Question

Please help

For the system shown below what are the coordinates of the solution that lies in quadrant II? write your answer in form (a,b) without using spaces.
x^2+4y^2=100
4y-x^2=-20

Answer 1

(6,4)

Explanation:

Just took the quiz

Answer 2

(-6,4)

Explanation:

The equations are:

`x^2+4y^2=100\\4y-x^2=-20`

Solving for x^2 of the 2nd equation and putting that in place of x^2 in the 2nd equation we have:

`4y-x^2=-20\\x^2=4y+20\\-------\\x^2+4y^2=100\\4y+20+4y^2=100`

Now we can solve for y:

`4y+20+4y^2=100\\4y^2+4y-80=0\\y^2+y-20=0\\(y+5)(y-4)=0\\y=4,-5`

So plugging in y = 4 into an equation and solving for x, we have:

`x^2=4y+20\\x=+-√(4y+20) \\x=+-√(4(4)+20) \\x=+-√(36) \\x=6,-6`

So y = 4 corresponds to x = 6 & x = -6

The pairs would be

(6,4) & (-6,4)

we see that (-6,4) falls in the 2nd quadrant, thus this is the solution we are looking for.