Question

Solve the following equation.
${(3-\frac{6y}{x+y})}^2+{(3+\frac{6y}{x-y})}^2=82,3x+7y=26$.

Answer 1

From first equation, we get
${\left\{\frac{3(x-y)}{x+y}\right\}}^2+{\left\{\frac{3(x+y)}{x-y}\right\}}^2=82$
Put $\frac{x+y}{x-y}=z$, we then have
$9(\frac{1}{z^2}+z^2)=82$
or $9z^4-82z^2+9=0$
or $(z^2-9)(9z^2-1)=0$
$\therefore z=\pm 3$ or $z=\pm \frac{1}{3}$
Taking $z=3$, we get $\frac{x+y}{x-y}=3$
or $x=2y$ .$\left(1\right)$
Also second given equation is
$3x+7y=26$ ..$\left(2\right)$
Solving $\left(1\right)$ and $\left(2\right)$, we get $x=4,y=2$
Similarly solving $z=-3$, that is $\frac{x+y}{x-y}=-3$ with $\left(2\right)$,
we shall get
$x=\frac{26}{17},y=\frac{52}{17}$
Again solving $z=\frac{x+y}{x-y}=\frac{1}{3}$ with $\left(2\right)$, we shall obtain
$x=-52,y=26$.
And finally solving $z=\frac{x+y}{x-y}=-\frac{1}{3}$, we shall get
$x=-\frac{26}{11},y=\frac{52}{11}$.