Video Transcripts: Multi-Step Equations

Why we do the same thing to both sides: Variable on both sides 

Alright, now we have a very interesting situation. On both sides of the scale, we have our  mystery mass and now I'm calling the mystery mass having a mass of Y. Just to show you that it  doesn't always have to be X. It can be any symbol as long as you can keep track of that symbol.  But all these have the same mass. That's why I wrote Y on all of them. And we also have a little 1 kilogram boxes on both sides of the scale. So the first thing I wanna do, we're gonna do step by  step and try to figure out what this mystery mass is. But the first thing I wanna do is, is, is, have  you think about, whether you can represent this algebraically? Whether with, with a little bit of  mathematic symbolry, you can represent what's going on this scale. Over here, I have three Ys  and three of these boxes and their total mass is equal to this one Y. And I think I have about let's see, I have 7 boxes right over here. So I'll give you a few seconds to do that. So let's think about  the total mass over here. We have 3 boxes and a mass Y. So they're going to have a mass of 3Y,  and then you have 3 boxes with a mass of 1 kilogram. So they're going to have a mass of 3  kilograms. Now over here, I have 1 box with a mass of Y kilograms. So that's going to be my 1Y  right over there. I could've written 1Y but that's, I don't need to. A Y is the same thing as 1Y. So I  have the Y kilograms right there. And I have seven of these, right? 1,2,3,4,5,6,7. Yup, seven of  these. So I have Y plus 7 kilograms on the right-hand side. And once again, it's balanced. The  scale is balanced. This mass, total mass is equal to this total mass. So we can write an equal  sign, right over there. So that's a good starting point. We were able to represent this situation  to this real-life simple situation. You know, back in the day, when people actually had to figure  out the mass of things if you were to go the jewelry store, whatever. They actually did had  problems like this. We were able to represent it mathematically. Now the next thing to do is,  what are some reasonable next steps? How can we start to simplify this a little bit? You know,  once again, I'll give you a few seconds to think about that. Well, the neat thing about algebra is  there's actually multiple paths that you could go down. You might say, well why won't we  remove 3 of these what, of these yellow blocks from both sides? That would be completely  legitimate. You might say, well, why won't we remove 1 of these Ys from both sides? That also  would be legitimate. And we could do it in either order. So let's just pick one of them. Let's say  that we've first want to remove, let's say that we first want to remove the, a Y from either sides. Just so that we feel a little bit more comfortable with all of our Ys sitting on the 1 side. And so  the best way, if we don't want all of our Ys to sit on the 1 side, we can remove, we can remove a  Y from each side. Remember, if you removed a Y from only 1 side, that would unbalance the  scale. The scale was already balanced whatever you have to do to the one side after you do the  other. So I'm gonna remove a Y, I'm gonna remove Y mass from both sides. Now what will that  look like algebraically? Well. I remove the Y from both sides. So I subtracted Y from the left hand side and I subtracted Y from the right-hand side. That's exactly what I did. The mass, it  had a massive Y. I don't know what that is but I did take it away. I lifted that block, that little,  that little block. And so, on the left-hand side, on the left-hand side, what am I left with? That  you can pick it up mathematically, you can even look up here, and look up here what you're left  with. If I had 3 of something, and I take away 1 of them, if I take away 1 of them, I'm left with 2  of that something. So I'm left with 2Y, right over here, you see it. I had 3, I got rid of 1, so I'm left with 2. And I still have those 3 yellow blocks. So I still have those 3 yellow blocks. On the right hand side, I had a Y, I took away the Y, and so now I have no Ys left. We see it visually right over 

