# Point of intersection

• Apr 10th 2008, 08:10 PM
MathNeedy18
Point of intersection
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?
• Apr 10th 2008, 08:12 PM
Mathstud28
Hmm
Quote:

Originally Posted by MathNeedy18
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?

I think you made a mistake...solving and setting them equal I got $\displaystyle \frac{-x}{2}+3=\sqrt{5-\frac{x^2}{4}}$
• Apr 10th 2008, 08:15 PM
MathNeedy18
Quote:

Originally Posted by Mathstud28
I think you made a mistake...solving and setting them equal I got $\displaystyle \frac{-x}{2}+3=\sqrt{5-\frac{x^2}{4}}$

I think I am doing the algebra wrong because when I solve that, the 1/4 cancel out. :(
• Apr 10th 2008, 08:21 PM
TheEmptySet
Quote:

Originally Posted by MathNeedy18
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)