Question

Solve the following equation.
$\frac{bx}{y+b}+\frac{ay}{x+a}=\frac{a+b}{2},\frac{x}{a}+\frac{y}{b}=2$.

Answer 1

From the second equation, we get
$\frac{x}{a}-1=1-\frac{y}{b}$ or $\frac{x-a}{a}=\frac{b-y}{b}=k$, say
Then $x=a(1+k)$ and $y=b(1-k)$ $\left(1\right)$
Substituting these values of x and y in first equation, we have
$\frac{ab(1+k)}{b(2-k)}+\frac{ab(1-k)}{a(2+k)}=\frac{a+b}{2}$
or $\frac{a(1+k)}{2-k}+\frac{b(1-k)}{2+k}=\frac{a+b}{2}$
or $[\frac{a(1+k)}{2-k}-\frac{a}{2}]+[\frac{b(1-k)}{2+k}-\frac{b}{2}]=0$
or $\frac{3ak}{2-k}-\frac{3bk}{2+k}=0$
or $k\left[\frac{a(2+k)-b(2-k)}{(2-k)(2+k)}\right]=0$
or $k\left[\right(a+b)k-2(b-a)=0$
$\therefore k=0$ or $k=\frac{2(b-a)}{b+a}$
Substituting these values of k in $\left(1\right)$, we get
$x=a(1+0)=a$ and $y=b(1-0)=b$
or $x=a\{1+\frac{2(b-a)}{b+a}\}=\frac{a(3b-a)}{a+b}$
and $y=b\{1-\frac{2(b-a)}{b+a}\}=\frac{b(3a-b)}{a+b}$
Hence the solutions are
$x=a,y=b$ or $x=\frac{a(3b-a)}{a+b}$, $y=\frac{b(3a-b)}{a+b}$.