here. And I still have 7 of the yellow blocks. So I still have 7 of the yellow blocks. And since I took the exact same mass from both sides of the scale, the scale is still going to be balanced. It was  balanced before, I took away the same thing from both sides. And so the scale is still, is still  balanced. So this is going to be equal to that. Now, now just trying to look a bit similar to what  we saw in that last video. But I will ask you, what can we do from this point? What can we do  from this point to simplify it further or so, even better, think of it so we could isolate the, these  Ys on the left-hand side. And I'll give you a few seconds to think about that. Well, if we want to  isolate these Ys on the left-hand side, these 2 Ys, the best way is to get rid of this 3, to get rid of  these 3 blocks. So why won't we do that? Let's take 3 blocks from this side but we can't just  take it from that side if we want to keep it balanced. We have to do it to this side too. We gotta  take away, we gotta take away 3 blocks. So we're subtracting 3 from that side, and subtracting  3 from the right side. So on the left-hand side, on the left-hand side, we're gonna be left with  just these 2 blocks of mass Y. So our total mass is now going to be 2Y. These 3 minus 3 is 0, and  you see that here. We're just left with 2Ys right over here, and on the right-hand side, we got rid of 3 of the blocks. So we only have 4 of them left. So you have 4 of them left. So you have 2 of  these Y masses is equal to 4 kilograms. And because we did the same thing to both sides, the  scale is still balanced. And now, well how do we solve this? And you might be able to solve this  in your head. I have 2 times something is equal to 4. And you can kind of think about what that  is, but if we want to stay true to what we've been doing before, let's think about it. I have 2 of  something, is equal to something else. What if I multiply both sides by 2? Oh, sorry, what if I  multiply both sides by 1/2 or another way is dividing both sides by 2. If I multiply this side by  1/2, if I essentially take away half of the mass, or I'll only leave half of the mass, Then, I'm only  gonna have 1 block here, and if I take away half of the mass over here, I'm gonna have to take  away 2 of these blocks right over there. And what I just did, you could say I multiplied both  sides by 1/2 or just for a sake of a little change, You could say I divided both sides by 2. And on  the left-hand side, I'm left with a mass of Y. And on the right-hand side, I'm left with a mass of 4  divided by 2 is 2. And once again I can still write this equal sign because the scale is balanced. I  did the exact same thing to both sides. I left half to what was on the left-hand side, and half of  it was on the right-hand side. It was balanced before, half of each side, so it's going to be  balanced again. But there, we've done it. We've solved something that's actually not so easy to  solve. Or it might not look so easy first. We figured out that our mystery mass Y is 2 kilograms.  And you can verify this, this is the really fun thing about algebra. Is it, once you get to this  point, you can go back and think about whether the original, the original problem you saw  made sense. Let's do that. Let's think about whether the original problem made sense. And to  do that, I want you to, I want you to calculate. Now that we know what the mass Y is it's 2  kilograms. What was the total mass on each side? Well, let's calculate it. We have 2 right, right  over here, this is 2 kilograms, I'll do that in purple color,so this is a 2, this is a 2, this is a 2. So we  had 6 kilograms plus these 3, we had 9 kilograms on the left-hand side. And on the right-hand  side, I had these 7 plus 2 here, 7 plus 2 is 9 kilograms. That's why it was balanced. Our mystery  mass, we had 9 kilograms, total on both sides. 

Intro to equations with variables on both sides 

Let's try to solve a more involved equation. So, let's say that we have 2x plus 3, 2x plus 3 is equal

to is equal to 5x minus 2. So this might look a little daunting at first. We have x's on both sides  of the equation. We're adding and subtracting numbers. How do you solve it? And we'll do it in  a couple of different ways. The the important thing to remember is, we just want to isolate an  

