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

C

Explanation:

Answer 2

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* 36+24`

`y=312`

At x=6:

we can plug x=6

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

`y=8* 36+24`

`y=312`

so, we get solution as

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