# How would you solve the following system of equations: x 2 </msup> + y =

How would you solve the following system of equations:
${x}^{2}+y=4\phantom{\rule{0ex}{0ex}}x+{y}^{2}=10$
I tried defining y in terms of x and then inserting in to the second equation:
$y=4-{x}^{2}\phantom{\rule{0ex}{0ex}}x+\left(4-{x}^{2}{\right)}^{2}=10$
Expand the second equation:
$x+16-8{x}^{2}+{x}^{4}=10$
Rearrange the terms:
${x}^{4}-8{x}^{2}+x+6=0$
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Savion Stanton
$y=4-{x}^{2}$
put it in 2nd equation
$x+\left(4-{x}^{2}{\right)}^{2}=10$
$x+16+{x}^{4}-8{x}^{2}=10$
${x}^{4}-8{x}^{2}+x+6=0$
${x}^{4}-{x}^{3}+{x}^{3}-7{x}^{2}-{x}^{2}+7x-6x+6=0$
${x}^{4}-{x}^{3}+{x}^{3}-{x}^{2}-7{x}^{2}+7x-6x+6=0$
${x}^{3}\left(x-1\right)+{x}^{2}\left(x-1\right)-7x\left(x-1\right)-6\left(x-1\right)=0$
$\left(x-1\right)\left({x}^{3}+{x}^{2}-7x-6\right)=0\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}x-1=0\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}x=1$
solve the above equation one of the value of $x=1$ and $y=3$

Dayanara Terry
To solve general equations of the form $a{x}^{4}+b{x}^{3}+c{x}^{2}+dx+e=0$ requires quartic formula.