Question

x^2+4y^2=4
y=x+1
Part I: What two shapes intersect in this problem? (2 points)


Part II: Use substitution to solve for the two values of x. (4 points)


Part III: Use the values of x you just found to get the corresponding values of y, representing the points of intersection of these two shapes. Write your answers in the form (x,y). (4 points)

Answer 1

what you can probably do is this:

\[y = (x+1)\]

you can plug in x+1 wherever you see y in the first equation.

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

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

Answer 2

\[(5x^{2}+8x+4) = 4\]

\[5x^{2}+8x = 0\]

do you know how you would continue?

Answer 3

@jh99

Answer 4

How did you solve it?

\[x(5x+8) = 0 \]

Answer 5

@photon336 I apologize for being late. Thank you for helping me out.
On the equation above, you solved for x?

Answer 6

x(5x+8) = 0

Answer 7

ideas?

Answer 8

Can we say 5x+8=0
?

Answer 9

yes we can

Answer 10

okay so step #1 solve for x

Answer 11

x=-5/8?

Answer 12

\[5x+8 = 0\]

\[\frac{ -8 }{ 5 } = x \]

Answer 13

We need to plug in that value to solve for y?

Answer 14

Also, I was wondering if the first equation is a circle..?

Answer 15

okay so next step let's plug this into the original equation

y = x+1

Answer 16

yep :) it is

Answer 17

We'll get the same results if we put it into the first equation, right? (Just to make sure)

Answer 18

\[(y = x+1) = \frac{ -8 }{ 5 }+1 = \frac{ -8+5 }{ 5 } = \frac{ -3 }{ 5 } = y \]

Answer 19

so we've got x = -8/5 and y = -3/5 now to see if this is true let's plug it into the circle equation

Answer 20

Do we need to plug it into a specific equation? It has to be the first one? The second one would be much easier!

Answer 21

so yeah we need to just do this to check. if we plug in x and y into the circle and get 4 then our equation is correct.

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

\[(\frac{ -8 }{ 5 })^{2}+4(\frac{ -3 }{ 5 })^{2} = 4\]

\[\frac{ 64 }{ 25 }+4*(\frac{ 9 }{ 25 }) = \frac{ 36 }{ 25 }+\frac{ 64 }{ 25 } = \frac{ 100 }{ 25 } = 4\]

Answer 22

Thank you so much for helping me out!
We can also just plug that in for y=x+1, right
?

again, thanks a lot.

Answer 23

wait before you go there is another point we need

Answer 24

note that (-8/5,-3/5) is only one point of intersection between the line and the circle.

Answer 25

I also want to show you how the graph looks and the intersection.

Answer 26

take a look

http://prntscr.com/biyjsb

Answer 27

now to find the second point.

notice how in the first equation we had:

\[(5x^{2}+8x) = x(5x+8) ~ x = 0 \]

x = 0 is another solution so we should plug into both equations:

\[(y+1) => (0+1) = 1 \]

and in

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

\[(0)^{2}+ 4(1)^{2} = 0+4 = 4 \]

Answer 28

Whenever you solve a system of equations if you find one set of points say x = 0 it must satisfy both equations if it doesn't then your answer is wrong.

Answer 29

For part I, you have to know that the first is an ellipse and the second is a line.

Answer 30

yeah also this is a circle in case you're interested.

\[x^{2}+4y^{2} = 4 | (x-h)^{2}+(y-k)^{2} = r^{2}| center~is~(h,k)|(0,0)\]

Answer 31

thank you so much guys!

Answer 32

@Photon336 wait, how can a circle be an ellipse?