foreign what's up everybody welcome to another video hope you're ready to flex those brain muscles in this video we're going to practice solving linear equations so I have seven examples here that we're going to go through and if you're trying to practice this stuff what I encourage you to do is pause the video pull out a pencil of paper and try some of these on your own I'll have all of these time stamped in the description below so you can skip around if you need to to get help with particular problems these are going to be a good variety of examples and progress in difficulty so let's go ahead and get started and since this is the first example we're about to do I'm going to talk a little bit about the general idea of what it is we're doing and sort of General strategy for how we're going to solve these linear equations so first of all what does it mean to solve a linear equation if we're given this equation where X is a variable and to solve it means we want to find all of the solutions so a solution to this equation is a number that we could plug in for X that makes the equation true another way that's often said is that it satisfies the equation so if I can take a number replace x with that number right and simplify the left hand side and it equals 19 and we get 19 equals 19 which is a true statement then that number is a solution to this equation so we want to find all of the solutions to this equation and with linear equations there are really three possibilities which are going to get into through some examples and I'm going to kind of summarize those for you there's either one solution there are infinitely many solutions where any real number will satisfy the equation or there's no solution so there are three possibilities and again we're going to summarize that here in a second once we work through some of these examples so now for the strategy now now that we've talked about what exactly it is we're doing the general strategy is that we want to eventually end up with x equals a number that's kind of our goal we want x equals some number right and how we get there is really two or three main things the first is using properties right we're going to Use the distributive property to get rid of parentheses we're going to move stuff around using commutative and associative properties that's the first thing second we're going to do a lot of combining like terms right so if we have like a 3x minus 2x we're going to combine that to be just X okay so if you're confused about those two things or you're not sure what those two things mean I'll link a video where we simplify expressions because that'll give you some good practice with that finally the last thing we're going to do is multiply divide add and subtract things to both sides we have to do the same thing to both sides right and that's sort of the key to solving these equations so the first thing I'm going to do to both sides of this equation because I want to isolate X because I'm going to add 17. and now the reason I'm adding a 17 and not dividing by 3 because you may say well this 3 is multiplied by the X why don't we divide by 3 is that generally I want to do the opposite of what's being done to X right in the opposite order so what I mean by that is when I plug something in for X it first gets multiplied by 3. then we subtract 17 and so I want to undo that and I want to reverse the order to undo that I want to First add 17 then divide by 3. that's not a hard and fast rule right but I find that it generally works better and leads to less mistakes and just less work overall so I'm going to add 17 first to both sides and what that does is minus 17 plus 17 simplifies to zero and so on the left hand side we're just left with 3x and on the right hand side we're left with 19 plus 17 which is 36. now I'm going to divide both sides by 3 right X is being multiplied by 3 I want to undo that to get X by itself so I'm going to divide 3 over 3 is 1 so we're just left with X on the left hand side and 36 over 3 is 12. now a lot of instructors including probably me actually would be fine if you just left your answer like this x equals 12. however this is still technically an equation right so what we've done here is we've rewrote this equation as x equals 12. these equations are equivalent in the sense that they have the same Solutions right so we've taken our equation and we've written it in a way where it's very easily easy to identify the solutions right there's one solution of this and it's 12. if I replace x with 12 I get 12 equals 12. but if we want to really identify the solutions then a lot of times what we want to do is we want to write the solution set which in this case since there's just one solution we can use our set notation these are these curly brackets and we can write the solution set as the set containing 12. so again it's just kind of a minor detail it really depends on your instructor and how picky they are but I figure I'd point this out this is the solution set we're listing all of the solutions to this equation since there's just one it's just 12 we can list it out and I'm going to continue to do that through this video write it out that way let's try another example this time we have parentheses so now we're in that situation where we really are going to want to use the distributive property first that is a pretty general rule that that works most of the time and is going to make your life easier is you really just want to get rid of all the parentheses however you can as soon as you can so we're going to Use the distributive property we're going to distribute this three and so what we get is 3x minus 3 times 6 that's 18 right equals 2x minus 5. so I haven't added subtracted multiply divide anything to both sides all I've done is Rewritten the left side using the distributive property now I can start doing stuff to both sides and what I'm going to do this time is subtract 2x from both sides and here there's not really a order that necessarily make things easier you could easily add 18 to both sides