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=1004y-x^2=-20

Answer:(-6,4)Step-by-step explanation:The equations are:[tex]x^2+4y^2=100\\4y-x^2=-20[/tex]Solving for x^2 of the 2nd equation and putting that in place of x^2 in the 2nd equation we have:[tex]4y-x^2=-20\\x^2=4y+20\\-------\\x^2+4y^2=100\\4y+20+4y^2=100[/tex]Now we can solve for y:[tex]4y+20+4y^2=100\\4y^2+4y-80=0\\y^2+y-20=0\\(y+5)(y-4)=0\\y=4,-5[/tex]So plugging in y = 4 into an equation and solving for x, we have:[tex]x^2=4y+20\\x=+-\sqrt{4y+20} \\x=+-\sqrt{4(4)+20} \\x=+-\sqrt{36} \\x=6,-6[/tex]So y = 4 corresponds to x = 6 & x = -6The 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.