x. Once you've isolated an x, you have x equals something. Or x equals something. You're done,  you've solved the equation. You can actually go back and check whether that works, So what  we're going to do is just do a bunch of operations on both sides of the equation, to eventually  isolate the x. But while we do those, I actually want to visualize what's occurring. Because I  don't want you just say, oh what are the rules or the steps of solving equations. And I forgot  whether this is a allowed or that isn't allowed. If you visualize what's happening, it'll actually be  common sense what's allowed. So let's visualize it. So we have 2x right here on the left-hand  side. So that's literally, that's x plus x. And then you have plus 3. Plus 3, I'll do it like this. So  that's equal to plus 1, plus 1, plus 1. That's the same thing as 3. I could've drawn 3 circles here as  well. Let do the same color. Plus 3. And then that is equal to 5 x's. Do that in blue. That is equal  to 5 x's. So, 1, 2, 3, 5, 6. And I want to make it clear. You would never actually have to do it this  way when you're solving the problem. You would just have to do the algebraic steps. But I'm  doing this for you so you can actually visualize what this equation is saying. the left-hand side is these two orange x's plus 3. The right-hand side is 5x minus 2. So minus 2, we could write as --  so let me do this in a different color, I'll do it in pink. So, minus 2, I'll do as minus 1 and minus 1.  Now, we want to isolate the x's on the same side of the equation. So, how could we do that?  Well, there's two ways of doing it. We could subtract these two x's from both sides of the  equation. And that would be pretty reasonable. Because then you'd have 5 x's minus the 2 x's.  You'd have a positive number of x's on the right-hand side. Or, you could actually subtract 5x  from both sides. And that's what's neat about algebra. As long as you do legitimate operations,  you will eventually get the right answer. So let's just start off subtracting 2x from both sides of  the equation. And what I mean there, I mean we're going to remove 2 x's from the left-hand  side. And if we were to move 2 x's on the left-hand side, we have to remove 2 x's the right-hand  side. Just like that. So what does that give us? We're subtracting 2 x's. 2 x's from the left. And  we're also going to subtract 2 x's from the right. Now, what does our left-hand side simplify to?  We have 2x plus 3 minus 2x. The 2 x's cancel out. So you're just left with -- you're just left with  the 3. And you see that over here. We took 2 of these x's away. We're just left with the plus 1,  plus 1, plus 1. And then on the right-hand side, 5x minus 2x. We have it over here. We have 5 x's  minus 2 x's. You only have 1, 2, 3, x's left over. 3 is equal to 3x. And then you have your minus 2  there. You have your minus 2. So, normally if you were to do the problem, you would just have  to write what we have here on the left-hand side. So what can we do next? Remember, we  want to isolate the x's. Well, we have all of our x's on the right-hand side right here. If we could  get rid of this negative 2, off of the right-hand side, then the x's will be alone. They'll be  isolated. So how can we get rid of this negative 2, if we visualize it over here. This negative 1,  this negative 1. Well, we could add 2 to both sides of this equation. Think about what happens  there. So, if we add 2, so I'm going to do it like this. Plus 1, plus 1. So you could literally see.  We're adding 2. And then we're going to add 2 to the left-hand side. 1 plus, 1 plus. What  happens? Let me do it over here as well. So we're going to add 2. We're going to add 2. So what  happens to the left-hand side? 3 plus 2 is going to be equal to 5. And that is going to be equal to 3x minus 2 plus 2. These guys cancel out. And you're just left with 3x. And we see it over here. 

