Question

Solve the following systems of equations.
$x^2+y^2=41,x+y=9$

Answer 1

Given that $x^2+y^2=41$ and $x+y=9$

now we have,

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

Now put the given value

$(9{)}^2=41+2xy$

$2xy=81-41=40$

$xy=20$

$(x-y{)}^2=(x+y{)}^2-4xy$

$(x-y{)}^2=(9{)}^2-4\times 20$

$(x-y{)}^2=81-80=1$

$(x-y{)}^2=1$

So, $(x-y)=1,-1$

$(x+y)=\mathrm{9...........}\left(a\right)$

$(x-y)=\mathrm{1........}\left(b\right)$

Now add $\left(a\right)$ and $\left(b\right)$, we have

$2x=10$

$x=5$

and $y=9-5=4$

For $(x-y)=-1$

$x=4$ and $y=9-4=5$