Question

Solve the following equations:
$\frac{x^2}{y}+\frac{y^2}{x}=\frac{9}{2}$,
$\frac{3}{x+y}=1$.

• A. $(5,3)$
• B. $(1,2)$
• C. $(2,1)$
• D. $(3,4)$

Answer 1

Answer: (B), (C)

$(1,2)$
$(2,1)$

Given, $\frac{3}{x+y}=1$

$\Rightarrow x+y=3$ .......(i)

Also given, $\frac{x^2}{y}+\frac{y^2}{x}=\frac{9}{2}$

$\Rightarrow \frac{x^3+y^3}{xy}=\frac{9}{2}\Rightarrow 2(x^3+y^3)=9xy$

Using $a^3+b^3=(a+b)(a^2+b^2-ab)$

$2(x+y)(x^2+y^2-xy)=9$

Substituting (i), we get

$2\left(3\right)(x^2+y^2-xy)=9$

$\Rightarrow 2(x^2+y^2-xy)=3xy$

$\Rightarrow 2({(x+y)}^2-3xy)=3xy$

$\Rightarrow 2(9-3xy)=3xy\Rightarrow 9xy=18$

$\Rightarrow xy=2$

$\Rightarrow y=\frac{2}{x}$

Substituting $y$ in (i), we have

$x+\frac{2}{x}=3\Rightarrow x^2-3x+2=0\Rightarrow x^2-2x-x+2=0\Rightarrow x(x-2)-1(x-2)=0\Rightarrow (x-1)(x-2)=0\Rightarrow x=1,2$

Now $y=\frac{2}{x}$

$x=1\Rightarrow y=\frac{2}{1}=2x=2\Rightarrow y=\frac{2}{2}=1$