1 Answer

Dan Snyder · Answer 1 · 2024-07-04T13:20:48+0000

Answer:

The solutions to the given system of equations are:

Explanation:

Given system of equations:

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

To solve the given system of equations, substitute the second equation into the first equation:

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

Add x² to both sides:

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

Subtract 4 from both sides:

$\implies x^2+2x-3=0$

Factor the quadratic equation:

$\implies x^2+3x-x-3=0$

$\implies x(x+3)-1(x+3)=0$

$\implies (x-1)(x+3)=0$

Apply the zero product property:

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

$(x+3)=0 \implies x=-3$

Substitute the found values of x into the second equation and solve for y:

$x=1 \implies y=2(1)+1=3$

$x=-3 \implies y=2(-3)+1=-5$

Therefore, the solutions to the given system of equations are:

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