza kept a record of pizza sales for the month of February the results are shown in the table below of all the pizzas sold in February what percent were plain deep dish pizzas all right so in other words right plain deep dish pizzas there's 200 of them what does that represent in terms of a percentage of all the pizzas made well the first thing that you would want to do is literally add up all these numbers you got to do it right i'm not going to pull out the calculator to do it but like literally I would do like 300 + 80 + 120 plus dot dot dot right i got to add them all up all the way to the 70 and they make the numbers come out nice here there are 1,000 i'm going to erase that write it right back in it's still going to be red there were I don't even know why I try there were 1,000 total pizzas sold at Market Street Pizza in the month of February 200 of them are plain deep dish pizzas so let's do our typical percentage right we have 200 / 1,000 if you want to you could reduce that very nicely to 20 / 100 and that is 20% by definition right never forget that a percent or a percentage is how many out of a 100 you have so 200 out of a,000 is the same as 20 out of 100 and that is by definition 20% all right let's take a look at a really ugly looking linear equation number 11 when solving -2 * the quantity 3x - 5 = 9x - 2 for x the solution is what all right so they love throwing kind of an ugly linear equation at you on the algebra 1 regions exam you already had to solve kind of linear equations with variables on both sides in 8th grade math so what makes algebra 1 and 8th grade a little bit different is that they're not scared to throw fractions at you decimals things like that so let's just kind of start going through this the way that we would normally go through it all right let me first write down the equation again so I'm going to get -2 * 3x - 5 = 9 x - 2 so first things first I'll distribute that -2 through those parentheses which will give me -6x + 5 is equal to 9 x - 2 now I am fully confident that almost all of you that are watching this video would have absolutely zero problem solving this if it didn't have nine halves there but had just some nice whole number you know like let's say it was nine instead of nine halves what makes this problem more challenging is the fact that there's a fraction thrown in well you still want to do what you've always done so you want to get all your x's on one side all of your normal numbers on the other side etc etc so to do that the first thing I'll do is I'll subtract a 9 x from both sides now I don't mind fractions okay but I know that many of you probably don't love them okay that is just what it is and I respect that right there's a lot of people that don't like fractions that's why you want to make sure that you've got your calculator handy right you just do and remember what we're doing here right is we're doing -6 - 9 all right but the calculator has absolutely no issue with that right -6 - 9 gives me -21 halves okay that is what it is now again if you like getting common denominators and all that business you can definitely do that but I'm going to get -21 x is equal to -2 uh sorry -21 x + 5 = -2 all right so now I'm going to you know I'm just going to subtract 5 from both sides and then I'm going to take it over here so I have more room so I'm going to have -21 x is equal to -7 all right now again you know if I'm teaching this back you know before calculators are a little bit better I'm probably talking to people about well let's multiply both sides of the equation by -2 over 21 etc etc totally can do that and I apologize to the math teachers who desperately want to see me do that right now but again the plain fact is right there's absolutely nothing that prevents you from doing something like this okay literally dividing both sides by -21 halves -21 halves is just a number right that's all it is it's -10.5 anyway so I'm going to come back over here and remember what I've got now is I've got -7 / 21 hs so let me do it i'm going to do -7 / negative and that's right i'm going to put another fraction in here whoops or I'm going to try to put another fraction in there oh come on now we're going to get it eventually i promise 21 halves enter that's not right that's weird give me just a sec and that is because I should have had a 10 there not a five man what is going on here who didn't catch that that's terrible okay I knew something looked wrong there you go plus 10 plus 10 yeah Joey my producer is supposed to catch all of my mathematical errors you're fired kidding all right so sorry about that folks h thank goodness I worked out the answers beforehand okay so we've got -12 also good that that's not one of the answers we've got -12 / -21 hs let's let's take two on the calculator um and I'm going to just do a little enter here and I'll delete there let's try that again ah there it is okay much better all right so we end up getting 87s again my apologies for that i cannot believe I distributed that -2 through the parenthesis and got still five again that's um that's horrible all the students watching this right now are like really yeah okay you know you try to talk be in front of a camera have lights in your face and do math at the same time it's not easy all right let's move on oh my goodness okay plus it's late well it's not late it's like 6:30 let's take a look at number 12 the expression x^ the 2 a + b is equivalent to which of the following all right so this is a great problem that's just testing a very basic exponent law all right and I'd like to remind you right that when you have something of a given base let's say x to the m * x to the n then we add the exponents x to the m + n now most of the time when you use exponent laws you sort of go in this direction right you see something like x^2* x to 5th and you get x to the 7th all right but equality is a two-way street meaning that you can go this direction but you can also go this direction in other words I can think of x to the 2 a + b as x to the a + a + b in other words what is 2 a 2 a is just a plus a right and now I'm going to kind of move in this direction and I'm going to say well then that is just x to the a time x to the a + b and that is choice three all right let's take a look at problem 13 the inputs and outputs of a function are shown in the table below x is 0 f ofx is 0625 x is 1 f ofx is.125 x is 2 f ofx is 0.25 x is 3 f ofx is.5 x is 4 f ofx is 1 and x is 5 f ofx is 2 this function can best be described as choice one linear choice two quadratic choice three exponential and choice four absolute value all right so in this particular function we really want to be looking at these outputs now one of the great things and this will pretty much always be the case in algebra 1 is that the inputs are just increasing by one unit at a time now the most important thing is that they're increasing by a constant amount so it could be that they're going up by twos by threes by fives whatever but always the same amount if they aren't this problem is completely out of the scope of algebra 1 all right but they're always going up by ones what is happening to the y values well it may not be obvious when you look at the first few but by the time you get down to 0.25 0.5 1 and two it should be pretty obvious that what's happening each time is that the y values keep being the previous y value multiplied by two all right and when that's the case when you get a constant multiplier right which sounds a lot like a geometric sequence then what that leads you to is an exponential function all right if we had been adding the same amount to get to the next one kind of like an arithmetic sequence then that would have been a linear quadratic and absolute value are a little bit harder to spot those are situations where you might see the inputs or sorry the outputs go down and then go back up but distinguishing between the absolute value and the quadratic would actually be a little bit difficult all right most of the time in a situation like this you're looking for either linear or exponential linear you are adding or subtracting the same amount to get to the next y exponential you are multiplying or dividing by the same amount all right let's take a look at number 14 stephanie is solving the equation x^2 - 12 is equal to 7 x - 8 her first step is shown below given x^2 - 12 = 7 x - 8 step one x^2 - 4 is equal to 7x which property justifies her first step well the important thing here is understanding what her first step was you know when we compare what we're given