in this video I'm looking at the application problems at the end of our 61 guided notes so let's take a look at at question number 15 so we had these two baseball teams interested in signing a a pitcher all right so one um team has already made an offer and then the agent was communicating to the other team and told them that they needed to double that offer and add an extra million per year and then they the team would be um presenting an offer that that they were interested in okay the player is Seeking a contract of $18 million per year so we want to find out what was that first team's offer so let's let X be the first team's offer we don't know what it is that's what we've got to figure out what we do know is that if they were to double that and add a million so double the offer and add and I'm just going to use one instead of 1 million So my answer will be have will be then in millions so if they double that that means 2 * X and add 1 million then that's supposed to be um an offer that would be what the player was interested in or the player's agent was interested in and we know the player is Seeking a contract of 18 million per year so this should equal 18 all right so then we can solve that so we can subtract 1 and get 2x = 17 and then divide by 2 and we wind up with 8.5 so the first team's off offer must have been $8.5 million and then if they double that and added one okay the other team if they wanted to to have the player sign with them would then be offering $18 million okay all right so on that problem I just let X represent the first team's offer and then I was able to look at what the other team's offer would be by Translating that double that and add an extra million Translating that into 2x + 1 and I knew that that would need to equal 18 so I was able to then just solve the equation so there may be two unknowns in a problem but if we can represent one unknown in terms of the other we can keep it where our equations will only have a single variable in them okay so the example I put here is if you know that one number is five more than another number and we let the first number be X then the second number could be represented by x + 5 because five more than than that first number right that's kind of what's going on in 16a so we have two quantities we have Donna's height and Christina's height right so we're going to let X repres present Christina's height in inches and if we know Donna Donna is 5 Ines taller than Christina then we know Donna's height would have to be whatever Christina's height is plus five okay right now let's try that with B so we have X representing Ariana's test grade then we're trying to come up with an expression for John's and rather than using a different variable like a y we could actually write an expression in terms of X for that and it comes from us taking this right here and transl ating it we want an expression for John's test grade well we know his test grade is Five Points less so we know it's Five Points less than Ariana's than twice Ariana okay so if Ariana's test grade is X twice that would be 2X and Five Points less than that would be 2x - 5 so 2x - 5 would be an expression could use to represent John's test grade okay all right now we've got um three more problems that I wanted to look at 17 and then the next one which I just numbered these wrong 16 I'm going to reum it to 18 and 19 okay so 17 and 18 are really the same exact problem in a way right there were a lot of mistakes in thinking on this one in class I think drawing a picture can help so the board is six feet long and we're going to cut it into two pieces but it didn't say two equal pieces in fact we know they can't be equal cuz one piece is going to be three times as long as the other piece so one piece has to be longer right so just cut your board just draw a mark That's not halfway but off to one side if I let X represent the length of the shorter piece so that would be this then I could write an expression for the length of the longer piece by thinking about the fact that the longer piece is three times as long as the shorter one okay so the longer piece is three times as long as the the other one so I could use 3x to represent the length of the longer piece and then the picture pretty much tells you what your equation would be because the shorter piece plus the longer piece would have to equal six okay I'm going to come over here to solve it so 4X would have to be six and if I divide by four that gives me that X would have to be um three Hales which is 1 and a half and the units on this were feet 1 and 1/2 feet so think about that that means this piece was 1.5 feet okay so then they want us to get the length of each piece so to get the longer piece I just need to do three times that three times the 1.5 and that would give me 4.5 ft okay so this would have to be 4.5 ft and then notice if you add 1.5 and 4.5 you you get the 6 feet that you're supposed to have so we kind of check it by doing that so they wanted the length of each piece you would you would be able to say one piece is 1.5 ft the other one is 4.5 ft okay now 18 is really just like that in a way okay so the height of the deck is going to be 92 in it calls for support post of two different heights okay the taller ones are 8 in longer than the shorter ones and the sum of the length should be the height of the deck