Question

Solve the following equations:
$\frac{x^3+y^3}{(x+y{)}^2}+\frac{x^3-y^3}{(x-y{)}^2}=\frac{43x}{8},5x-7y=4$.

Answer 1

$5x-7y=\mathrm{4......}\left(i\right)$

$\frac{x^3+y^3}{{(x+y)}^2}+\frac{x^3-y^3}{{(x-y)}^2}=\frac{43x}{8}$

$\frac{(x+y)(x^2+y^2-xy)}{{(x+y)}^2}+\frac{(x-y)(x^2+y^2+xy)}{{(x-y)}^2}=\frac{43x}{8}$

$\frac{x^2+y^2-xy}{x+y}+\frac{x^2+y^2+xy}{x-y}=\frac{43x}{8}$

$\frac{x^2+y^2+2xy-3xy}{x+y}+\frac{x^2+y^2-2xy+3xy}{x-y}=\frac{43x}{8}$

$\frac{{(x+y)}^2-3xy}{x+y}+\frac{{(x-y)}^2+3xy}{x-y}=\frac{43x}{8}$

$x+y-\frac{3xy}{x+y}+x-y+\frac{3xy}{x-y}=\frac{43x}{8}$

$3xy(\frac{1}{x-y}-\frac{1}{x+y})=\frac{43x}{8}-2x$

$\frac{6x{y}^2}{x^2-y^2}=\frac{27x}{8}$

$\frac{6x{y}^2}{x^2-y^2}-\frac{27x}{8}=0$

$3x(\frac{2y^2}{x^2-y^2}-\frac{9}{8})=0$

$\Rightarrow x=0$

substituting $x$ in $\left(i\right)$

$\Rightarrow y=-\frac{4}{7}$

Also $\frac{2y^2}{x^2-y^2}-\frac{9}{8}=0$

$\frac{2y^2}{x^2-y^2}=\frac{9}{8}$

$16y^2=9x^2-9y^2$

$25y^2=9x^2$

$y^2=\frac{9}{25}{x}^2$

$\Rightarrow y=\pm \frac{3}{5}x$

susbstituting $y$ in $\left(i\right)$

(1) $y=\frac{3}{5}x$

$5x-7.\frac{3}{5}x=4$

$\frac{4x}{5}=4$

$x=5$

$\Rightarrow y=3$

(2) $y=-\frac{3}{5}x$

$5x-7(-\frac{3}{5}x)=4$

$\frac{46x}{5}=4$

$x=\frac{10}{23}$

$\Rightarrow y=-\frac{6}{23}$

So the values of $x$ are $0,5,\frac{10}{23}$ and corresponding values of $y$ are $-\frac{4}{7},3,-\frac{6}{23}$