i.e the original equation to step one what we notice is that the8 is no longer on the right side of the equation meaning that we must have added 8 to both sides of the equation and whenever you add or subtract the same number or expression from both sides of an equation you are using the addition property of equality the addition property of equality and that's it all right let's take a look at number 15 what is the sum of 8 * the 3 and 3 well look right don't make this harder than it is if they said to you "What is 8x plus x?" If they said "What is 8x plus x?" I would hope that the vast majority of you even as early as possibly sixth or seventh grade would say "Well 8x + x is equal to 9x." That's what it is it's not 9x squared it's not 9 * 2x it's I have 8 x's and I add another x I have nine x's if I have eight oranges and add an orange I get nine oranges eight llamas add a llama I get nine llamas if I have eight square roots of three and then I add a square root of three I get nine square roots of three right we're just keeping track of how many square roots of three we have so it's the exact same thing right 8 square 3 + 1 3 will be 9 3 don't make those harder than they have to be okay let's keep going ah I like this problem the dot plots below represent test scores for 20 students on a math test let me make this a little bit smaller so that we can kind of see them while reading the text the mode for this maths test is 80 and the median is 85 which dot plot correctly represents this data okay so let's start with the first piece of information the mode for this math test is 80 and recall that the mode is the value that occurs the most all right so given that we look at our four dot plots and we say well which of these four dot plots has the highest bar if you will or the most dots above 80 well choice one does and so does choice three right so both choice one and choice three have modes equal to 80 now the second piece of the information that they tell us is that the median is 85 now the median is the number right in the middle okay and that can be a little bit tricky on a dot plot let's take a look at number three though for a moment right if we look at number three what we notice is that this graph is completely symmetric across the value of 80 means it's exactly the same on this side as it is on this side and that means by default graph three the median is also equal to 80 right the median is also equal to 80 in choice three so it's really got to be choice one because choice one is the only one that has a mode of 80 right and possibly a median of 85 now we could actually prove that it's got a median of 85 by sort of counting from this direction and counting from this direction and we'd find that yeah the median is 85 but just by process of elimination we know that choice three has both a mode and a median of 80 and therefore it's only choice one that could possibly be the right answer all right let me go back to our larger screen and let's take a look at number 17 a function is graphed on the set of axes below the domain of this function is what all right so it's pretty obvious from the choices but to remind you right functions are all about taking inputs which are x values and converting them into outputs which are y values all right the domain is the set of all the inputs all the x values now the x values right basically go from here to the right okay so what is that x value well that x value is -2 all right so obviously we're getting all the x values that are greater than -2 but should it be this one all x x greater than -2 or should it be this one all values of x x greater than or equal to -2 these two by the way are out because they include x values that are out here and those aren't part of the graph all right well here's the key notice this open circle right and open circles are a way in math kind of like dashed lines of saying that all right the function goes all the way up to this point but doesn't include this point so the open circle means that the value x= 2 is not sorry x= -2 is not included in other words it's only x greater than -2 and not x greater than or equal to -2 okay so that's the last one on that page let's keep going question number 18 which ordered pair is a solution to the equation y -1 = 2 * x + 1/4 well this problem when I first looked at it I thought to myself oh they're testing the point slope version of a line but they aren't really because the point slope version of line would tell you that the point - 1/4 one lies on this line and that's not one of the choices so for me at least the easiest thing to do in this problem is to rewrite this in slope intercept form and this time I'm going to try to distribute the two through the parentheses correctly so let's see if Kirk can get it right on the second time here we go so I'm just going to like rewrite this equation y -1 = 2 * x + 1/2 right so 2 * 1/4 is 1/2 then I'm going to add one to both sides and I think I'm going to go decimal at this point because I don't know I like them better 1/2 + 1 is well 1 and a half or 1.5 okay so now to me there's a couple ways you can do this problem there's no way that just gives you the right answer immediately because there's an infinite number of points that lie on this line i think the best case scenario is that you just start substituting points in and seeing what you get but one thing to note right this form is a positive 2 * x plus 1.5 plus a positive number so if I look at choice 1 which is 0.75 0 there's no way that's correct 2 * 0.75 plus a positive number can't be equal to zero it just can't right likewise if I look at choice 3 and choice 4 2 * 2.5 plus a positive number can't be -6.5 you can't get a negative out of that and the same thing with number four 2 * 4 + 1.5 isn't 9.5 i honest I I don't even know why they did that so the correct answer is going to be choice 2 and let me show you really quick right i mean if I just simply do y = 2 * 1.25 + 1.5 that ends up being 2.5 + 1.5 and that's equal to 4 all right alternatively I suppose you could put this into your calculator and generate a table but that gets a little bit dicey because the x values of almost all of the choices are decimal so that gets a little bit trickier and and a little bit messy i do think one of the keys here though is to rearrange this so that it looks like that i think this is just easier to work with than that all right let's take a look at a unit conversion problem number 19 elena's fastest time for the 50 m dash is 7 seconds she wants to know how fast this is in inches per minute which expression can Elena use for a correct conversion all right well number one she's looking for a speed right she's looking for a rate a speed in inches per minute right so we have to start with me/ second in other words we have to start with 50 m divided by 7 seconds speed is always distance per time all right I say that and then all the runners out there that are like "Yeah but I think about my speed in terms of the minutes per mile that I run." What are you going to do anyway so you know that immediately eliminates choice one and choice two and now we're going with choice three and choice four now here's the key with all unit conversions okay with all unit conversions what you want to make sure to do is have the units that you want right we eventually want this to be in inches per minute right so we want inches in the numerator we want minutes in the denominator and we can't have meters and seconds left so they have to cancel right but what will cancel meters in the numerator will be meters in the denominator not meters in the numerator in fact if I have meters in the numerator times meters in the numerator I would have me squared i'd have like an area kind of unit right so I'm looking for units to cancel and if I look at choice three here I have meters in the numerator and also meters in the numerator those wouldn't cancel right now the seconds would cancel that would be good and they would leave me with minutes in the denominator which is what I want but they would also leave me with inches in the denominator and that wouldn't be correct on the other hand if I look at the correct choice which is choice four what I see is meters here cancel with meters here seconds here cancel with seconds here and I'm left with inches per minute which is my proper units so choice four all right let's take a look at number 20 the table below shows the highest temperatures recorded in August for several years in one town the inter quartortile range of this data or sorry of