We have the left-hand side is 1 plus 1 plus 1 plus 1 plus 1. We have 5 1's, or 5. And the right-hand  side, we have the 3 x's, right over there. And then we have the negative 1, negative 1. Plus 1,  plus 1, negative 1, these cancel out. They get us to 0. They cancel out. So we're just left with 5 is  equal to 3x. So we have 1, 2, 3, 4, 5 is equal to 3x. Let me clear everything that we've removed,  so it looks a little bit cleaner. These are all of the things that we've removed. Let me clear that  out. And then let me clear that out, like that. Edit. Clear. So now we are just left with 1, 2, 3, 4, 5. Actually, let me move this over. So I could just move this over right over here. We now have 1, 2, 3, 4, 5. These are the two that we added here, is equal to 3x. These guys canceled out. That's  why we have nothing there. Now, to solve this, we just divide both sides of this equation by 3.  And this is going to be a little hard to visualize over here. But if we divide over here both sides  by 3, what do we get? We divide the left by 3. We divide the right by 3. The whole reason why  we divided by 3 is because the x was being multiplied by 3. 3 is the coefficient on the x. Fancy  word, it literally just means the number multiplying the variable. The number we're solving, the  variable we're solving for. So these 3's cancel out. The right-hand side of the equation is just x.  The left-hand side is 5/3. So 5/3, we could say is is equal to 5/3. And this is different than  everything we've seen so far. I now have the x on the right-hand side, the value on the left-hand side. That's completely fine. This is the exact same thing as saying 5/3 is is equal to x is the same thing as saying x's equal to 5/3. Completely equivalent. Completely equivalent. We sometimes  get more used to this one, but this is completely the same thing. Now, if we wanted to write  this as a mixed number, if we want to write this as a mixed number, 3 goes into 5 one time with  remainder 2. So it's going to be 1 2/3. So it's going to be 1 2/3. So we could also write that x is  equal to 1 2/3. And I'll leave it up for you to actually substitute back into this original equation.  And see that it works out. Now, to visualize it over here, you know, how did he get 1 2/3, let's  think about it. Instead of doing 1, I'm going to do circles. I am going to do circles. Actually, even  better, I'm going to do squares. So I'm going to have 5 squares on the left-hand side. I'll do it in  this same yellow color right here. So I have 1, 2, 3, 4, 5. And that is going to be equal to the 3 x's. x plus x plus x. Now, we're dividing both sides of the equation by 3. We're dividing both sides of  the equation by 3. Actually, that's where we did it up here, we divided both sides by 3. So how  do you do that right-hand side's pretty straightforward. You want to divide these 3 x's into 3  groups. That's 1, 2, 3 groups. 1, 2, 3. Now, how do you divide 5 into 3? And they have to be  thought through even groups. And the answer tells us. Each group is going to be 1 2/3. So, 1 2/3. So it's going to be 2/3 of this, the next one. And then we're going to have 1 2/3. So this is 1/3.  We're going to need another. Another 1, so this is 1 1/3. We're going to need 1 more 1/3, so this  is going to be right here. And then we're left with 2/3 and 1. So we've broken it up into 3 groups.  This right here. Let me make it clear. Let me make it clear, this right here is 1 2/3. 1 2/3. And  then this right here, this 1/3. That's another 1/3, so that's 2/3, and then that's 1 right there. So  that's 1 2/3. And then finally this is 2/3 and this is 1, so this is 1 2/3. So when you divide both  sides by 3 you get 1 2/3. Each section, each bucket, is 1 2/3 on the left-hand side. On the left hand side, or 5/3. And on the right-hand side we just have an x. So it still works. A little bit  harder to visualize with fractions. 

Equations with variables on both sides: 20-7x=6x-6 

We have the equation 20 minus 7 times x is equal to 6 times x minus 6. And we need to solve for

x. So the way I like to do these is we just like to separate the constant terms, which are the 20  and the negative 6 on one side of the equation. I'll put them on the right-hand side. And then  we'll put all the x terms, the negative 7x and the 6x, we'll put it all on the left-hand side. So to  get the 20 out of the way from the left-hand side, let's subtract it. Let's subtract it from the left 

