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

Respuesta :

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 = -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.