To solve for the greatest possible value of [tex]\( x \)[/tex] where there is some solution [tex]\( (x, y) \)[/tex] to the given system of equations, we will follow these steps:
1. Write the system of equations:
[tex]\[
\begin{cases}
y = x^2 + 8x - 33 \\
y = 2x - 6
\end{cases}
\][/tex]
2. Set the equations equal to each other:
Since both expressions equal [tex]\( y \)[/tex], we can set the right-hand sides of these equations equal to each other:
[tex]\[
x^2 + 8x - 33 = 2x - 6
\][/tex]
3. Rearrange the equation:
Get all terms on one side to form a standard quadratic equation:
[tex]\[
x^2 + 8x - 33 - 2x + 6 = 0
\][/tex]
Simplify by combining like terms:
[tex]\[
x^2 + 6x - 27 = 0
\][/tex]
4. Solve the quadratic equation:
We can solve the quadratic equation [tex]\( x^2 + 6x - 27 = 0 \)[/tex] using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = 6 \)[/tex], and [tex]\( c = -27 \)[/tex]:
[tex]\[
x = \frac{-6 \pm \sqrt{6^2 - 4 \cdot 1 \cdot (-27)}}{2 \cdot 1}
\][/tex]
[tex]\[
x = \frac{-6 \pm \sqrt{36 + 108}}{2}
\][/tex]
[tex]\[
x = \frac{-6 \pm \sqrt{144}}{2}
\][/tex]
[tex]\[
x = \frac{-6 \pm 12}{2}
\][/tex]
This yields two solutions:
[tex]\[
x = \frac{-6 + 12}{2} = \frac{6}{2} = 3
\][/tex]
and
[tex]\[
x = \frac{-6 - 12}{2} = \frac{-18}{2} = -9
\][/tex]
5. Identify the greatest [tex]\( x \)[/tex] value:
The solutions for [tex]\( x \)[/tex] are [tex]\( 3 \)[/tex] and [tex]\( -9 \)[/tex]. The greatest possible value of [tex]\( x \)[/tex] is therefore:
[tex]\[
\boxed{3}
\][/tex]
