# Solve the system of equations: 3x - y = -2, 2x^2 - y = 0

Solve the system of equations: $3x-y=-2,2{x}^{2}-y=0$
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odgovoreh
Step 1
Given: The system of equations
3x−y=−2
$2{x}^{2}-y=0$
To find: The solution of given system of equation.
Step 2
Explanation:
Let
3x−y=−2 ...(i)
$2{x}^{2}-y=0$ ...(ii),
From equation (i),
3x−y=−2,
$⇒y=3x+2$
Substituting y=3x+2 in equation (ii),
$2{x}^{2}-y=0$
$⇒2{x}^{2}-\left(3x+2\right)=0$
$⇒2{x}^{2}-3x-2=0$
$⇒2{x}^{2}-4x+x-2=0$
$⇒\left(x-2\right)\left(2x+1\right)=0$
$⇒x=2,-\frac{1}{2}$
Now,
For $x=2,y=3×2+2=8$,
For $x=-\frac{1}{2},y=3×\left(-\frac{1}{2}\right)+2=-\frac{3}{2}+2=\frac{1}{2}$
Hence x=2, y=8 and $x=-\frac{1}{2},y=\frac{1}{2}$ are solution of given system of equations.
Result: x=2, y=8 and $x=-\frac{1}{2},y=\frac{1}{2}$ are solution of given system of equation.