x^2 + 4y^2=20 and x+2y=6
I solved for y and set them equal to each other but the x's cancel out. What else could I do?
$\displaystyle x^2 +(2y)^2=20$ and $\displaystyle 2y=6-x $
subbing into the first equation we get
$\displaystyle x^2+(6-x)^2=20$
$\displaystyle x^2+36-12x+x^2=20 \iff 2x^2-12x+16=0 $
$\displaystyle 2(x^2-6x+8)=0 \iff 2(x-4)(x-2)=0$
so x=2 or x=4 sub these values back into the linear equation
$\displaystyle 2y=6-2 \iff y=2$
$\displaystyle 2y=6-4 \iff y=1$
so we get the two points (2,2) and (4,1)