Solving tricky proportions

- [Instructor] We have the proportion x minus nine over 12 is equal to two over three, 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 minus nine 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 minus nine, three times x minus nine 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 three x minus 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 three x 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 minus nine is eight. 8/12 is the same thing as 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 minus nine. And on your right-hand side, 2/3 times 12, well, 2/3 of 12 is just eight. And you could do the actual multiplication, 2/3 times 12 over one. 12, 12 and three, so 12 divided by three is four. Three divided by three is one. So it becomes two times four over one, which is just eight. And then you add nine 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 three, 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 eight right over here on the left-hand side. If we wanna get rid of this eight on the left-hand side in the denominator, we can multiply the left-hand side by eight. 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, similar-larly, if (laughs) I, if I wanna get this x plus one out of the denominator, I could multiply by x plus one 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 eights cancel out, and this x plus one cancels with that x plus one right over there. And you are left with, you are left with x plus one, x plus one times seven, and I could write it as seven times x plus one, is equal to five times eight, is equal to five times eight. 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 seven times x plus one is equal to five times eight. And now we can go and solve the algebra. So distributing the seven, we get seven x plus seven is equal to 40. And then subtracting seven from both sides, so let's subtract seven from both sides, we are left with seven x is equal to 33. Dividing both sides by seven, we are left with x is equal to 33 over seven. 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 5/7, And we're done.