The system is: x^2+2y^2+3xy=12, y^2−3y=4, Turn x^2+2y^2+3xy into (x+y)^2+y^2+xy, but it's a dead end from here.

fluerkg

fluerkg

Answered question

2022-10-22

The system is:
x 2 + 2 y 2 + 3 x y = 12
y 2 3 y = 4
Turn x 2 + 2 y 2 + 3 x y into ( x + y ) 2 + y 2 + x y, but it's a dead end from here. Any help?

Answer & Explanation

Layton Leach

Layton Leach

Beginner2022-10-23Added 15 answers

Solve 12 y 2 3 y 4 = 0 for y
Set those two values one by one in the first equation
Chelsea Pruitt

Chelsea Pruitt

Beginner2022-10-24Added 5 answers

The system consists the two equations x 2 + 2 y 2 + 3 x y = 12 and y 2 3 y = 4..
From the second equation we get y = 1 , y = 4.. Then from the first equation we find x .

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