Question

Solve the following system of equations. Show all work and solutions.

Answer 1

Answer:

System of equations

$\large \begin{cases}y=2x^2+6x+1\\y = -4x^2+1\end{cases}$

To solve the given system of equations, use the substitution method:

$\large\begin{aligned}2x^2+6x+1 & = -4x^2+1\\2x^2+6x+1+4x^2-1 & = 0\\6x^2+6x & = 0 \\6x(x+1) & = 0\\\implies x+1 & = 0 \implies x=-1\\\implies 6x & = 0 \implies x=0\end{aligned}$

Therefore the x-values of the solutions are:

        $\large \begin{aligned}x & =-1\\x & =0\end{aligned}$

Substitute the found values of x into the second equation to find the y-values:

$\large \begin{aligned}x & =-1 & \implies &-4(-1)^2+1 =-3 & \implies & (-1,-3)\\x & =0 & \implies & -4(0)^2+1 =1 & \implies & (0,1)\end{aligned}$

Therefore, the solutions of the system of equations are:

(-1, -3) and (0, 1)

Answer 2

Answer: (0, 1), (-1, -3)

Given two equation's:

1. y = 2x² + 6x + 1
2. y = −4x² + 1

Solve them simultaneously:

$\sf 2x^2 + 6x + 1 = -4x^2 + 1$

$\sf 2x^2 + 4x^2 = 1 - 1$

$\sf 6x^2 + 6x = 0$

$\sf 6x(x + 1) = 0$

$\sf 6x = 0, \ x + 1 = 0$

$\sf x = 0, \ x = -1$

(a) When x = 0,     y = −4(0)² + 1 = 1

(b) When x = -1,    y = -4(-1)² + 1 = -3

Solution: (x, y) → (0, 1), (-1, -3)