Equalities of a rational numbers

Question

I have the equalities $$\frac a {1 + a + b } = \frac x {1+x+y}$$ $$\frac b {1 + a + b } = \frac y {1+x+y}$$ How can I show that $a = x$ and $b=y$?

Answer 1

Dividing term by term for $\frac b {1 + a + b }\neq 0$ and $\frac y {1+x+y}\neq 0$ we obtain

$$\frac{\frac a {1 + a + b }}{\frac b {1 + a + b }} = \frac{\frac x {1+x+y}}{\frac y {1+x+y}}\implies\frac ab=\frac xy \implies ay=bx$$

then from

$$\frac a {1 + a + b } = \frac x {1+x+y} \implies a+ax+ay=x+ax+bx \implies a=x$$

$$\frac b {1 + a + b } = \frac y {1+x+y}=b+bx+by=y+ay+by\implies b=y$$

Answer 2

Hint: from given equation we get $$x:a = y:b$$

Now mark this ratio with say $t$. So $x=at$ and $y=bt$. Now put this in first (or second) equation.

Does it help?

Answer 3

Add both sides of $$\frac a {1 + a + b } = \frac x {1+x+y}$$

$$\frac b {1 + a + b } = \frac y {1+x+y}$$ to get $$\frac {a+b} {1 + a + b } = \frac {x+y} {1+x+y}$$ which implies $$a+b = x+y$$

Subtract both sides to get $$\frac {a-b} {1 + a + b } = \frac {x-y} {1+x+y}$$ This and $$a+b = x+y$$ imply $$a-b = x-y$$ Solve for $x$ and $y$ to get $x=a$ and $y=b$