Answer:
The pair is (-8, 2
Step-by-step explanation:
Allow me to rewrite the system shown below:
[tex]x^{2} + 4y^{2} = 80[/tex] <=> [tex]x^{2} = 80 - 4y^{2}[/tex]
and y = [tex]\frac{1}{32} x^{2}[/tex] (1)
Substitute [tex]x^{2} = 80 - 4y^{2}[/tex] into equation (1), we have:
y = [tex]\frac{1}{32}[/tex] ([tex]80 - 4y^{2}[/tex] )
<=> 32y = 80 - [tex]4y^{2}[/tex]
<=> [tex]4y^{2}[/tex] + 32y -80 = 0
<=> y = 2 and y = -10
When y = 2 we have: x= 8 or x = -8
When y = -10, we have no x
Because the coordinates of the solution that lies in quadrant II, so we delete the value of x = 8
So the pair is (-8, 2)
