Question

Solve the following systems of equations.
$x^4+y^4=82,xy=3$

Answer 1

$x^4+y^4=82⟶\left(1\right)$ $xy=3⟶\left(2\right)$

From $\left(1\right)$

$\Rightarrow x^4=y^4=82$

$\Rightarrow {(x^2+y^2)}^2-2x^2{y}^2=82$

${(x^2+y^2)}^2=2\times 3^2=82$ $(\because xy=3)$

${(x^2+y^2)}^2=82+2\times 9$

$=82+18$

$=100$

$\Rightarrow x^2+y^2=\sqrt{100}=10$

$x^2+y^2={(x+y)}^2-2xy=10$

${(x+y)}^2-2\times 3=10$ $(\because x+y=3)$

${(x+y)}^2=10+6=16$

$\Rightarrow x+y=4$

and $xy=3$ $\left(from\left(2\right)\right)$

$\therefore (4-y)y=3$

$4y-y^2=3$

$y^2-4y+3=0$

$y^2-y-3y+3=0$

$y(y-1)-3(y-1)=0$

$\Rightarrow (y-3)(y-1)=0$

When $y=1\Rightarrow x\times 1=3\Rightarrow y=3$

When $y=3\Rightarrow x\times 3=3\Rightarrow x=1$

Hence, solved.