To solve the system of equations given by:
[tex]\[ y = -x^2 + 4 \][/tex]
and
[tex]\[ y = 2x + 1 \][/tex]
we need to find the values of [tex]\(x\)[/tex] and [tex]\(y\)[/tex] that satisfy both equations simultaneously.
Step 1: Set the Equations Equal to Each Other
Since both equations equal [tex]\(y\)[/tex], we can set them equal to each other to solve for [tex]\(x\)[/tex]:
[tex]\[ -x^2 + 4 = 2x + 1 \][/tex]
Step 2: Rearrange the Equation
To solve for [tex]\(x\)[/tex], we rearrange the equation to bring all terms to one side:
[tex]\[ -x^2 - 2x + 4 - 1 = 0 \][/tex]
Simplifying gives us:
[tex]\[ -x^2 - 2x + 3 = 0 \][/tex]
For easier handling, let's multiply through by [tex]\(-1\)[/tex]:
[tex]\[ x^2 + 2x - 3 = 0 \][/tex]
Step 3: Factor the Quadratic Equation
Now, we factor the quadratic equation:
[tex]\[ x^2 + 2x - 3 = (x + 3)(x - 1) = 0 \][/tex]
Step 4: Solve for [tex]\(x\)[/tex]
Set each factor equal to zero to find the solutions for [tex]\(x\)[/tex]:
[tex]\[ (x + 3) = 0 \][/tex]
[tex]\[ x = -3 \][/tex]
and
[tex]\[ (x - 1) = 0 \][/tex]
[tex]\[ x = 1 \][/tex]
So, the solutions for [tex]\(x\)[/tex] are [tex]\(x = -3\)[/tex] and [tex]\(x = 1\)[/tex].
Step 5: Solve for [tex]\(y\)[/tex]
Now, we substitute each [tex]\(x\)[/tex] value back into one of the original equations to find the corresponding [tex]\(y\)[/tex] values. We will use the first equation [tex]\(y = -x^2 + 4\)[/tex]:
1. For [tex]\(x = -3\)[/tex]:
[tex]\[ y = -(-3)^2 + 4 \][/tex]
[tex]\[ y = -9 + 4 \][/tex]
[tex]\[ y = -5 \][/tex]
2. For [tex]\(x = 1\)[/tex]:
[tex]\[ y = -(1)^2 + 4 \][/tex]
[tex]\[ y = -1 + 4 \][/tex]
[tex]\[ y = 3 \][/tex]
Step 6: Present the Solutions
The solutions to the system of equations are the pairs [tex]\((x, y)\)[/tex]:
[tex]\[ (x, y) = (-3, -5) \][/tex]
and
[tex]\[ (x, y) = (1, 3) \][/tex]
So, the system of equations has two solutions: [tex]\((-3, -5)\)[/tex] and [tex]\((1, 3)\)[/tex].
