Starting with $\frac{X+A}{a}=\frac{B}{b}$ and $X-A=B$, derive $A=\frac{a-b}{a+b}X$ and $B=\frac{2b}{a+b}X$.

Question

I've tried a few approaches, to no avail. Thanks in advance.

Answer 1

The first equation gives $b(X+A)=aB$ and substituting $B=X-A$ gives $b(X+A)=a(X-A)$.

Re-arranging gives $X(b-a)=-A(a+b)$ so $A=\frac{b-a}{-(a+b)}X=\frac{a-b}{a+b}X$ and then $B=X-A=X-\frac{a-b}{a+b}X=\frac{a+b}{a+b}X-\frac{a-b}{a+b}X=\frac{2b}{a+b}X$.