# Equation with variables on both sides: fractions

Common Core Math:

## Video transcript

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.