hand side. But this is an equation, anything you do to the left-hand side, you also have to do to  the right-hand side. If that is equal to that, in order for them to still be equal, anything I do to  the left-hand side I have to do to the right-hand side. So I subtracted 20 from the left, let me  also subtract 20 from the right. And so the left-hand side of the equation, 20 minus 20 is just 0.  That was the whole point, they cancel out. Don't have to write it down. And then I have a  negative 7x, it just gets carried down. And then that is equal to the right-hand side of the  equation. I have a 6x. I'm not adding or subtracting anything to that. But then I have a negative  6 minus 20. So if I'm already 6 below 0 on the number line, and I go another 20 below that,  that's at negative 26. Now, the next thing we want to do is we want to get all the x terms on the left-hand side. So we don't want this 6x here, so maybe we subtract 6x from both sides. So let's  subtract 6x from the right, subtract 6x from the left, and what do we get? The left-hand side,  negative 7x minus 6x, that's negative 13x. Right? Negative 7 of something minus another 6 of  that something is going to be negative 13 of that something. And that is going to be equal to 6x minus 6x. That cancels out. That was the whole point by subtracting negative 6x. And then we  have just a negative 26, or minus 26, depending on how you want to view it, so negative 13x is  equal to negative 26. Now, our whole goal, just to remember, is to isolate the x. We have a  negative 13 times the x here. So the best way to isolate it is if we have something times x, if we  divide by that something, we'll isolate the x. So let's divide by negative 13. Now, you know by  now, anything you do to the left-hand side of an equation, you have to do to the right-hand  side. So we're going to have to divide both sides of the equation by negative 13. Now, what  does the left-hand side become? Negative 13 times x divided by negative 13, that's just going to be x. You multiply something times x, divide it by the something, you're just going to be left  with an x. So the left-hand side just becomes an x. x is equal to negative 26 divided by negative  13. Well, that's just positive 2, right? A negative divided by a negative is a positive. 26 divided by 13 is 2. And that is our answer. That is our answer. Now let's verify that it really works. That's the fun thing about algebra. You can always make sure that you got the right answer. So let's  substitute it back into the original equation. So we have 20 minus 7 times x-- x is 2-- minus 7  times 2 is equal to 6 times x-- we've solved for x, it is 2-- minus 6. So let's verify that this left hand side really does equal this right-hand side. So the left-hand side simplifies to 20 minus 7  times 2, which is 14. 20 minus 14 is 6. That's what the left-hand side simplifies to. The right hand side, we have 6 times 2, which is 12 minus 6. 12 minus 6 is 6. So they are, indeed, equal,  and we did, indeed, get the right answer. 

Equation with variables on both sides: fractions 

We have the equation 3/4x plus 2 is equal to 3/8x minus 4. Now, we could just, right from the  get go, solve this the way we solved everything else, group the x terms, maybe on the left-hand side, group the constant terms on the right-hand side. But adding and subtracting fractions are  messy. So what I'm going to do, right from the start of this video, is to multiply both sides of  this equation by some number so I can get rid of the fractions. And the best number to do it 

by-- what number is the smallest number that if I multiply both of these fractions by it, they  won't be fractions anymore, they'll be whole numbers? That smallest number is going to be 8.  I'm going to multiply 8 times both sides of this equation. You say, hey, Sal, how did you get 8?  And I got 8 because I said, well, what's the least common multiple of 4 and 8? Well, the smallest number that is divisible by 4 and 8 is 8. So when you multiply by 8, it's going to get rid of the  fractions. And so let's see what happens. So 8 times 3/4, that's the same thing as 8 times 3 over  4. Let me do it on the side over here. That's the same thing as 8 times 3 over 4, which is equal to 8 divided by 4 is just 2. So it's 2 times 3, which is 6. So the left-hand side becomes 8 times 3/4x is 6x. And then 8 times 2 is 16. You have to remember, when you multiply both sides, or a side, of  an equation by a number, you multiply every term by that number. So you have to distribute  the 8. So the left-hand side is 6x plus 16 is going to be equal to-- 8 times 3/8, that's pretty easy,  the 8's cancel out and you're just left with 3x. And then 8 times negative 4 is negative 32. And  now we've cleaned up the equation a good bit. Now the next thing, let's try to get all the x  terms on the left-hand side, and all the constant terms on the right. So let's get rid of this 3x  from the right. Let's subtract 3x from both sides to do it. That's the best way I can think of of  getting rid of the 3x from the right. The left-hand side of this equation, 6x minus 3x is 3x. 6  minus 3 is 3. And then you have a plus 16 is equal to-- 3x minus 3x, that's the whole point of  subtracting 3x, is so they cancel out. So those guys cancel out, and we're just left with a  negative 32. Now, let's get rid of the 16 from the left-hand side. So to get rid of it, we're going  to subtract 16 from both sides of this equation. Subtract 16 from both sides. The left-hand side  of the equation just becomes-- you have this 3x here; these 16's cancel out, you don't have to  write anything-- is equal to negative 32 minus 16 is negative 48. So we have 3x is equal to  negative 48. To isolate the x, we can just divide both sides of this equation by 3. So let's divide  both sides of that equation by 3. The left-hand side of the equation, 3x divided by 3 is just an x.  That was the whole point behind dividing both sides by 3. And the right-hand side, negative 48  divided by 3 is negative 16. And we are done. x equals negative 16 is our solution. So let's make  sure that this actually works by substituting to the original equation up here. And the original  equation didn't have those 8's out front. So let's substitute in the original equation. We get 3/4-- 3 over 4-- times negative 16 plus 2 needs to be equal to 3/8 times negative 16 minus 4. So 3/4 of  16 is 12. And you can think of it this way. What's 16 divided by 4? It is 4. And then multiply that  by 3, it's 12, just multiplying fractions. So this is going to be a negative 12. So we get negative  12 plus 2 on the left-hand side, negative 12 plus 2 is negative 10. So the left-hand side is a  negative 10. Let's see what the right-hand side is. You have 3/8 times negative 16. If you divide  negative 16 by 8, you get negative 2, times 3 is a negative 6. So it's a negative 6 minus 4.  Negative 6 minus 4 is negative 10. So when x is equal to negative 16, it does satisfy the original  equation. Both sides of the equation become negative 10. And we are done. 

