Question

Solve the following equations:
$y(y^2-3xy-x^2)+24=0,x(y^2-4xy+2x^2)+8=0$.

• A. $(2,4)$
• B. $(2,6)$
• C. $(3,4)$
• D. $(4,7)$

Answer 1

Answer: (B)

$(2,6)$

$y(y^2-3xy-x^2)+24=0y^2-3xy-x^2=-\frac{24}{y}.......\left(i\right)x(y^2-4xy+{2x}^2)+8=0y^2-4xy+{2x}^2=-\frac{8}{x}......\left(ii\right)$

susbtrscting $\left(ii\right)$ from $\left(i\right)$

$\Rightarrow xy-3x^2=(\frac{8}{x}-\frac{24}{y})x(y-3x)=8\left(\frac{y-3x}{xy}\right)x^{2}y(y-3x)=8(y-3x)x^{2}y(y-3x)-8(y-3x)=0(x^{2}y-8)(y-3x)=0y=3x,\frac{8}{x^2}$

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

(1) $y=3x$

$3x({\left(3x\right)}^2-3x(3x)-x^2)+24=0-3x^3=-24x^3=8\Rightarrow x=2y=3x\Rightarrow y=6$

(2) $y=\frac{8}{x^2}$

$\frac{8}{x^2}({\left(\frac{8}{x^2}\right)}^2-3x\left(\frac{8}{x^2}\right)-x^2)+24=0\frac{64}{x^4}-\frac{24}{x}-x^2+3x^2=0x^6-12x^3+32=0x^6-4x^3-8x^3+32=0x^3(x^3-4)-8(x^3-4)=0(x^3-4)(x^3-8)=0x^3=4,8\Rightarrow x=\sqrt[3]{34},2y=\frac{8}{x^2}x=2\Rightarrow y=2x=\sqrt[3]{34}\Rightarrow y=2\sqrt[3]{34}$

So the values of $x$ are $2,\sqrt[3]{34},2$ and the corresponding values of $y$ are $6,2\sqrt[3]{34},2$