these data is which of the following all right so this is probably the first one that you honest to goodness want to use the statistical abilities of your calculator so let's take a look at that on the TI Inspire all right so I've got my TI Inspire and what I'm going to do is I'm going to add a page and I'm going to add a list and spreadsheet now of course one of the really unfortunate things about any of this type of review is that if you're not using the TI Inspire or the TI 83 or 84 Plus let's say you're using a Casio or Huelet Packard or Desmos I don't know whatever right it's going to be harder because it's going to be different buttons that you push etc etc the math is more or less the same what I'm going to be doing in column A is I am going to be putting in those temperatures i don't care about the years whatsoever i want to know the inter cortile range of the temperatures so let me put them in i'm going to put in 86 78 i want to make sure I get them right 84 95 81 77 88 and 93 and I do want to make sure that they're all correct all right but once I've done that once I've really looked over my data and made sure they're correct then on the TI Inspire I'm going to go to menu menu is my key to almost everything on here right i'm going to go to statistics that's choice four i'm going to go to statistical calculations and I want one variable statistics cuz that's all I've got my one variable is temperature so I hit enter number of lists one don't worry about that hit enter and all of my stuff is right there now I'm looking for my interquartile range and that won't just naturally show up here it's probably not long before it does but it's not right now the interquartile range is the difference between the first quartile and the third quartile or I should really say the difference between the third quartile and the first quartile q3 minus Q1 all right so if I just kind of go down look for them min max there's my Q1 and there's my Q3 so my first quartile value is 79.5 my third quartile value is 90.5 i just need to find the difference between the two let me get rid of this so let me write that down q1 is 79.5 q3 is 90.5 IQR and this is one of the formulas they give you but I'm hoping that since you've seen this a few years running you can then just do this calculation 79.5 and that gives me an inter quartile range of 11 awesome yep and that's the last one on that page let's keep going we're almost done with the part ones but we've got a little bit more work to do number 21 the function f ofx= x^2 is multiplied by k where k is less than -1 which graph could represent g ofx= k times f ofx all right now this one's a little bit tricky all right so we should all feel pretty comfortable that the graph of f ofx= x^2 looks something like this yeah loosely speaking it's a parabola that's open upward and as a side note it goes to the point 1 one ah the red it's so inconsistent all right so that's f ofx= x^2 now if I take any function whatsoever and I multiply the overall function by a number less than zero a number that's negative it's going to take that function and flip it across the x-axis all right so as long as k is less than zero right then what I'm going to get is I'm going to get something that looks like this now we know not only is K less than zero K is actually less than -1 one thing I do know right away though is that it's not going to be choice one it's not going to be choice three okay it can't be those two so it's got to either be choice two or choice four and now how do I know which one it is well right since k is less than -1 it would have to be a number like -2 or -3 or it could be even something like -1.5 right the important thing though is that its absolute value is bigger than one and what that means is not only will it reflect across the x-axis but it'll stretch out and become thinner than y = x² and for that we end up getting choice 4 now if you really had to what you could do is you could actually graph something like y = -2x^2 or y = -3x^2 one of the tricky things though is making sure that you interpret k less than -1 correctly because it could be really easy understandable to think oh less than -1 so like - 1/2 or - 1/3 but those numbers are actually greater than -1 right they're not less than -1 so just be careful on that all right let's take a look at problem 22 which graph is the solution to the inequality 6.4 - 4x is greater than or equal to -2.8 all right well this gives us a great um opportunity to solve a linear inequality in one variable so let's take a look i've got 6.4 - 4x is greater than or equal to -2.8 well the way that we solve a inequality is not that different than the way we solve an equation first thing I'm going to do is subtract 6.4 from both sides and that's going to give me -4x is greater than or equal to 7.2 if I've got my math correct and I think I do yes no 9.2 man I just I just cannot cannot do the arithmetic in my head right now 9.2 okay that's better now the one place that solving a linear inequality really truly differs from solving a linear equation is when you divide or multiply both sides of the equation by a negative number and that's what we're going to do right here and recall mostly it's division when we divide both sides of an inequality by a negative number the inequality symbol flips all right it flips and we won't get into why that is right now but if I divide both sides by -4 then what happens is it becomes x is less than or equal to and then I do 9.2 / -4 and we get 2.3 okay now what does that look like well it should be all the numbers that are equal to 2.3 or to the left of them because it's less than and that's choice four right now why is it not choice three why is it choice four that's because we have the equality right we have the the equal to as well as the less than all right excellent let's take a look at number 23 the number of fish in a pond is eight more than the number of frogs awesome the total number of fish and frogs in the pond is at least 20 if x represents the number of frogs which inequality can be used to represent this situation all right awesome so let's just make sure we got this right we know that the number of fish h there we go sometimes the pen just lags the number of fish in a pond is eight more than the number of frogs and x represents the number of frogs so obviously if x is equal to frogs which is a little bit weird then x + 8 must be equal to the number of fish now it says that the total number of fish and frogs in the pond is at least 20 right meaning that it could be 20 it could be 21 it could be 22 etc etc etc so when I add the x to the x + 8 I have to be equal to or greater than 20 right so I can symbolize all of that by saying x + x + 8 is greater than or equal to 20 i can now combine these two x's well to get two x's and I get 2x + 8 is greater than or equal to 20 excellent all right let's take a look at 24 which graph below represents a function that is always decreasing over the entire interval -3 is less than x is less than 3 all right so understand what this problem says or is asking is if we go on each one of these graphs starting at -3 and traveling to the right until we get to positive3 we always want to be going downhill we always want to be decreasing so let's just take a look at the first graph if I start here at -3 and I go to positive3 right then as I travel I'm always actually increasing the same thing is through true for choice three right if I start here at -3 and I go to positive3 I am always going up always always always travel from left to right right always always let x increase let's take a look at choice two well here right i increase and then I decrease doesn't look right so that one's not right either so it's got to be choice four but just to be clear about it right if I go here to -3 which is right here positive3 which is right here then it is always decreasing from -3 to 3 okay i think that is it for part one so let's just take a momentary break while I get a drink of lemonade gives me a little bit of calorie and a lot of hydration it's tough right when you're doing math in front of a camera and you've got a mic on you because number one I'm trying not to slurp right cuz that'll get caught on the mic trust me i mean you know all sorts of things do uh and I don't want to slurp i also don't want to spill the lemonade all over myself and I'm a I'm a very clumsy person so that is that's highly likely all right let me take one more drink i'm I'm tempting fate right now um I've been told Nico had caught that but Yeah he did nice