Question

Solve the following systems of equations.
$\frac{x^2}{y}+\frac{y^2}{x}=18,x+y=12$

Answer 1

$\frac{x^2}{y}+\frac{y^2}{x}=18⟶\left(1\right),x+y=12⟶\left(2\right)$

$\Rightarrow \frac{x^3+y^3}{xy}=18$

$\Rightarrow \frac{(x+y)(x^2+y^2-xy)}{xy}=18$

$\Rightarrow \frac{12(x^2+y^2-xy)}{xy}=18$ $(\because x+y=12)$

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

$\frac{2[{\left(12\right)}^2-3xy]}{xy}=3$

$2[144-3xy]=3xy$ $(\because x+y=12)$

$288-6xy=3xy$

$\Rightarrow 288=9xy$

$32=xy$

$\Rightarrow y=\frac{288}{9x}=\frac{32}{x}$

From $\left(2\right)$

$\Rightarrow x+\frac{288}{9x}=12$

$\Rightarrow 5x^2+288=60x$

$5x^2-60x+288=0$

$9x^2+288=108x$

$9x^2-108x+288=0$

$x^2-12x+32=0$

$x=\frac{12\pm \sqrt{144-32\times 4}}{2}$

$x=\frac{12\pm \sqrt{16}}{2}$

$\Rightarrow x=\frac{12+4}{2}=\frac{16}{2}=8$ and $x=\frac{12-4}{2}=4$

Hence, solved.