Answer:
Given the following system: [tex]10x^2-y=48[/tex] and [tex]2y =16x^2+48[/tex]
we can write it as:
[tex]y=10x^2-48[/tex] .......[1]
Substitute this in equation [tex]2y =16x^2+48[/tex] we get;
[tex]2(10x^2-48)=16x^2+48[/tex] or
[tex]20x^2-96=16x^2+48[/tex]
Subtract [tex]16x^2[/tex] from both sides of an equation:
[tex]20x^2-96 -16x^2= 16x^2+48-16x^2[/tex] or
[tex]4x^2-96=48[/tex] or
[tex]4x^2=144[/tex]
Simplify:
[tex]x^2=36[/tex]
⇒ x = 6 , -6
Now, substituting these x values in equation [1];
for x =6
[tex]y=10x^2-48[/tex]
[tex]y=10(6)^2-48[/tex] or
[tex]y= 10\cdot 36 -48[/tex] =360-48 = 312
Similarly for, x = -6
[tex]y=10x^2-48[/tex]
[tex]y=10(-6)^2-48[/tex] or
[tex]y= 10\cdot 36 -48[/tex] =360-48 = 312
So, we have the solution of the given following system; (6,312) and (-6,312).
Also shown in graph below;