nice we've got some local uh Red Hook students who are watching our video tonight i'm not going to call them out because I I don't want them to hate me um [Laughter] um so we are now going to move on to part two now part two problems on the algebra 1 exam are a conundrum for me some of them require very little work some of them require an inordinate amount of work but either way they're worth only two points right so you can get some partial credit you can get full credit or you can lose all credit right it's very very easy to get one out of two points on them so you know you want to make sure that you are sort of dotting all your eyes crossing all of your tees and really kind of kind of making sure that you you get everything done and done correctly so let's take a look at problem 25 and again let me like zoom out a little bit so that we can see the graph and the text a little bit better and let's take a look at 25 the graph below models Sally's drive to the store state an interval when Sally is traveling at a constant speed explain your reasoning now it is really really critical okay when you have a problem like this to look at the graph and understand what they're telling you along the x-axis is time minutes along the y ais is speed in miles per hour okay the reason that's so critical is sometimes the x-axis is time almost always time and the y-axis is distance from home or you know distance from school or something like that but here the y-axis is actually speed right so if you think about this right right right at the zero right my speed is zero and then I'm speeding up to 10 mph and then I'm speeding up to 35 mph and then I'm hanging out at 35 mph all through this interval and then I'm slowing down until I'm at a speed of 0 miles hour after 10 minutes all right so everywhere in here where the graph is flat right I'm moving at a constant speed so as long as you state a time interval that is somewhere between 5 minutes and 9 minutes you're fine so there's a lot of different answers here in fact I would argue there's an infinite number of correct answers to the first part of this problem right you could go very broad so you could say whoops five minutes to 9 minutes that's sort of the broadest interval but you could also say 6 minutes to 8 minutes you could be like 6.9 minutes to 7.1 minutes right as long as it's between five and 9 minutes you are absolutely fine now I would love for you to be able to explain your reasoning by just saying man look at the graph you know but you can't right so I think that the best thing that you can say in this case explain your reasoning is from five to 9 minutes the speed is always and let me Just make sure I got it right yep 35 miles per hour you know that's it right it's It's constant that's what constant means constant means not changing all right let's go back to fitting the width that looks good let's go on to the second one 26 graph the function f(x) = x^2 + 4x + 3 and there's another part of this problem state the equation of the axis of symmetry of f(x) so to me at least this seems like a problem that's a little bit too big for a two-pointer but hey it is what it is maybe they think it's okay for it to be a two-pointer at this point because you can use your calculator so let's talk about how we're going to use our calculator so what I'm going to do is I'm going to add a new page and I'm going to add a graph okay and what I want now let me just get to the page so I don't have to kind of keep flipping back and forth um what I'm going to do is I'm going to put in the equation for f ofx right here all right so my equation is x^2 + 4x + 3 now it's kind of nice because it does give me the graph but I don't want to just like oh let me just kind of like wing it right what I want is an as accurate of a graph as I can so I want a table of values now you can get to the table on the TI Inspire very quickly by hitting CtrlT or you can go to the menu and you can go to table and you can go to split screen table all right now I can already tell by looking at the graph that the majority of it is sort of if you will left of the y ais okay but what would be great is to have a little kind of xy table here and let me go back here right so let me let me throw some things down i've got my x i've got my y i don't even think the -58 is going to fit on there but let me write it down anyway -4 comma 3 -3 comma 0 -2 comma 1 -1 comma 0 uh 1 comma oops sorry 0 comma 3 and 1 comma 8 all right great that should be enough to get the job done right we can also see that beautiful symmetry 8 3 0 1 038 let's plot some points um I can't get in five comma 8 the y- axis is not big enough but I can go -4 3 um -3 0 -21 1 0 and 03 all right i just want to do as good of a job as I can passing through all the points i want to make sure I put arrows on the end right i want to do that because they didn't say just graph it over the following values of x and there we have it nice graph of our parabola nothing more that I have to do because it's the only graph there i don't have to label it with its equation i'm done now they say state the equation of the axis of symmetry well if you recall the formula sheet actually has a formula for the axis of symmetry you could totally go get that that formula by the way is x=b over 2 a and you could totally use that formula to state the axis of symmetry or right you could recognize that the axis of symmetry is the vertical line that splits the parabola into two mirror image portions right that's the the the vertical line that if I flip the parabola across it it would map onto itself but I have to state its equation right and what is true about every point along that vertical line is that the xcoordinate is equal to -2 so the axis of symmetry is x= -2 and again you would get that if you just plugged in the numbers here right so watch that watch this just really quick right that would be -4 over 2 * 1 which would be -4 / 2 and that would be x= -2 right but since I already have the graph I should feel very comfortable with just sketching the axis of symmetry on you don't have to do that though and then just writing down its equation x= -2 all right excellent let's keep going here we go 27 the function f ofx is shown in the table below state an appropriate value for m in the table so that f ofx remains a function and explain your reasoning i actually really like this problem all right because it gets at what at gets to the heart of what a function is right a function which is a very abstract idea especially for eth 9th and tth graders right a function is a rule that for every input i.e a value of x there is exactly one output so let me suggest a value for m that wouldn't wouldn't allow f ofx to remain a function let's say I said m was equal to zero okay now again this is an incorrect answer well if m was equal to zero then an input of zero would have an output of six and would also have an output of 9 and that wouldn't be allowed right let's say I said m was equal to six well that would be a problem now because then an input of six would have an output of five and it would also have an output of 9 so what's the real key here m can be any value you want as long as it's not a value of x that is already shown all right it you could put down m equals pi you could put down m= -1000 i would encourage you to be creative on this you know you you you really could a lot of people would probably just be like well let me just make m equal to 7 right so let me put that down because it's a it's a pretty obvious choice right seven is not already an x value now I also have to explain my reasoning all right so what is my reasoning well a lot of students will just simply say x values can't repeat in a function and that would be great right so you could say x values cannot repeat in a function i would love to have the more technical definition which is simply that for a given x value there can be only one yvalue and therefore x values cannot repeat in a function all right that's the word cannot it almost looks like the word carrot from where I'm standing but it is the word cannot anyway man my writing here we go number 28 solve x^2 + 8x = 33 4x completing the square now I really think this one shouldn't be a two-point problem i love the problem i just don't think it should be two points and we're going to see why in just a moment so now we have to solve a quadratic equation and they're very very specific we must do it by completing the square we cannot do it by factoring we cannot do it by the quadratic formula to be honest you could