Question

Please help

Answer 1

Answer:

(-6,4)

Step-by-step 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:

$x^2=4y+20\\x=+-\sqrt{4y+20} \\x=+-\sqrt{4(4)+20} \\x=+-\sqrt{36} \\x=6,-6$

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.