To solve the given system of equations:
[tex]\[
\left\{\begin{array}{l}
y = x^2 - 6x + 5 \\
y = 2x - 7
\end{array}\right.
\][/tex]
we need to find the points [tex]\((x, y)\)[/tex] where the two equations intersect.
1. Step 1: Set the equations equal to each other because at the points of intersection the [tex]\(y\)[/tex]-values from both equations must be equal:
[tex]\[
x^2 - 6x + 5 = 2x - 7
\][/tex]
2. Step 2: Rearrange the equation to set it to zero:
[tex]\[
x^2 - 6x + 5 - 2x + 7 = 0 \\
x^2 - 8x + 12 = 0
\][/tex]
3. Step 3: Factor the quadratic equation:
[tex]\[
x^2 - 8x + 12 = (x - 2)(x - 6) = 0
\][/tex]
4. Step 4: Solve for [tex]\(x\)[/tex]:
[tex]\[
x - 2 = 0 \quad \text{or} \quad x - 6 = 0 \\
x = 2 \quad \text{or} \quad x = 6
\][/tex]
5. Step 5: Substitute the [tex]\(x\)[/tex]-values back into either original equation to find the corresponding [tex]\(y\)[/tex]-values. We can use the linear equation [tex]\(y = 2x - 7\)[/tex] because it is simpler:
For [tex]\(x = 2\)[/tex]:
[tex]\[
y = 2(2) - 7 = 4 - 7 = -3
\][/tex]
For [tex]\(x = 6\)[/tex]:
[tex]\[
y = 2(6) - 7 = 12 - 7 = 5
\][/tex]
6. Step 6: Write the solutions as ordered pairs and sort by the [tex]\(x\)[/tex]-values:
[tex]\[
(2, -3) \quad \text{and} \quad (6, 5)
\][/tex]
Thus, the solution to the system of equations is:
[tex]\[
([2, 6], [-3, 5])
\][/tex]