do it by either one of those methods but even if you do it completely correctly using one of those two methods you would at best get one out of two points because it says you must do it by completing the square all right that being said let's talk about how you do completing the square let me rewrite the equation x^2 + 8x is equal to 33 and maybe make that look like a plus all right so how do we complete the square well as long as the coefficient on x² is equal to 1 which it should be in algebra 1 all right then what I do is I look at the coefficient that's multiplying x and I find half of it so I take it and I divide it by two and I get four and then I take that number and I square it i multiply it by itself and I get 16 now what do I do with that 16 well I'm going to add it to both sides of the equation and it is really critical that you add it to both sides of the equation now if you do that correctly you take B you divide it by two and then you square it and you add it on the tromial that you get over here is what's known as a perfect square tromial meaning that it is a binomial that has been squared and specifically it's going to look something like this come on you know the nine's going to show up there it is now it's an ugly nine one more time there we go 49 now what's inside of this parenthesis well it will always be x plus half that b right so if that had been let's say -8 I would have had x - 4 here if it had been -10 I would have had x - 5 here if it had been positive 20 I would have had x + 10 here it's always x plus or minus whatever the half the b is quantity squared now we can solve this equation by inverses so I'm going to take the square root of both sides as my first step well as my first step from here on out all right so I'll get x + 4 and keep in mind when I take the square root of 49 I'm going to get both positive and -7 right so I've got x + 4 is equal to pos 7 and then I subtract 4 from both sides and I get x = 3 and I get x + 4 = -7 and I subtract 4 from both sides and I get x=1 and those are my two solutions i personally think this is a little bit too much work for a two-pointer primarily because there are so many different places where you can make a mistake you can really know the math quite well but let's say that you did everything right but you just had x + 4 = pos7 and you get x= 3 right you've done a lot of great work you know completing the square you know taking half of b squaring it adding on all of that stuff you just forgot the plus or minus right or let's say you add 16 to one side but you forget to add the 16 to the other side there's a lot of little places or you know you you pull a kirk and you do 7 minus 4 and you get four instead of three you know what I mean and you get the -1 i mean you've done so many things right you've just pulled a kirk and you've done simple arithmetic wrong so I like that i think I think we should coin that term pulling a kirk uh actually maybe not anyway um I don't really want you know mistakes associated with my name but hey who knows let's take a look at 29 if f ofx = -3x - 5 / 2 algebraically determine the value of x when f ofx = -22 all right now the number one mistake that students will make in this problem is they'll go "This thing's easy they just want me to put -22 in for x and figure out what f ofx is equal to but that's actually not what they want they want to know the value of x that you would plug in here to get an output of -22 in other words you need to solve the following equation -3x - 5 all / 2 = -22 all right so it would be so tempting so tempting to substitute -22 in for x but keep in mind they want you to determine the value of x right which would be weird if they gave you x= -22 and then asked you to determine the value of x all right but let's solve this equation and let's do it by inverses right so take a look at x it's been multiplied by -3 then we've subtracted 5 then we've divided by two and we've gotten -22 let's just undo all of that in the opposite order so the last thing that we did was divide by two so the first thing I'm going to do is multiply by two that's going to cancel and that's going to give me -3x - 5 is equal to -44 then I'm going to add five to both sides okay that's going to give me -3x is equal to -39 and then I'm going to divide by 3 on both sid -3 on both sides and I'm going to get x is equal to 13 all right and that's it but this is actually kind of a tricky problem because again you know you got very used to in algebra 1 being given a function and sort of asked what what is f of five what is f of seven what is f of -10 right so it'd be very very understandable to see this problem read it too quickly and be like I'm going to put -22 in for x and figure out what f ofx is but I want the x that gives me a y value of -22 all right number 30 let's take a look rationalize the denominator of the fraction below express the solution in simplest form okay so this is a very sort of like technical piece right rationalize the denominator this word rationalize right the denominator right now is the square root of two that is an irrational number and for historical reasons that we aren't going to even touch right now it is standard to not have an irrational number in the denominator of a fraction now there's nothing really technically wrong with it but again for historical reasons we tend to not want them there so how do we get rid of the irrational number in the denominator well the way that you do it is actually quite simple whatever that number is so in this case it's the square of two what I'm going to do is I'm going to multiply both the numerator and the denominator by that square root now it is very important that you multiply both the numerator and the denominator by the same number because as long as you do that you're actually just multiplying the overall fraction by the number one right any number divided by itself except for zero is equal to one so the square of two divided by the square of two is equal to one and I can always multiply by the number one now why do I do that well in the top we just get 4 * the of two so nothing special there but anytime I multiply a square root by itself I'm squaring it and I'm getting rid of the square root square of two * square of two is just two and now I have rationalized the denominator the only issue here is it's not in simplest form all right so what do I do at this point well I can think of this as 4 / 2 * 2 over 1 and 4 over 2 is just 2 so I just get 2 * the And that is simplest form okay let's keep going part three let's take another break we're getting there folks we're moving through this thing if you were with me for the geometry review a week ago well that would be weird unless you're a teacher then it's completely okay for you to be both teaching geometry and algebra 1 but if you're taking both the algebra 1 exam and the geometry exam well I feel for you uh and it would definitely be a little strange i would suggest that you drink more lemonade as well all right so part three problems part three problems are lengthy problems they're worth four points each lots of lots of opportunity for uh for I almost said for spare credit for partial credit i don't know what spare credit would be it's like ah there's some credit just hanging out over there for partial credit you know on these problems but often times you know these part three questions are a couple part two questions or maybe even three of them lumped together which seems manifestally unfair anyway let's take a look at our first one in problem 31 Alex had $1.70 in nickels and dimes on his desk there were 25 coins in all write a system of equations that could be used to determine both the number of nickels n and the number of dimes d that Alex had and then we'll get to the second part in a second but notice the first part of this problem simply says to write a system of equations that could be used to solve for the number of nickels n and the number of dimes d that Alex had now when you do systems of equations doing ones with money are classics right and I hope you did enough of these that you're like I know exactly how to do this but if you didn't right that's why we're here tonight to review you know what what what you should have should know and what you will have to do on Wednesday now one thing that's really critical here is that N is the number of nickels and D is the number of dimes it is not the amount of money that you have in nickels and the amount of money that you have in dimes literally it's like okay I I take some change