To solve the system of equations:
[tex]\[
\begin{array}{c}
y = 2x^2 - 9x + 7 \\
y = 2x - 2
\end{array}
\][/tex]
we can set the two equations equal to each other since they both equal [tex]\( y \)[/tex]. This gives:
[tex]\[
2x^2 - 9x + 7 = 2x - 2
\][/tex]
Next, we will move all terms to one side of the equation to set it to 0:
[tex]\[
2x^2 - 9x + 7 - 2x + 2 = 0
\][/tex]
Combine like terms:
[tex]\[
2x^2 - 11x + 9 = 0
\][/tex]
Now, we solve the quadratic equation [tex]\( 2x^2 - 11x + 9 = 0 \)[/tex] using the quadratic formula:
[tex]\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
For our equation [tex]\( 2x^2 - 11x + 9 = 0 \)[/tex], [tex]\(a = 2\)[/tex], [tex]\(b = -11\)[/tex], and [tex]\(c = 9\)[/tex]. Substituting these values into the quadratic formula, we get:
[tex]\[
x = \frac{-(-11) \pm \sqrt{(-11)^2 - 4 \cdot 2 \cdot 9}}{2 \cdot 2}
\][/tex]
Simplify inside the square root:
[tex]\[
x = \frac{11 \pm \sqrt{121 - 72}}{4}
\][/tex]
[tex]\[
x = \frac{11 \pm \sqrt{49}}{4}
\][/tex]
[tex]\[
x = \frac{11 \pm 7}{4}
\][/tex]
This results in two potential solutions for [tex]\( x \)[/tex]:
[tex]\[
x = \frac{11 + 7}{4} = \frac{18}{4} = \frac{9}{2}
\][/tex]
[tex]\[
x = \frac{11 - 7}{4} = \frac{4}{4} = 1
\][/tex]
Now, let's check which of these possible values of [tex]\( x \)[/tex] match the given options:
A) [tex]\(-1\)[/tex] \\
B) [tex]\(\frac{7}{2}\)[/tex] \\
C) [tex]\(\frac{9}{2}\)[/tex] \\
D) [tex]\(\frac{11}{2}\)[/tex]
The value [tex]\( x = 1 \)[/tex] is not listed in the given options. However, [tex]\( x = \frac{9}{2} \)[/tex] is indeed one of the provided options.
Therefore, the correct answer is:
[tex]\[
\boxed{\frac{9}{2}}
\][/tex]
