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## Algebra (all content)

### Course: Algebra (all content)>Unit 2

Lesson 12: Old school equations with Sal

# Linear equations 4

Solving linear equations with variable expressions in the denominators of fractions. Created by Sal Khan.

## Video transcript

Welcome to the presentation on level four linear equations. So, let's start doing some problems. So. Let's say I had the situation-- let me give me a couple of problems-- if I said 3 over x is equal to, let's just say 5. So, what we want to do -- this problem's a little unusual from everything we've ever seen. Because here, instead of having x in the numerator, we actually have x in the denominator. So, I personally don't like having x's in my denominators, so we want to get it outside of the denominator into a numerator or at least not in the denominator as soon as possible. So, one way to get a number out of the denominator is, if we were to multiply both sides of this equation by x, you see that on the left-hand side of the equation these two x's will cancel out. And in the right side, you'll just get 5 times x. So this equals -- the two x's cancel out. And you get 3 is equal to 5x. Now, we could also write that as 5x is equal to 3. And then we can think about this two ways. We either just multiply both sides by 1/5, or you could just do that as dividing by 5. If you multiply both sides by 1/5. The left-hand side becomes x. And the right-hand side, 3 times 1/5, is equal to 3/5. So what did we do here? This is just like, this actually turned into a level two problem, or actually a level one problem, very quickly. All we had to do is multiply both sides of this equation by x. And we got the x's out of the denominator. Let's do another problem. Let's have -- let me say, x plus 2 over x plus 1 is equal to, let's say, 7. So, here, instead of having just an x in the denominator, we have a whole x plus 1 in the denominator. But we're going to do it the same way. To get that x plus 1 out of the denominator, we multiply both sides of this equation times x plus 1 over 1 times this side. Since we did it on the left-hand side we also have to do it on the right-hand side, and this is just 7/1, times x plus 1 over 1. On the left-hand side, the x plus 1's cancel out. And you're just left with x plus 2. It's over 1, but we can just ignore the 1. And that equals 7 times x plus 1. And that's the same thing as x plus 2. And, remember, it's 7 times the whole thing, x plus 1. So we actually have to use the distributive property. And that equals 7x plus 7. So now it's turned into a, I think this is a level three linear equation. And now all we do is, we say well let's get all the x's on one side of the equation. And let's get all the constant terms, like the 2 and the 7, on the other side of the equation. So I'm going to choose to get the x's on the left. So let's bring that 7x onto the left. And we can do that by subtracting 7x from both sides. Minus 7x, plus, it's a minus 7x. The right-hand side, these two 7x's will cancel out. And on the left-hand side we have minus 7x plus x. Well, that's minus 6x plus 2 is equal to, and on the right all we have left is 7. Now we just have to get rid of this 2. And we can just do that by subtracting 2 from both sides. And we're left with minus 6x packs is equal to 6. Now it's a level one problem. We just have to multiply both sides times the reciprocal of the coefficient on the left-hand side. And the coefficient's negative 6. So we multiply both sides of the equation by negative 1/6. Negative 1/6. The left-hand side, negative 1 over 6 times negative 6. Well that just equals 1. So we just get x is equal to 5 times negative 1/6. Well, that's negative 5/6. And we're done. And if you wanted to check it, you could just take that x equals negative 5/6 and put it back in the original question to confirm that it worked. Let's do another one. I'm making these up on the fly, so I apologize. Let me think. 3 times x plus 5 is equal to 8 times x plus 2. Well, we do the same thing here. Although now we have two expressions we want to get out of the denominators. We want to get x plus 5 out and we want to get this x plus 2 out. So let's do the x plus 5 first. Well, just like we did before, we multiply both sides of this equation by x plus 5. You can say x plus 5 over 1. Times x plus 5 over 1. On the left-hand side, they get canceled out. So we're left with 3 is equal to 8 times x plus five. All of that over x plus 2. Now, on the top, just to simplify, we once again just multiply the 8 times the whole expression. So it's 8x plus 40 over x plus 2. Now, we want to get rid of this x plus 2. So we can do it the same way. We can multiply both sides of this equation by x plus 2 over 1. x plus 2. We could just say we're multiplying both sides by x plus 2. The 1 is little unnecessary. So the left-hand side becomes 3x plus 6. Remember, always distribute 3 times, because you're multiplying it times the whole expression. x plus 2. And on the right-hand side. Well, this x plus 2 and this x plus 2 will cancel out. And we're left with 8x plus 40. And this is now a level three problem. Well, if we subtract 8x from both sides, minus 8x, plus-- I think I'm running out of space. Minus 8x. Well, on the right-hand side the 8x's cancel out. On the left-hand side we have minus 5x plus 6 is equal to, on the right-hand side all we have left is 40. Now we can subtract 6 from both sides of this equation. Let me just write out here. Minus 6 plus minus 6. Now I'm going to, hope I don't lose you guys by trying to go up here. But if we subtract minus 6 from both sides, on the left-hand side we're just left with minus 5x equals, and on the right-hand side we have 34. Now it's a level one problem. We just multiply both sides times negative 1/5. Negative 1/5. On the left-hand side we have x. And on the right-hand side we have negative 34/5. Unless I made some careless mistakes, I think that's right. And I think if you understood what we just did here, you're ready to tackle some level four linear equations. Have fun.