Point of intersection

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April 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?
April 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 $\frac{-x}{2}+3=\sqrt{5-\frac{x^2}{4}}$
April 10th 2008, 08:15 PM
MathNeedy18

Quote:

Originally Posted by Mathstud28

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

I think I am doing the algebra wrong because when I solve that, the 1/4 cancel out. :(
April 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?

$x^2 +(2y)^2=20$ and $2y=6-x$

subbing into the first equation we get

$x^2+(6-x)^2=20$

$x^2+36-12x+x^2=20 \iff 2x^2-12x+16=0$

$2(x^2-6x+8)=0 \iff 2(x-4)(x-2)=0$

so x=2 or x=4 sub these values back into the linear equation

$2y=6-2 \iff y=2$

$2y=6-4 \iff y=1$

so we get the two points (2,2) and (4,1)

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