# Simple algebra proportion question

I have this exercise below that I need to find the x , y for.

$\displaystyle \frac{12}{X} = \frac{8}{Y} = \frac{4}{3}$

Now, I got to X + Y = 15

I did $\displaystyle \frac{12 + 8}{X + Y} = \frac{4}{3}$ And cross multiplication

But from some reason I can't get to x or y...can someone please tell me why?

tnx

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The numbers $X$ and $Y$ are already separated, why combine them?

Flip the fractions over. You will be near the end. For example, I am sure that from $\frac{X}{12}=\frac{3}{4}$ you can find $X$. Don't let $Y$ distract you.

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thanks allot buddy. solved it :) @André Nicolas – user804968 Apr 14 '13 at 19:27

Hint: Solve the equations $$\frac{12}{X}=\frac{4}{3}\quad\text{and}\quad\frac{8}{Y}=\frac{4}{3}$$separately. Also, note that in general $$\frac{12}{X}+\frac{8}{Y}\neq\frac{12+8}{X+Y}.$$For an example why this equality doesn't hold, take the following: $$1=\frac{2}{2}=\frac{1+1}{1+1}\overset{\ast}{=}\frac{1}{1}+\frac{1}{1}=1+1=2.$$The $\ast$ indicates where the flaw occurs.

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But the OP is correct: $\rm\ \dfrac{a}{b} = \dfrac{c}d\:\Rightarrow\: \dfrac{a+c}{b+d} = \dfrac{a}b,\$ i.e. the mediant of two equal fractions must equal them (since it lies between them).the – Math Gems Apr 14 '13 at 19:24
@MathGems: Fair enough; I've added "in general" to my post. – Clayton Apr 14 '13 at 19:27
@Clayton thank you – user804968 Apr 14 '13 at 19:27
@user804968: You're welcome. – Clayton Apr 14 '13 at 19:28