Question

What are the solutions of the following system?

Answer 1

The correct answer is: (6, 312), (-6, 312)

Explanation:

This is a system of two square equations with two unknowns x and y

We will solve this system in the next way:

2 y = 16 x² + 48

First we will divide this equation with number 2, the both sides and get:

y = 8 x² + 24 => 8 x² - y = - 24

10 x² - y = 48 and 8 x² - y = - 24

Now we will subtract the second equation from the first and get, the both sides:

10 x² - 8 x² - y - (-y) = 48 - ( - 24) => 2 x² = 72 => x² = 72/2 = 36

x = √36 => x₁ = 6 and x₂ = - 6

now we will replace this solutions in this equation 8 x² - y = - 24 and get:

8 · 6² - y = - 24 => 8 · 36 - y = - 24 => 288 - y = - 24 => y₁ = 288 + 24 = 312

8 · (-6)² - y = - 24 => 8 · 36 - y = - 24 => 288 - y = - 24 => y₂ = 288 + 24 = 312

We have that y₁ = y₂ = 312

We have two pairs of solutions:

( 6, 312) and ( - 6, 312)

God with you!!!

Answer 2

x=2.83 y=88

hope this helped you hon :)

Explanation:

2y-16x^2=48

-2y+10x^2=48

y cancels out

-6x^2=48

divide six by both sides

x^2=8

Find the square root

x=2.83

plug x=2.83 to one of the equatiom

2y=16(2.83)^2+48

2y=16(8)+48

2y=128+48

2y=176

divide by 2

y=88