Question

Solve by using system of equations.

Answer 1

Answer:

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

Step-by-step explanation:

Given quadratic equations:

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

Solve by the method of substitution by substituting the first equation into the second equation:

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

Add 4x² to both sides:

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

$\implies 6x^2+6x+1=1$

Subtract 1 from both sides:

$\implies 6x^2+6x+1-1=1-1$

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

Factor out the common term 6x:

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

Apply the zero-product property:

$\implies 6x=0 \implies x=0$

$\implies (x+1)=0 \implies x=-1$

Substitute the found values of x into one of the equations and solve for y:

$\textsf{When } \: x = 0\implies y=-4(0)^2+1=1$

$\textsf{When } \: x = -1\implies y=-4(-1)^2+1=-3$

Therefore, the solutions to the given system of equations are (0, 1) and (-1, -3).

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Answer 2

$2x^2 +6x+1=-4x^2 + 1 \\ \\ 6x^2 + 6x=0 \\ \\ x^2 + x=0 \\ \\ x(x+1)=0 \\ \\ x=0, -1 \\ \\ \\ x=0 \implies y=1 \\ \\ x=-1 \implies y=-3$

So, the solutions are (x,y)={(0,1), (-1,-3)}