To solve the system of equations:
[tex]\[ y = 2x - 2 \][/tex]
[tex]\[ y = x^2 - x - 6 \][/tex]
we need to find the values of [tex]\(x\)[/tex] and [tex]\(y\)[/tex] that satisfy both equations simultaneously. Here is a step-by-step solution:
1. Since both equations are equal to [tex]\(y\)[/tex], we set the expressions for [tex]\(y\)[/tex] equal to each other:
[tex]\[
2x - 2 = x^2 - x - 6
\][/tex]
2. Rearrange the equation to form a standard quadratic equation:
[tex]\[
x^2 - x - 6 - 2x + 2 = 0
\][/tex]
[tex]\[
x^2 - 3x - 4 = 0
\][/tex]
3. Factor the quadratic equation:
[tex]\[
x^2 - 3x - 4 = (x - 4)(x + 1) = 0
\][/tex]
4. Set each factor equal to zero and solve for [tex]\(x\)[/tex]:
[tex]\[
x - 4 = 0 \quad \text{or} \quad x + 1 = 0
\][/tex]
[tex]\[
x = 4 \quad \text{or} \quad x = -1
\][/tex]
5. Substitute these values of [tex]\(x\)[/tex] back into the first equation [tex]\( y = 2x - 2 \)[/tex] to find the corresponding [tex]\(y\)[/tex]-values:
For [tex]\(x = 4\)[/tex]:
[tex]\[
y = 2(4) - 2 = 8 - 2 = 6
\][/tex]
For [tex]\(x = -1\)[/tex]:
[tex]\[
y = 2(-1) - 2 = -2 - 2 = -4
\][/tex]
6. Therefore, the solutions to the system of equations are:
[tex]\[
(x, y) = (4, 6) \quad \text{and} \quad (-1, -4)
\][/tex]
So the solutions to the system of equations are [tex]\((-1, -4)\)[/tex] and [tex]\( (4, 6)\)[/tex].
