Question

Can someone help me with substitution??

x^2+4y^2=4
y=x+1

Use the substitution method to solve for the two values of x. I got (0,1) and (-8/5, -3/5). But I feel like that isn't correct.

Answer 1

first, you should isolate the x in the first equation

Answer 2

so, x^2= 4y^2-4?

Answer 3

it has to be in slope intercept form

Answer 4

first one is an ellipse, second is a line. how many points of intersection are possible?

Answer 5

two?

Answer 6

0, 1, 2

Answer 7

substitute x+1 for y in the first equation and solve for x.

Answer 8

x^2 + 4(x+1)^2 = 4

Answer 9

Got it! I'm solving it.

Am I correct so far?

x^2+4*(x+1)^2=4

x^2+4x^2+16=4
-16 -16
x^2+4x^2= -12?

Answer 10

no...
(x+1)(x+1) = x^2 + 2x + 1

4(x+1)^2 = 4(x^2 + 2x + 1)

Answer 11

Oh goodness! I'm so lost! D;

Answer 12

x^2 + 4(x+1)^2 = x^2 + 4(x^2 + 2x + 1) = x^2 + 4x^2 + 8x + 4 = 4

so

5x^2 + 8x = 0 => x(5x + 8) = 0 => x = 0, x = -8/5

those are the x values of the points of intersection. plug those into y = x+1 to find the y values.

Answer 13

So y=1 and y= -0.6?