out of my pocket i've got you know five dimes and I've got three nickels you know or something like that okay it is not the amount of money that you have it's how many of each coin you have now one equation in our system should be easy there are 25 coins in all and what that means is if I take the number of nickels and I add the number of dimes I have to get 25 not 25 cents just 25 coins right now the other piece of information they tell me is that I have a total of $1.70 in nickels and dimes so how do I incorporate that well first let's talk about how much money we have in nickels to figure that out what I would do is take the number of nickels I have and multiply by how much a nickel is worth which is 0.05 so 05 * n is the amount of money that I have in nickels then if I add to it plus10 * D I have the amount of money I have in dimes and I put those together and I get 1.70 right so the amount of money I have in nickels plus the amount of money I have in dimes is equal to the total amount of money and this is all you want to do for that first part of the problem because now let's take a look at the second part of the problem use your system of equations to algebraically determine both the number of nickels and the number of dimes that he had okay great well this is a classic system of equations that you solve using a process known as elimination now some of you might be able to might want to do substitution on this but generally speaking right what I'm going to do is I'm going to try to eliminate either n or d by getting them to cancel out when I add the two equations and what I choose to do in this problem is I'm going to take this first equation and I'm going to multiply both the left side and the right side by.05 you could also multiply by.10 but the first thing I'm going to do is I'm going to take this equation and I'm going to multiply both sides by.05 and I have to multiply both sides of the equation by it all right let me just kind of get my my my cheat sheet where I've worked out all the numbers already just to save ourselves a little bit of time and what that's going to be then isn minus.05d is equal to -1.25 it is very critical that you distribute the.05 05 and you multiply both sides of the equation by.05 huge error to only multiply the left or the right that's like unbalancing the equality scale now what I'm going to do is I'm going to write this equation right underneath it so I'm going to write down 05N +.1 D is equal 1.70 and now what I notice is that I've got opposites additive opposites.05N and positive.05N so when I add these two equations together those two will cancel when I add these two together I will get 0.05D and when I add these two together I will get 0.45 it's called the method of elimination because you're always looking to eliminate one of the two variables and then be left with an equation that has only one variable in it now I can divide both sides by 05 of course you can do that on your calculator if you don't feel comfortable doing it in your head and that tells me I have nine dimes now I also have to figure out how many nickels I have i can take that nine dimes and I can substitute it into this equation or this equation to figure out how many nickels I have but quite frankly I know I have 25 coins in all so this is going to be the easiest equation to do it with and in fact I can really just say well the number of nickels I have is 25 minus 9 and that means I have 16 nickels all right and that is a nice system of equations problem let's move on to the next part four all right finally linear regression problem number 32 the table below shows the average heart rate X and the calories burned Y for seven men on an Olympic rowing team during a 1-hour workout class so you know you got your average heart rate and then your calories burned the first part of the problem asks us to write the linear regression equation that models these data rounding all values to the nearest 10th all right great well we have to go back to our TI Inspire or whatever calculator you're using there is literally no way to do this problem without using your calculator so let's do it let's go into the calculator now I actually already have this i'm going to open a document in here because I didn't want you to have to sit around waiting for me to like enter all the data but on the TI Inspire you open up a spreadsheet again all right i like to label the columns so the first column the X column was the heartbeat the second column was the calories burned so I've got you know 135 heartbeat 725 calories etc now I want the linear regression equation just like before when we did stats I'm going to hit menu all right i'm going to go to statistics choice 4 i'm going to go to statistical calculations and linear regression choice three all right so I'll just hit choice three now this is why I like to label my sort of my two columns because remember what I want is my heartbeat as X i want my calories as Y and I don't really have to deal with anything else in this table just that i hit okay and I get a bunch of stuff right here right here it says regression equation m * x + b where m is 9.11777 etc and b is 527.55 etc now you want to watch out because right you have to round to the nearest 10th it's a four-point problem and if you botch one of the rounding you lose a point if you botch both of them you lose two points right so you could just see yourself rounding to the nearest hundredth and at best you're now two out of four which is a shame right but you know all that said we can now write down our equation with our parameters rounded to the nearest 10th 9.1x minus 527.6 six all right so that is the best line that fits this data let's take a look at the second part of the problem state the correlation coefficient rounded to the nearest 10th well I'll tell you this right the correlation coefficient which is also known as the R value is never in practical purposes rounded to the nearest tenth it is almost always rounded to the nearest hundredth if not thousandth right but they want it rounded to the nearest tenth so let's take a look at our calculator output again right here I need to go down a little bit and there is my r value right 900223 maybe the reason they wanted it to the nearest 10th is because to the nearest 100th it would have been 0.90 and that would have been confusing I don't know but to the nearest 10th This thing is r= 0.9 all right and the last part of the problem state what the correlation coefficient suggests about the linear fit of these data what does it suggest about the linear fit well there's two things to note one r is positive and what that means is that it is a positive fit it's a positive correlation and two the R value is pretty close to one right which means it's a pretty strong fit so what I want to say is that there is a strong positive fit and that's it there's a strong positive fit and I'm done with that problem all right so a lot of this problem was more about terminology than anything else but isn't that a case in on on many tests right maybe not as much on math but sometimes all right here comes our second quadratic 33 using the quadratic formula solve x^2 + 4x - 3 = 0 express your solution in simplest radical form and I kind of wish they had said solutions but uh you know what are you going to do right quadratic equations almost have more than one solution in fact they almost always have two solutions but uh let's not quibble all right now the quadratic formula right is definitely one that's on that formula sheet so if you see something that says use the quadratic formula and you have any doubt about your memory of the quadratic formula look at the sheet all right not surprisingly I've seen the quadratic formula now for a few years so I'm just going to write it down.b plus or minus b ^ 2 - 4 a c all / 2 a all right now maybe you know they would make this a little more difficult put some numbers on the other side of the equation i don't know stuff like that but they they really set us up here so remember a is the number multiplying x^ squar b is the number multiplying x and c is the constant of the quadratic equation so in this case a is equal to 1 b is equal to 4 and c is equal to -3 and I now need to just be careful in my substitution so let's take a look we're going to get x = b which is -4 plus or minus the of 4^ 2 - 4 * 1 * -3 all over 2 * 1 now the next step always when you do the quadratic formula is simplifying what is under the square root symbol that is known as the discriminant you don't particularly need to