How to solve for $x$ when $x$ is in both the numerator and denominator I'm terrible at solving equations. I can do simple things like $x=\frac{y}{2}$ but for example I'd like to solve for $x$ in the following equation
$$a=\frac{x(f+g)h+fgh2}{-x}$$
For example I tried these steps
Divide by $-x$:
$$a(-x)=x(f+g)h+fgh2$$
Now subtract $fgh2$:
$$a(-x)-fgh2=x(f+g)h$$
Now divide by $(f+g)h$:
$$\frac{a(-x)-fgh2}{(f+g)h}=x$$
But that's doesn't really help because $x$ is on both sides. I tried moving $x$ to the left by making it $=0$ but I was lost after that. Of course I'd like the solution but more than that I'd like to know how to think so I can solve the next one.
 A: Let me see if I can clarify.
When you re-arrange terms, you want all the x's to one side and all non-x's to the other.
Continuing from where you start
$$
a(-x)=x(f+g)h+fgh^2
$$
to get
$$
a(-x) - x(f+g)h = fgh^2$$
Now collect the coefficients of $x$
$$
x (-a -(f+g) h) = fgh^2$$
Now you can solve for $x$
$$
x = \frac{fgh^2}{(-a -(f+g) h)} = -  \frac{fgh^2}{(a +(f+g) h)}$$
A: You could divide every term on the right side of the equation by $x$ to get
$$a = \frac{(f+g)h+\frac{1}{x}fgh^2}{-1}.$$
Can you see where to go from here?
A: Forget this question for a moment: how would you solve the equation below?
$$2x + 3 = 4x + 5$$
A: $$a=\frac{x(f+g)h+fgh2}{-x}$$
Think of $(f+g)h$ as some number $3$. Think of $fgh2$ as some other number $5$. You can think of $a$ as a number also, say $7$. How would you solve it then?
$$7=\frac{3x+5}{-x}$$
Multiply both sides as you did by $-x$
$$-7x=3x+5$$
$$-7x-3x=5$$
Only in your problem, at this point, you factor out the x:
$$(-7-3)x=5$$
$$x=\frac{5}{-7-3}$$
and then simplify. You can do your problem similarly.
