Question

Solve the following systems of equations.
$x^2+y^2-xy=13,x+y-\sqrt{xy}=3$

Answer 1

$x^2+y^2-xy=13...\left(i\right)$

$x+y-\sqrt{xy}=\mathrm{3....}\left(ii\right)$

Simplifying equation (i)

$x^2+y^2-xy=13$

or, $x^2+y^2+2xy-2xy-xy=13$

or, $(x+y{)}^2-2xy-xy=13$ $[a^2+b^2+2ab=(a+b{)}^2]$

or, $(x+y{)}^2-3xy=\mathrm{13...}(iii)$

Simplifying equation (ii)

$x+y-\sqrt{xy}=3$

or, $x+y=\sqrt{xy}+\mathrm{3....}\left(iv\right)$

Putting value of $(x+y)$ in equation (iii)

$(\sqrt{xy}+3{)}^2-3xy=13$

or, $xy+9+6\sqrt{xy}-3xy=13$

or, $6\sqrt{xy}-2xy=13-9$

or, $6\sqrt{xy}-2xy=4$

or, $4-6\sqrt{xy}+2xy=0$

or, $(\sqrt{2xy}{)}^2-2\times \sqrt{2xy}\times \frac{3}{\sqrt{2}}+(\frac{3}{\sqrt{2}}{)}^2-(\frac{3}{\sqrt{2}}{)}^2+4=0$

or, $(\sqrt{2xy}-\frac{3}{\sqrt{2}}{)}^2-\frac{9}{4}+4=0$ $[a^2-2ab+b^2=(a-b{)}^2]$

or, $(\sqrt{2xy}-\frac{3}{\sqrt{2}}{)}^2=\frac{9}{4}-4$

$=\frac{9-16}{4}$

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

Since, square of a real number is never negative

Hence, the given pair of equations will have

no real solution.