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

Enter all the solutions, separated by commas.### Response: Let's start by multiplying both sides of the equation by $x-4$.
That will get rid of the fraction and give us $x(x-4)+45=-10(x-4)$.
Expanding both sides gives us $x^2-4x+45=-10x+40.$
Now let's move all the terms to one side of the equation and get $x^2+6x+5=0$.
We can solve this equation by factoring it.
Yes, the left side of the equation factors into $(x+5)(x+1)$.
So now we have $(x+5)(x+1)=0$.
In order for this equation to be true, one of the factors on the left side must be equal to zero.
That's right. So either $x+5=0$ or $x+1=0$.
Solving for $x$ in each of these equations gives us $x=-5$ or $x=-1$.