okay so the longer one plus the shorter one would have to equal the height of the deck okay so if I wind up letting the shorter one be X and again this would be now in this case it would be in inches so I'm going to let X represent the length of the shorter one then if I know the taller ones are 8 in longer than the shorter then I know the longer ones would be x + 8 and I know the sum of these lengths should equal the height of the deck so then my equation becomes 2x + 8 would have to be 92 and then I can just use the um equation solving process to finish this so I'm going to subtract eight okay that would give me 2x would be equal to um what is that 84 and then I would divide by two me move it up a little bit 2x would equal 84 so if I divide by two x would equal 42 and the units would be inches since this was inches okay and that would be the shorter post and then the long one would be eight more than that so I would just add eight to that and get 50 in and then my check would be that those two added up do equal the 92 in they're supposed to so my answer would be 42 in and 50 in now I wanted to look at question number 19 I gave you a question similar to this on the test you know we were doing problem solving on our test let me get a little more paper here okay we were doing problem solving on our test and we were able to just use any Common Sense procedure we wanted okay all right so let's do this one using a common sense procedure like you would have done on the last test so we have a total amount of $137 in tips all right it's composed of $1 and $5 bills so we could do a a trial and error approach trying to figure out our job is to get how many of each type of Bill did she get and so we know some things about it so you might just on this one do a trial and error approach where you might just be trying some stuff like um all right well let's see what if what if what if she did 20 had $25 bills okay well if she had $25 bills it tells us that she got 53 more ones than fives so 53 + 20 that would be 73 so then let's calculate how much that would be and see how close it is to the 137 so 73 * the 1 + 20 * 5 would equal $173 so it's too much so I need fewer fives and more ones so let me drop this down maybe to 10 10 fives if it were 10 fives then it would be 63 ones since she got 53 more singles than fives okay so then I could calculate how much that would turn out to that would turn out to be by doing 63 * 1 + 5 * the 10 that would be 50 + 63 so that would be $113 which is too little so that means I need more fives so let me go to 15 fives okay if I did 15 fives then 15 + 53 or 68 would be the number of ones and if she had 68 1es 68 * 1 would be the amount how much that would be worth 155 so 15 * 5 would be how much that would be let's see what that turns out to be so that's 68 + 75 and 68 + 75 that turns out to be 143 so we're getting close but it's too much remember our goal is 137 so let me do fewer fives okay 53 added to that would be 67 okay 53 more did I do 53 more yeah that's 68 okay and then that would be $67 plus $70 and 67 + 70 finally gives us the 137 so we could do that kind of guess and check approach so this was too high so we jumped down this was too low so we needed more fives this was too high but close so I need to fewer fives and I just went down a little bit and wound up getting this was the right one and then the question on that one was how many of each did she get so she got 67 ones and 145 okay so that would be an an valid approach right that would be a trial and error approach but now we've been working on using some algebra to help us with things so we're going to try that here okay so if we're going to use an algebra approach for this what we would do is a little bit different okay we might let X represent maybe how many fives she got okay so let's do that so let's let X be the number of fives okay then remember how it said she got 53 more singles than fives then x + 53 would be the number of singles of ones and then we know the total amount of money she received in tips was 137 so if she had x5s they're worth $5 each and then she had x + 53 ones they're worth $1 each so that should total to be $137 and now we have an equation that we can solve so 5x + x + 53 has to be 137 the these are like terms so 6X + 53 would be 137 so I could take away the 53 and see I'm not thinking about the context anymore okay once I've assigned and you know represented the quantities with x and x + 3 and written this equation right here then I can let the context leave my mind and just do that algebraic process real quick to get to the end of this and so 6X winds up being 137 minus 53 which is 84 and when I divide both sides by six I wind up with X equaling 14 then I have to go back and look what that 14 was so remember we had 14 x being the number of fives so she would have gotten 14 fives and then we know that she had 53 more ones so we would add 53 to that and get 67 ones and they wanted how many of each she got so you would say 14 fives and 67 ones okay so that last problem I wanted you to look at the the one the approach that's trial and error based more like our problem solving before we did any algebra and then what your approach could be once you get used to mathematizing a situation and using those variables to come up with an equation that connects the quantities in a problem