know that term for algebra 1 for algebra 2 maybe algebra 1 not as much be careful with this right especially if b is negative now it's not negative here so that's a good thing right but if it is negative you have to be very careful and I would highly suggest just taking your calculator out right let me add a calculator page and literally without the square root figure out what that is remember that's 4^ 2 - 4 * 1 * -3 i'm literally typing it in pretty much exactly i can't seem to move the calculator pretty much exactly the way it is right there right so I hit enter and that's 28 so 28 is the number under the square root all right so I'll get -4 plus or minus the of 28 all over 2 now that will get you two out of four points all right getting that far the rest of it is simplest radical form all right now what that means is I've got to look at the square root and I've got to simplify it by simplifying any perfect squares that go into 28 now the largest perfect square that goes into 28 is simply the number four right so if you think about it right I can take the of 28 and I can break it into the of 4 * the 7 and that becomes 2 * the of 7 all right so I'm going to replace the of 28 with 2 * the of 7 now what's interesting is I looked at the state's official documentation in terms of like this problem and they actually said you get full credit for just that which I find fascinating because it is simplest radical form but it's not simplest form simplest form would mean that I would then go on and do the following -4 / 2 plus or - 2 / 2 7 and that would be -2 + or minus the 7 so this to me is simplest form but since they said simplest radical form technically this is an okay place to stop as well but that is not all right let's keep going here we go problem number 34 solve the following system of equations algebraically for all values of x and y all right so a system of equations we've already seen one right we saw the one where we had the uh the um the nickels and the dimes and that was two linear equations in this case what we have is we have the equation of a quadratic function also known as a parabola and the equation of a linear function whose graph is a line okay now I could graph the parabola graph the line and find out where they intersect to solve this system except they want me to solve algebraically and in this case the easiest way to do it is by doing what's called substitution so what I'm going to do is I'm going to substitute one of these two into the other equation and just because I like to have x squares on the left side of the equations and x to the 1st on the right side of the equation what I'm going to do is I'm going to take this x^2 - 7 x + 12 and I'm going to substitute it in for that y so I'm going to begin by saying x^2 - 7 x + 12 is equal to 2x - 6 a lot of teachers understandably will call this setting the two equations equal to each other and that is a great way to put it especially when both of them are y equals now how do I solve this well I've got a quadratic equation now we've already solved a quadratic equation this evening by completing the square and by using the quadratic formula right now you could do either one of those but I'm going to take advantage of this and I'm going to solve it by factoring and using what I call the zero product law so I'm going to move this 2x - 6 over to the other side of the equation to get everything equal to zero that's going to be my next step so I'm going to subtract a 2x and subtract a 2x that's going to give me x^2 - 9x + 12 is equal to -6 and then I'm going to add a six and add a 6 and that's going to give me x^2 - 9x + 18 is equal to zero all right great now again you can solve this using the quadratic formula you can e even if you're gutsy solve it by using completing the square i think I would avoid that given that you'd have to take half of an odd number and square it so that would get not so nice but one could do it i'm going to do it by factoring and using the zero product law so all right I'm thinking to myself I want two numbers that have a product of 18 and a sum of -9 and those two numbers are -3 and -6 right -3 * -6 is pos8 and -3 + -6 is -9 and that means that this tromial factors into x - 6 and x - 3 or x - 3 and x - 6 it doesn't matter all right so now I'm going to take x - 6 and set it equal to zero get x= 6 and I'm going to take x - 3 and set it equal to zero that's not a six that's a zero and I'll get x= 3 so those are the two xcoordinates where this line intersects this parabola but to properly solve the system I need to not just get the x values I also have to get the y values i can do that by taking those x values and either substituting them into this equation or substituting them into this equation and I would just humbly suggest that the easier of the two equations to work with is y = 2x - 6 so we take this and we substitute it into y = 2x - 6 and we'll get y = 2 * 6 - 6 that's y = 12 - 6 and that's y = 6 so when x is = 6 y is equal to 6 not that that's confusing at all here we will also substitute it in so 2 * 3 - 6 that'll be 6 - 6 and that'll be y = 0 that is a great way of sort of like showing your work if you wanted to you could summarize your answers as two coordinate pairs 66 and 30 but you don't have to do that as long as it's very clear what y values go with each x value all right awesome and we are to our last problem so let's take one more break drink just a little more lemonade we're cruising along we're probably going to be finishing up this review in a nice solid 2 hours which is great feel like the geometry review took us like 3 and 1/2 hours i don't think it actually did but there's the slurping all right got to slurp all right part four you know you're getting close to the end when you hit part four cuz there's absolutely just one part four problem it's worth six points so it's going to be pretty involved but again it's not likely to just be like "Here's a really hard equation solve it for six points that's not going to be the case right it's going to be a problem that they're going to kind of walk you through right in like little subp parts all right so let's take a look at problem number 35 anna plans to spend $30 on balloons and party hats for her daughter's birthday party including tax balloons cost $2 each and party hats cost $1.50 each the number of party hats Anna needs is twice as many as the number of balloons if X represents the number of balloons and Y represents the number of party hats write a system of equations that can be used to represent this situation wow i got to say they're really knocking it out of the ballpark on this particular test in terms of systems of equations right we saw the system of equations that we had to create right with the nickels and the dimes then we have the system of equations that we just saw solved with the line and the parabola and now we're going to create a system of equations that involves I guess hats and balloons now keep in mind right that X represents the number of balloons and Y represents the number of party hats so Anna plans to spend $30 in total the balloons cost $2 each and party hats cost 1.50 each woohoo all right party hats yeah all right now that being said right what are we going to do i think the first thing I'm going to do is actually deal with the money part right very similar to the nickels and the dimes right what I'm going to do is I'm going to take the number of balloons which is X and I'm going to multiply it by $2 to figure out how much money I'm spending on balloons so that's going to be 2 * X then I'm going to add to it $1.50 50 times the number of party hats right which is y and when I add those two together I get 30 so 2x + 1.50 y or just 1.5 y whichever is equal to 30 easy and that's actually a very very similar to that nickel and dime problem right where I had 0.05 time the nickels plus10 * the dimes was equal to I think it was $1.70 something like that so where does my second equation come from well that comes from the fact that the number of party hats that Anna needs is twice as many as the number of balloons all right and that is a direct translation so the number of party hats which is Y is going to be twice the number of balloons now I I really think with all due respect to Anna that she's probably got you know sort of her priorities mixed up you know I this just means that like every kid's going to get a hat but then only half of them are going to get a balloon see I think it really should be that the number of balloons is twice the number of hats so that