Question

Please help Medal! For the system shown below what are the coordinates of the solution that lies in Quadrant 2?
Write your answer in the form (a,b)
x^2+4y^2=100
4y^2-x^2=-20

Answer 1

\[8y^2 = 80; y^2 = 10; y =\+/-sqrt(10)\]
plug in 10 for y^2
\[x^2+4(10) = 100; x^2 = 60; x = +/- \sqrt(60)\]
In quadrant II, y is positive and x is negative

So the coordinates are

\[(-\sqrt(60), \sqrt(10))\]

Answer 2

Do you know how I got
\[8y^2 = 80\]?

Answer 3

no i dont

Answer 4

Add the two equations together.

\[4y^2+4y^2 = 8y^2\]
\[x^2-x^2 = 0\]
\[100+ (-20) = 80\]

Do you see?

Answer 5

I just added like terms.

Answer 6

ok yea i see it

Answer 7

wait omg i wrote one of the equations wrong!

Answer 8

my mistake this is the correct equations
x^2+4y^2=100
4y-x^2-20

Answer 9

@DoYourHomework

Answer 10

Ok

Answer 11

Wait. Is it 4y-x^2 = -20?

Answer 12

yes

Answer 13

Ok.

Answer 14

my keyboard isnt working properly

Answer 15

Add the two equations to get.
\[4y^2+4y = 80\]
Divide equation by 4
\[y^2+y = 20\]
Bring 20 on other side and solve for y.
\[y^2+y-20 = (y+5)(y-4); y = -5 and y = 4 \]
Since we are looking at the second quadrant, y must be positive. So, y = 4
Plug y into either equation

\[4(4)-x^2 = -20; -x^2 = -36; x^2 = 36; x = +/-6\]
Since x is negative in quadrant 2 the coordinates are

(-6,4)

Answer 16

woa thats great

Answer 17

idk how you could solve it quick

Answer 18

Idk, I've seen a lot of these problems before. Just apply what you learned in class that day to solve your homework problems. It's dumb advice, but it's true.

Answer 19

*your homework problems for that day.

Answer 20

yea

Answer 21

well thank you

Answer 22

np!