Solving an equation for X I am really stuck on a very simple question! I am sorry, it's been 3 years since I did this.
The equation is: $$\frac{5}{x+1}-\frac{1}{2x-1}=0$$
The problem I'm having is I have been taught to get rid of the fractions first by multiplying out, and then clearing the subsequent brackets, but doing this introduces lots of terms of $x$ on both sides which I then seem to get lost on, but I don't know if I can find a common denominator for this since they are different (+/-) signs?
Any help in how to proceed would be much appreciated, even if no answer. 
KR, 
Q
 A: Multiply both sides of the equation with $(x+1)(2x-1)$. Then write out what you got and I will explain what you did wrong.
A: You can also take one of the elements and place it on the right side. In this equation, let's place $-\frac{1}{2x-1}$ to the right side.
You will get the following equation: $\Large \frac{5}{x+1} = \frac{1}{2x-1}$. We can now apply the following rule: $\frac{a}{b}=\frac{c}{d} \rightarrow ad=bc$, so we get this the following equation:
$5(2x-1) = 1(x+1) \iff 10x-5=x+1 \iff 9x=6 \iff x=\frac23$
A: Both of the other answers are completely right, but I think they seem to leave out the explanation of how they work. 
We have
$$\frac{5}{x+1}-\frac{1}{2x-1}=0$$
What we want to do is combine these two fractions into one fraction. So we multiply 
$\frac{5}{x+1}$ by $\frac{2x-1}{2x-1}$ and $\frac{1}{2x-1}$ by $\frac{x+1}{x+1}$. As we can see, we're multiplying both of these fractions by 1. This leaves us with
$$\frac{(10x-5)-(x+1)}{(x+1)(2x-1)} = 0$$
Now we multiply both sides by $(x+1)(2x-1)$ and simplify, which gives us
$$9x-6=0$$
I'm sure at this point you can solve for $x$.
