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Video transcript

we have the proportion x minus 9 over 12 is equal to 2 over 3 and we want to 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 want to cross multiply they want to say hey 3 times X minus 9 is going to be equal to 2 times 12 and that's completely legitimate you would get let me write that down so 3 times X minus 9 3 times X minus 9 is equal to 2 times 12 so it would be equal to 2 times 12 and then you can distribute the 3 you'd get 3x 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 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 work 17 minus 9 is 8 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 want to 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 9 and on your right hand side 2/3 times 12 well 2/3 of 12 is just 8 and you could do the actual multiplication 2/3 times 12 over 1 12 12 and 3 so 12 divided by 3 is 4 3 divided by 3 is 1 so it becomes 2 times 4 over 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 could multiply both sides by 12 and most sides by 3 and then that would be functionally equivalent to cross multiplying let's do one more so here another proportion and this 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 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 want to 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 I if I want to if I want to get this X plus 1 out of the denominator I could multiply by X plus 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 8 cancel out and this X plus 1 cancels with that X plus 1 right over there and you are left with you are left with X plus 1 X plus 1 times 7 and I can write it as 7 times X plus 1 is equal to 5 times 8 is equal to 5 times 8 notice this is exactly what you have done if you were to cross multiplied cross multiplication is just a shortcut of multiplying both sides by both the denominators we have 7 times X plus 1 is equal to 5 times 8 and now we can go and solve the algebra so distributing the 7 we get 7 X plus 7 is equal to 40 and then subtracting 7 from both sides so let's subtract 7 from both sides we are left with 7 X is equal to 33 dividing both sides by 7 we are left with X is equal to 33 over 7 if we want to write that as a mixed number this is the same thing let's see this is the same thing as 4 + 5 7 s and we're done