add 5 right you can really do whatever you want I'm choosing to subtract 2x from both sides because I notice that first of all that cancels on the right hand side here right 2x minus 2x gives me 0. so on the right hand side I'm just left with negative 5. but also as I notice that 3x minus 2x is just X which is really nice if I'm left with just X then that kind of tells me okay really now I just have one step which is add 18 to both sides if I were to subtract 3x from both sides that works as well but I'd be left with minus X on the right hand side and I'd later have to deal with that it just adds an additional step right so you can do that and that's fine as well there's a lot of different ways to get to the correct answer but now I just have to add 18 to both sides the minus 18 and the plus 18 are going to cancel out so on the left hand side we have X on the right hand side we have minus 5 plus 18 that's positive 13. and so this linear equation has one single solution and that is 13. so I'm going to write out the solution set here and now we're going to move on to the next example so again I see a parenthesis I'm going to immediately deploy the distributive property so what I always do I want to get rid of all the parentheses expand out all the terms and see what I'm working with so 5 times x minus 5 times 2 that's 10. and now it's looking like on the left hand side I can actually combine some like terms before I even do anything to both sides I can say well 5x minus 2x is 3x so I'm going to go ahead and combine those on the left hand side and I get 3x minus 10 equals 3x plus 7. and you may already have a funny feeling looking at this like something doesn't look right and if you do then that's true something doesn't look right right so let's investigate what that is 3x minus 10 equals 3x minus 7. you may notice well wait a minute if I subtract 3x from both sides what I get is 3x minus 3x 3x minus 3x I actually get negative 10 equals 7 which doesn't make any sense my variable terms have canceled and I get something that just isn't true negative 10 doesn't equal 7 right so in this case this is what we call a contradiction which means that this linear equation has no Solutions right there are no solutions to this linear equation and we could have probably identified this here when we had 3x minus 10 equals 3x plus 7. there isn't a number that I can multiply by three and then subtract 10 and that's the same as multiplying it by 3 and then adding 7. that doesn't make sense right there's no numbers that that satisfy that equation now we can see it even more clear when we subtract the 3x from both sides and end up with this not true statement so anytime this is the case and you're canceling stuff out and you end up with something absurd like negative 10 equals seven then the linear equation has no Solutions and the solution set we can actually write as the empty set it's essentially a zero with a little slash through it empty set right because there are no Solutions awesome let's try the next example seven times p plus two okay so again first thing we want to do is get rid of these parentheses we're going to apply the distributive property here and so we're going to have 7 times P plus 7 times 2 that's 14. minus 4p and I'm just going to rewrite what I had on the right hand side 3p Plus 14. and now what I see is that I can again combine like terms 7 times P minus 4 times P that's going to combine to be 3 times P plus 14. equals 3 times P plus 14 and you may have a funny feeling about this as well and say wait a minute we have the same thing on both sides right so at this point if you notice this then you can actually be done right because we can think about is what can I plug in for P that would make this true and it turns out that any real number would satisfy this equation right if I take any real number multiply by 3 and add 14 it's going to be the same as that real number multiplied by 3 with 14 added to it right any real number will satisfy this equation but if you don't notice that initially then what you're eventually going to get is something like this you're going to be like okay well let's subtract 14 from both sides 3p equals 3p then maybe you subtract 3 p from both sides and you get 0 equals zero or you divide both sides by three and you get P equals P right regardless of how you do it you're going to end up with something that's always true again we could have noticed that here at this step that this is an equation that's true for all values of P for all real numbers but at whatever point you notice it is fine eventually if you do notice it what you can say is that the solution set is the set of all real numbers there are infinitely many solutions to this linear equation so the solution set you could maybe write it as this as like all real numbers I prefer to write it as an integral from negative Infinity to positive Infinity in interval notation right so it really depends on your instructor and what they prefer but I would consider both of these to be correct personally and now we've gone through examples of all three cases so let's kind of summarize what we just did right because we've noticed there are three possibilities when we're solving these linear equations the verses that we end up with something like x equals four x equals negative two right x equals a number and in this case we're going to have one solution and that's just going to be whatever number we replace x with to make this true right then we saw another case where we ended up with something that was always true 2 equals two zero equals zero three p equals three p right stuff like that always true true for all real numbers there's infinitely many solutions in these cases and we can write the solution set as the interval from negative Infinity to positive Infinity finally we have these weird cases where we get something that is absurd 0 equals two right x minus one equals x stuff like that there are no real numbers that will satisfy those equations and in those cases we have no solution and