To solve the equation
[tex]\[
\frac{x^2 + 9}{x + 9} = \frac{90}{x + 9},
\][/tex]
we start by noting that [tex]\( x \neq -9 \)[/tex] to avoid division by zero. Let's proceed step-by-step.
1. Cross-multiply to eliminate the denominators:
[tex]\[
(x^2 + 9) \cdot (x + 9) = 90 \cdot (x + 9).
\][/tex]
2. Notice that both sides have the factor [tex]\( x + 9 \)[/tex]. We can cancel out [tex]\( x + 9 \)[/tex] from both sides, provided [tex]\( x \neq -9 \)[/tex]:
[tex]\[
x^2 + 9 = 90.
\][/tex]
3. Move 90 to the left side and simplify:
[tex]\[
x^2 + 9 - 90 = 0.
\][/tex]
4. Simplify the equation:
[tex]\[
x^2 - 81 = 0.
\][/tex]
5. Factor the quadratic equation:
[tex]\[
(x + 9)(x - 9) = 0.
\][/tex]
6. Solve for [tex]\( x \)[/tex]:
[tex]\[
x + 9 = 0 \quad \text{or} \quad x - 9 = 0.
\][/tex]
[tex]\[
x = -9 \quad \text{or} \quad x = 9.
\][/tex]
7. Check for extraneous solutions:
We must check each potential solution in the original equation.
- For [tex]\( x = 9 \)[/tex]:
[tex]\[
\frac{9^2 + 9}{9 + 9} = \frac{90}{9 + 9}
\][/tex]
[tex]\[
\frac{81 + 9}{18} = \frac{90}{18}
\][/tex]
[tex]\[
\frac{90}{18} = 5 \quad \text{and} \quad \frac{90}{18} = 5.
\][/tex]
The equation holds true.
- For [tex]\( x = -9 \)[/tex]:
Substituting [tex]\( x = -9 \)[/tex] into the original equation results in division by zero, which is undefined. Thus, [tex]\( x = -9 \)[/tex] is an extraneous solution.
Therefore, the valid solution to the equation is:
[tex]\[
x = 9.
\][/tex]