Equation with the variable in the denominator 

So I have the equation 7 minus 10/x is equal to 2 plus 15/x. And so this isn't the type of equation  that you might think that you're used to solving. But I'll give you a few moments to see if you  can solve it on your own. Well, what we'll see is we can do a quick multiplication of both sides to actually simplify this to a form that we are more used to looking at. So what's probably 

bothering you, because it's bothering me, is these x's that we have in the denominators right  over here. We're like, well, how do we deal with that? Well, whenever we see an x in the  denominator, the temptation is to multiply it by x. But we can't just multiply one of the terms  by x. We have to multiply the entire side by x. So we could multiply this entire side by x. But we  can't just multiply the left-hand side by x. We'd also want to multiply the right-hand side by x.  And so what will that give us? Well, we distribute the x. We get x times 7 is 7x. And then x times  negative 10/x, well, that's just going to be negative 10. So you get negative 10 right over there.  So the left-hand side simplifies to 7x minus 10. And then your right-hand side, once again,  distribute the x. x times 2 is 2x. x times 15/x, well, x times something over x is just going to be  the something. x times 15/x is just going to be 15-- plus 15. So now we've simplified this to a  linear equation. We have the variable on both sides. So we just have to do some of the  techniques that we already know. So the first thing that I like to do is maybe get all my x's on  the left-hand side. So I want to get rid of this 2x right over here. So I subtract 2x from the right hand side. Now, and I always remind you, I can't do that just to the right-hand side. If I did it just to the right-hand side, it wouldn't be an equality anymore. You have to do that to the left-hand  side as well. And so we are left with-- let me get that pink color again. On the left-hand side, 7x,  7 of something minus 2 of something, well, you're going to have 5 of that something, minus 10.  These two x's negate each other. And you're left with equals 15. Now we can get rid of this  negative 10 by adding 10 to both sides. You know, I like that green color when I do stuff to both  sides. So I can add 10 to both sides. And I'm left with 5x-- these negate each other-- is equal to  25. And this is the home stretch. You see where this is going. We can divide both sides by 5. And  we are left with x is equal to 5. Now let's verify that this actually worked. So let's go back to the  original equation. We have 7 minus 10/5. This needs to be equal to-- I'm just taking our 5 and  substituting it back here. This needs to be equal to 2 plus 15/5. So this is 7 minus 10/5. This is just 2. It needs to be equal to 2 plus 15/5, which is just 3. So 2 plus 3, 7 minus 2 is 5, 2 plus 3 is 5, 5 is  indeed equal to 5. And we are done.