we can write no solution or we can write the solution set as the empty set so again keep in mind these three possibilities if you end up with something like two equals two or zero equals two it doesn't necessarily mean you did something wrong I would definitely double check your work but it probably just means that it's one of those cases where there's infinitely many solutions or no solution all right there's three examples left let's try this example here so again now we see two sets of parentheses you're really going to want to get rid of these parentheses before we start doing anything to both sides so this left hand parenthesis I think of this negative as Distributing right one way to think of it is this is a little invisible one here and we have negative one times everything in these parentheses that's perfectly fine as long as you remember that that negative has to go to everything in the parentheses so when we expand this at 6 minus t minus 2. here we have another application of the distributive property 5 times 3T gives us 15 T minus 4 times 5 which is 20. can we combine any like terms before we continue I think we can 6 minus two right six minus 2 gives us four and so we have on the left hand side is 4 minus t and on the right hand side we have 15 t minus 20. all right now we have a few different ways we can continue I'm just going to go ahead and add T to both sides I'm going to add T to both sides and so on the left hand side the t's will cancel I'll be left with just four and on the right hand side I'll have 15 t plus T that's 16 T and I still have this minus 20. and now all I have to do is add 20 to both sides right and I can isolate the T term here and so on the left I have 24 on the right I have 16 t and then I can divide both sides by 16. and let me draw a little arrow here I don't want to run out of room what I have here is 16 over 16 that's 1. so I have t by itself and what does t equal well 24 over 16. now let's think about how we can reduce this if we could simplify this 24 over 16. what we want to think about is common factors between 24 and 16. so 24 and 16 both share a common factor of 8. I think that's the greatest common factor between the two numbers right so I'm going to divide 24 by 8 and that's going to give me three and I'm going to divide 16 by 8 and that's going to give me two and another way to clearly see this maybe this will help you see it a little bit clearer as if we rewrite 24 as eight times three and rewrite 16 as 8 times 2. now we can see the eights canceling right and what we end up with is three over two and so we end up with t equals three over two this is our answer fully simplified and so we can write our solution set as the set containing three over two awesome let's try the next example all right now we have sets of parentheses brackets all kinds of stuff going on so again on the right hand side it's fairly straightforward consistent with what we've been doing which is distributive property right 2 times 3 2 times Z and so what we end up with is 6 minus 2z on the left hand side it's a little trickier because we have these brackets and we have these parentheses and then we have this negative 2 out here that we're going to distribute but maybe we're not sure when to distribute it what I like to do is work from inside out so the first thing I'm going to do is deal with what's inside the brackets I'm going to leave this negative 2 out here just chilling I'm going to put the bracket here and the first thing I'm going to do is apply the distributive property inside the bracket so I'm going to say well this is 5 minus 2z minus one remember that negative goes to everything when we distribute and there's a minus 4 out here and this is all equal to 6 minus 2z it got a little crooked here but hopefully you can see what's going on so again I like to work from inside out because now what I can do is distribute this negative 2 to everything and in fact before I do that I can actually combine like terms within the bracket here because I see a 5 and a minus 1. so maybe I'll do that first I'll do five minus one hopefully I'll see that I'll Circle them in blue here five minus one I'm going to combine those that's going to give me four and I still have minus two Z and there's a minus four and we rewrite this a little nicer here this is all equal to 6 minus 2z all right now I can distribute this negative 2 and again I could have done it initially you just have to be really careful is that the negative 2 goes to the 5 and then it goes out in front here and it's going to become plus two times all of this right so it's just a little bit different it works either way I'm just doing what makes the most sense for me okay so negative 2 times 4 is going to give us negative eight and then negative 2 times negative 2z so we have a negative and a negative so that's going to become a positive for Z and then at the end we have minus 4 and this is all equal to 6 minus two Z oops let me write that z a little bit better my 2 and Z look similar that's why I do the line through the Z all right minus 8 minus four I see like terms I see like terms on the left hand side so we always want to combine these before we start adding and subtracting to both sides right we always want to make sure we have like terms combined here so let's see I can write this as minus 12. so minus 12 plus 4z and again if you wrote 4z minus 12 that's totally fine it's the same thing there's a lot of different ways to write these expressions and now I think I have everything combined and now I can start doing stuff to both sides so look at all this work we did before we even started doing the sides right and this is really what we want to do is get something like this because now I can say well I'll add 2z to both sides right because doing that it's going to get rid of the minus 2z on the right here and so on the left I'm going to have negative 12 plus 6z equals six and now I can add 12 to both sides and now what I end up with is what well minus 12 plus 12 that's 0 and so I just have 6z