Solving proportions 2 exercise examples | Algebra Basics | Khan Academy

• [Instructor] We have the proportion ( x - 9 ) over ( 12 ) is equal to ( \frac{2}{3} ), and we wanna solve for the ( x ) that satisfies this proportion. Now, there's a bunch of ways that you could do it. A lot of people, as soon as they see a proportion like this, they wanna cross-multiply. They wanna say, "Hey, three times ( x - 9 ) is going to be equal to two times ( 12 )." And that's completely legitimate. You would get, let me write that down.

So three times ( x - 9 ), three times ( x - 9 ) is equal to two times ( 12 ). So it would be equal to two times ( 12 ). And then you can distribute the three. You'd get ( 3x - 27 ) is equal to ( 24 ). And then you could add ( 27 ) to both sides, and you would get, let me actually do that. So let me add ( 27 ) to both sides, and we are left with ( 3x ) is equal to, is equal to, let's see, ( 51 ). And then ( x ) would be equal to ( 17 ). ( x ) would be equal to ( 17 ). And you can verify that this works. ( 17 - 9 ) is ( 8 ). ( \frac{8}{12} ) is the same thing as ( \frac{2}{3} ). So this checks out.

Another way you could do that, instead of just straight up doing the cross-multiplication, you could say, "Look, I wanna get rid of this ( 12 ) in the denominator right over here. Let's multiply both sides by ( 12 )." So if you multiply both sides by ( 12 ), on your left-hand side, you are just left with ( x - 9 ). And on your right-hand side, ( \frac{2}{3} ) times ( 12 ), well, ( \frac{2}{3} ) of ( 12 ) is just ( 8 ). And you could do the actual multiplication, ( \frac{2}{3} ) times ( \frac{12}{1} ). ( 12, 12 ) and ( 3 ), so ( 12 ) divided by ( 3 ) is ( 4 ). ( 3 ) divided by ( 3 ) is ( 1 ). So it becomes ( \frac{2 \cdot 4}{1} ), which is just ( 8 ).

And then you add ( 9 ) to both sides. So the fun of algebra is that as long as you do something that's logically consistent, you will get the right answer. There's no one way of doing it. So here you get ( x ) is equal to ( 17 ) again. And you can also, you can multiply both sides by ( 12 ) and both sides by ( 3 ), and then that would be functionally equivalent to cross-multiplying.

Let's do one more. So here, another proportion, and this time the ( x ) is in the denominator. But just like before, if we want, we can cross-multiply. And just to see where cross-multiplying comes from, that it's not some voodoo, that you still are doing logical algebra, that you're doing the same thing to both sides of the equation, you just need to appreciate that we're just multiplying both sides by both denominators.

So we have this ( 8 ) right over here on the left-hand side. If we wanna get rid of this ( 8 ) on the left-hand side in the denominator, we can multiply the left-hand side by ( 8 ). But in order for the equality to hold true, I can't do something to just one side. I have to do it to both sides. Similarly, similarly, if (laughs) I, if I wanna get this ( x + 1 ) out of the denominator, I could multiply by ( x + 1 ) right over here. But I have to do that on both sides if I want my equality to hold true.

And notice, when you do what we just did, this is going to be equivalent to cross-multiplying. Because these ( 8s ) cancel out, and this ( x + 1 ) cancels with that ( x + 1 ) right over there. And you are left with, you are left with ( (x + 1) ) times ( 7 ), and I could write it as ( 7(x + 1) ), is equal to ( 5 \times 8 ), is equal to ( 5 \times 8 ). Notice, this is exactly what you have done if you would've cross-multiplied. Cross-multiplication is just a shortcut of multiplying both sides by both the denominators.

We have ( 7(x + 1) ) is equal to ( 5 \times 8 ). And now we can go and solve the algebra. So distributing the ( 7 ), we get ( 7x + 7 ) is equal to ( 40 ). And then subtracting ( 7 ) from both sides, so let's subtract ( 7 ) from both sides, we are left with ( 7x ) is equal to ( 33 ). Dividing both sides by ( 7 ), we are left with ( x ) is equal to ( \frac{33}{7} ). And if we wanna write that as a mixed number, this is the same thing, let's see, this is the same thing as ( 4 \frac{5}{7} ), and we're done.