### Instruction: Find all real values of $x$ that satisfy $x + \frac{45}{x-4} = -10.$

Enter all the solutions, separated by commas.### Response: To solve for $x$, I want to get rid of the fraction and have a polynomial equation.
To do that, I can multiply both sides of the equation by $x-4$, which is the denominator of the fraction.
However, I have to be careful not to introduce any extraneous solutions by doing this, since multiplying by $x-4$ is not valid when $x=4$, as that would make the fraction undefined.
So, I will multiply both sides by $x-4$ and then check if $x=4$ is a solution or not later.
Multiplying both sides by $x-4$, I get
\begin{align*}
(x-4)(x + \frac{45}{x-4}) &= (x-4)(-10) \\
x^2 -4x + 45 &= -10x + 40 \\
x^2 + 6x + 5 &= 0
\end{align*}
This is a quadratic equation, which I can try to factor or use the quadratic formula to solve.
I notice that the constant term, 5, is the product of two factors that add up to the coefficient of $x$, 6.
So I can try to find those factors and write the equation as a product of two binomials.
The factors of 5 that add up to 6 are 1 and 5, so I can write
\begin{align*}
x^2 + 6x + 5 &= 0 \\
(x+1)(x+5) &= 0
\end{align*}
Now, by the zero product property, I know that either $x+1=0$ or $x+5=0$, or both.
Solving for $x$, I get $x=-1$ or $x=-5$ as possible solutions.
However, I have to remember that I multiplied by $x-4$ at the beginning, so I have to check if these solutions make the original equation true or not.
If $x=-1$, then the original equation becomes
\begin{align*}
-1 + \frac{45}{-1-4} &= -10 \\
-1 - 9 &= -10 \\
-10 &= -10
\end{align*}
This is true, so $x=-1$ is a valid solution.
If $x=-5$, then the original equation becomes
\begin{align*}
-5 + \frac{45}{-5-4} &= -10 \\
-5 - 5 &= -10 \\
-10 &= -10
\end{align*}
This is also true, so $x=-5$ is also a valid solution.
Therefore, the two solutions are $x=-1$ and $x=-5$.
I also have to make sure that $x=4$ is not a solution, since I excluded that possibility when I multiplied by $x-4$.
If $x=4$, then the original equation becomes
\begin{align*}
4 + \frac{45}{4-4} &= -10 \\
4 + \frac{45}{0} &= -10
\end{align*}
This is undefined, since I cannot divide by zero.
So $x=4$ is not a solution, as expected.
Therefore, the final answer is $x=-1,-5$.