What are the solutions of the following system?
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Answer:
x=2.83 y=88
hope this helped you hon :)
Step-by-step 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
Answer:
The correct answer is: (6, 312), (-6, 312)
Step-by-step 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!!!