Question

In the system shown below, what are the coordinates of the solution that lies in quadrant II? pic attached

Answer 1

y positive x negative

Answer 2

right

Answer 3

i figured -2,1 but it doesnt seem right and i keep doing it over an over an it comes out a mess

Answer 4

If you substitute the second equation into the first, you have a quadratic in y: 4y^2 + 32y - 80 = 0. Which simplifies to y^2 + 8 - 20 = 0 Did you get that far?

Answer 5

no

Answer 6

okay, before going on, I should explain how we did that. The second equation becomes 32y = x^2. We can now take 32y and put that in place of x^2 in the first equation. We get 32y + 4 y^2 = 80

Answer 7

okay got that like substitution

Answer 8

Continuing on, y^2 + 8y - 20 = 0 -> (y - 2)(y + 10) = 0 So, we get as solutions for y, both 2 and -10. But we want a quadrant II solution, so we choose positive y which is the value 2.

Answer 9

correct

Answer 10

Using the second equation, 2 = x^2/32 or x^2 = 64 and x is +8 or -8, but we have to have a negative x, so our solution is (-8, 2)

Answer 11

thankyou!

Answer 12

You're welcome!