every kid could have two balloons but what do I know you know I'm only a parent anyway so here is our system you know I've overthought this problem here is our system of equations and that's the first part of the problem and it's going to be worth two out of the six points all right blank space taylor Swift would be so happy okay so let's take a look at the second part of the problem graph your system of equations on the set of axes below awesome great my system of equations that's on the other page all right now what were my two equations my two equations were 2x + 1 y is equal to I just want to make sure i believe it's 20 oh my goodness no it's 30 it's good that I checked i would have gotten all the way through this problem and been like why isn't this working and then the other one is y = 2x let me kind of zoom out a little bit so that we can see a bit more of this all right that's better now one thing that's really nice about this and I'm going to go even a little bit further out there we go is that you know on the x-axis we have the number of balloons on the y-axis we have the number of party hats that's good because that's what the two variables represent and what's really nice is that the two scales are the same and they're all scaled in one units so the first of these two that I'm going to graph is the easy one or at least what I consider to be the easy one which is y= 2x right that should be simple all I want to do is plot points right literally I want to plot points where y is twice x you know so if I did that over here just did a little table literally I would have things like well 0 0 cuz 0 is twice 0 weird but true uh 510 20 etc right so those are points like here right 510 here 10 20 here right and you definitely then want to use a straight edge to draw this line okay but yeah and you could always fill in more points if you wanted i'm going to kind of freehand it just so I'm not wasting your time and then I definitely want to label this y = 2x okay and again please please please make sure to do it with a straight edge it's already probably bad enough that it's going to throw off my final answer now the harder one 2x + 1.5 y = 30 as is that is a hard hard thing to graph okay so what I want to do is I actually want to rearrange that to get it into y equals form now it doesn't have to look like y= mx plus b because I'm going to use my calculator to generate a table of values um let me just zoom in a little bit so that you can see me do the work on this a bit more all right and I'm going to actually erase this so that I can kind of manipulate this now remember I want to get y equals so the first thing that I'm going to do is I'm going to subtract 2x from both sides so I'm just going to have 30 - 2x and I can write it that way or I can write it as -2x + 30 and then I'm going to divide by 1.5 and I'm going to divide remember all of this by 1.5 now you could distribute that division but I'm just going to leave it like this now the reason I'm going to leave it like that is because I can all right now I know that sounds stupid but I can literally put it into my calculator just like this all right so let me do that go into my calculator i'm going to add a page add graphs all right and I just want to put that in exactly like this right so I'm going to have 30 minus 2x in the top and 1.5 in the bottom and I'm going to hit enter now I don't see anything come up on the screen and that's okay all right i just want this for a table of values that's the only reason so I'm going to go into my menu i'm going to go down to table split screen table all right and now let's whoop oh oh boy sometimes this thing just goes a little nuts so let's actually create an XY table we'll put it right here xy for zero i have 20 now my numbers for 1 and two are horrible but for three I have 16 for 6 I have 12 for 9 I have 8 12 I get four and 15 I get zero and that'll be plenty okay so now let's graph that table of values all right 020 is right here again let me uh let me go out a little bit so 0 20 3 and 16 6 and 12 yeah that's 6 and 12 uh 9 and 8 12 and 4 come on and 15 0 so that is this particular line and now I'm just going to label it 2x + 1 y = 30 maybe draw a little line okay great so now I graphed my system now again right we we could even generate a table for y= 2x in our calculator i'm hoping you don't have to do that but if you do you know go ahead and do it so next part of the problem and let me zoom in because now it's kind of hard to see again here we go state the coordinates of the point of intersection of your lines all right well this should be easy enough my intersection point is right there that's why actually using a straight edge is so very critical that's at the point 6 12 really come on there we go 6 12 all right so I have my intersection point and then finally right remember this is a six-point problem explain what each coordinate means in the context of the problem right what was the context of the problem right you know here we had Anna she's buying balloons x she's buying party hats why for some reason she's buying twice as many party hats as balloons which is going to cause fights okay mistake but it is what it is so what do we say well we're going to say Anna must buy Six balloons got to make sure yep x are the balloons again I'm just questioning it and 12 party hats and that's it all right so that's the last problem there is only one of those right we go here and we again have just scrap paper and please do keep that in mind right on the test you know you've got some nice pieces of scrap paper you've got some pieces of graph paper multiple pieces of graph paper that will not be scored will not be scored right but they're there in case you need them you know in case you get done with the test early and you need to doodle i don't know you know uh but the blank the blank sheet of paper is actually really quite helpful if you need to do kind of some scratch work and things like that this is kind of a nice sheet to have right and of course you don't also want to forget that after all of that you've also got your formula sheet with that quadratic formula axis of symmetry and slope formulas oh and also that arithmetic sequence formula that we used all right let's go all the way back to the beginning and wrap this review up all right so you know for many of you this will be one of the most highstakes math tests that you will take it certainly is probably the highest stakes math test that you'll have taken up to this point right you have to pass a region's math exam in order to graduate from high school in New York State right and typic typically the easiest one to pass is algebra 1 okay now you've got 3 hours for this test you've got a calculator at your disposal that can do a lot of things right especially with fractions and decimals and things like that we reviewed tonight how you can enter equations into the graphing utility and create a table of values that will help you with graphing right and you know you've got a lot of time on this test most students will finish this test not all but most students will finish this test in under two hours right and then you've got all this extra time to really go through the test and make sure you've done the best job that you can on it don't leave any multiplechoice questions blank right there is no penalty for guessing on one so make sure to take good educated guesses especially if you can eliminate some choices and realize that if you've been working hard this year you're going to pass this test you're going to do better than just passing this test you're going to do well on this test right you're going to get an 85 you're going to get a 95 whatever you're going to do well i have full faith and confidence in anyone that's watching this video tonight that you can master this test and you can like send yourself off on summer vacation knowing that yeah next year it's geometry but that you mastered algebra 1 all right I really want to thank all of you for joining me tonight you know I know that like we we sent out the incorrect you know test form originally and again I apologize for that but you know I hope that you were with me that you were working through this test and that you review hard for W Wednesday's exam get a good night's sleep eat a good breakfast and be ready to take that test because you can handle it for now I just want to thank you for joining me again for the e-math instruction regions review for algebra 1 my name is Kirk Wiler and until I see you again keep thinking and keep solving problems