Question

How would I find the intersection points for this system of equations?

(I'll put the equations in the comments)

Answer 1

\[\frac{ x ^{2} }{ 4 }+y ^{2}=1\]\[y=x+1\]

Answer 2

Since y = x + 1... then can I plug it into the other equation?

\[\frac{ x ^{2} }{ 4 }+(x+1)^{2}=1\]

Answer 3

thats what i was thinking

Answer 4

I'm just not sure how to solve for x from there /:

Answer 5

multiply out the brackets, times through by 4 and solve the equation, im trying to do it also

Answer 6

iv got my answer -8/5

Answer 7

5x^2 +8x+4=4
5x^2 + 8x=0
5x + 8 =0
5x = -8
x = -8/5

Answer 8

x=-8/5 y=-3/5

Answer 9

I'm not quite sure how you got that... when I mutiplied out the parenthesis I got \[\frac{ x ^{2} }{ 4 }+x^{2}+2x+1=1\] and I moved the 4 to the other side of the equation to get \[2x ^{2}+2x+1=4\]

Answer 10

But I don't know if I'm doing it right either...

Answer 11

you have to times through the whole equation by 4 so x^2/4 + x^2+2x+1=1
times everything my 4 so it becomes x^2 +4x^2 + 8x + 4 = 4

Answer 12

Ohhh I understand now

Answer 13

And how did you turn 5x^2 + 8x = 0 into 5x + 8 = 0?

Answer 14

divide by x

Answer 15

Oh okay

Answer 16

And how did you find y? Did you plug the x value into the y = x + 1 equation?

Answer 17

yes

Answer 18

And how do I find the other intersection? There are two.

Answer 19

since we only got 1 answer from finding x there is only one intersection

Answer 20

otherwise we would have gotten a polynomial answer not a linear

Answer 21

I graphed the equations on a calculator though, and it shows two intersections. The first equation is a circle and the other one is a line.

Answer 22

x=0 is also an answer, as we divided through by x

Answer 23

When did we get that? I mean, it's right, but when did we get that?

Answer 24

dividing through by x also means that x=0