and 6c is equal to 6 plus 12. that's 18 and so now when I divide both sides by 6 what I'm going to get is z equals three and so if I want to write the Solutions in a set I will write to the set containing three because there is one solution and that's three so this is my solution set all right so this last example I wanted to include some fractions and I have a full video on solving linear equations involving fractions so if you want more practice a link should pop up right now and you can click that to see some more examples but generally what we want to do is we want to clear the fractions as soon as we can and we can actually do it at this step right before we even do anything we can clear the fractions and we do that by multiplying both sides of the equation by the least common denominator is typically what we do any common denominator will work but we like to do the least because then we have to do less simplifying later so what is the least common denominator well we have one-fifth three-fifths and one-tenth so our two distinct denominators are five and ten right so the least common denominator is the least common multiple of the denominators well what are multiples of five we can list some of them out here five ten fifteen twenty twenty five right and they continue to go and what are multiples of 10 well 10 20 30 right and they continue to go and we want to find the smallest number that's in both of these lists the smallest one I see is ten ten is in both of these lists and it's the smallest one that's in both of these lists you'll notice that there are many others like 30 is in both of these lists but really we want the smallest one because that's going to be the least amount of simplifying later and there's another way you could figure out least common denominator I chose this way because I think it's the most intuitive the other way I think I talk about in my other video that I linked but this is going to work fine the least common denominator is 10 so we can multiply the entire left hand side I like to put parentheses here to make it clear the entire left hand side is being multiplied by 10. and the entire right hand side is being multiplied by 10. again I think these parentheses are really helpful because this 10 like for example on the right hand side is going to have to distribute to every single term we're now using the distributive property here on the left hand side since we have one-fifth times n minus 5 in parentheses there's really no Distributing we can combine the 10 and the one-fifth and that can simplify right so I'll just write it for now as 10 over 5. but hopefully you can see that that is equal to 2 right because they share a common factor similar to what we did a couple of examples ago 10 times three-fifths I'm going to write this as 3 times 10 over 5. n right you could write 30 over 5 or maybe you would you could just immediately write it as six right that's totally fine but I want to kind of show exactly each step of this since we're still learning so 10 times 1 10 is 10 over 10. plus 1 times 10 is just 10. so I kind of wrote it all out now I'm going to simplify everything right 10 over 10 is pretty easy that's just one so that can actually go away because one n is the same thing as n right so that can actually go away here's what we do here 10 over 5 right again we're looking for common factors 10 over 5 we can divide 5 by 5 and that's one and we can divide ten by five and that's two another way to think about it is 5 goes into ten two times so 10 over 5 is equal to two totally fine so we can replace this with two over one as well which is just two and now let's see what we have here two times n minus five equals three times two that's six times n plus n plus ten and again if you went initial if you went from this first step directly to here if you're like good at multiplying whole numbers and fractions in your head and simplifying totally fine I just wanted to show each step here I think this is helpful especially when the fractions get a lot more complicated to write them out as factors in the numerator and denominator and it's easy to see how Stuff reduces and cancels now let's apply the distributive property One Last Time 2 times n minus two times five that's ten and then we can actually I'm going to go ahead and combine like terms as well so 6n plus n is going to combine to be 7 and plus 10. and now we're in this comfortable step we've been here a lot before right we have a variable term and a constant term on each side so first I'm going to subtract 2N from both sides and again when usually when we have two variable terms I always tend to subtract the one with the smaller coefficient so I can end up with a variable term with a positive coefficient that's just my preference so 2N minus 2N is zero so we're left with negative 10 on the left hand side and on the right hand side we have 5 and Plus 10. now I can subtract 10 from both sides we are almost done this is the home stretch plus 10 minus 10 those cancel minus 10 minus 10 that is minus 20. so I have minus 20 equals 5n and now when I divide both sides by 5 to isolate n to get n by itself I get let me draw an arrow here n right five over five cancels equals minus 20 over five five goes into 24 times so I can write that as negative four and so our solution set is negative four and one thing I haven't talked about yet in this video is maybe you're unsure maybe you're like I wonder if I made a mistake or not if this is right how could we check our answer well we can take this negative 4 and we can substitute it back in for n everywhere we're seeing n in the initial equation plug in negative 4 and see if you're left with a true statement like 2 equals two for example if you are then you've done it correct and this is in fact a solution if you aren't then go back and check your work and see where you made a mistake hopefully this video was helpful make sure to like subscribe leave a comment if you have any questions most importantly keep flexing those brain muscles and I'll see y'all later