Question

What are the solutions of the following system? { 10x^2-y=48 and 2y=16x^2+48

A) (2 sqrt3, 120) and (-2 sqrt3, 120)
B) (2 sqrt3, 120) and (-2 sqrt 3, -72)
C) (6, 312) and (-6, 312)
D) (6, 312) and (-6,-264)

Answer 1

we are given

$10x^2-y=48$

$2y=16x^2+48$

Firstly, we will solve for y

$y=8x^2+24$

now, we can plug that into first equation

$10x^2-(8x^2+24)=48$

now, we can simplify it

$10x^2-8x^2-24=48$

$2x^2-24=48$

Add both sides by 24

$2x^2-24+24=48+24$

$2x^2=48+24$

$2x^2=72$

Divide both sides by 2

$x^2=36$

now, we can solve for x

$x=-6,x=6$

now, we can find y-value

$y=8x^2+24$

At x=-6:

we can plug x=-6

$y=8(-6)^2+24$

$y=8\times 36+24$

$y=312$

At x=6:

we can plug x=6

$y=8(6)^2+24$

$y=8\times 36+24$

$y=312$

so, we get solution as

C) (6, 312) and (-6, 312)..........Answer

Answer 2

Answer:

C

Step-by-step explanation: