To solve the system of equations:
[tex]\[
\begin{array}{l}
y = -x^2 + 4 \\
y = 2x + 1
\end{array}
\][/tex]
we need to find the values of [tex]\(x\)[/tex] and [tex]\(y\)[/tex] that satisfy both equations simultaneously. Here's the step-by-step solution:
1. Set the equations equal to each other since they both equal [tex]\(y\)[/tex]:
[tex]\[
-x^2 + 4 = 2x + 1
\][/tex]
2. Rearrange the equation to set it to zero:
[tex]\[
-x^2 - 2x + 4 - 1 = 0
\][/tex]
[tex]\[
-x^2 - 2x + 3 = 0
\][/tex]
3. Multiply through by -1 to make the coefficient of [tex]\(x^2\)[/tex] positive:
[tex]\[
x^2 + 2x - 3 = 0
\][/tex]
4. Factor the quadratic equation:
[tex]\[
x^2 + 2x - 3 = (x + 3)(x - 1) = 0
\][/tex]
5. Solve for [tex]\(x\)[/tex]:
[tex]\[
x + 3 = 0 \quad \text{or} \quad x - 1 = 0
\][/tex]
[tex]\[
x = -3 \quad \text{or} \quad x = 1
\][/tex]
6. Substitute the values of [tex]\(x\)[/tex] back into either original equation to find the corresponding values of [tex]\(y\)[/tex]. We use [tex]\(y = 2x + 1\)[/tex] for simplicity:
- For [tex]\(x = -3\)[/tex]:
[tex]\[
y = 2(-3) + 1 = -6 + 1 = -5
\][/tex]
- For [tex]\(x = 1\)[/tex]:
[tex]\[
y = 2(1) + 1 = 2 + 1 = 3
\][/tex]
7. Write the solutions as ordered pairs:
The solutions to the system of equations are [tex]\((-3, -5)\)[/tex] and [tex]\((1, 3)\)[/tex].
So, the solutions are:
[tex]\[
\{ (-3, -5), (1, 3) \}.
\][/tex]
