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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

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  • $\begingroup$ To find the common denominator, first you factor each denominator completely. Here, each denominator is already factored completely. Then you determine which factors are "missing" from each denominator that would make all the denominators look the same. The fraction with $(x + 1)$ in the denominator is missing a factor of $(2x - 1)$, while the fraction with $(2x-1)$ in the denominator is missing a factor of $(x + 1)$. If these were in the denominators, then all of the denominators would be the same and would look like $(x + 1)(2x -1)$, so this is the common denominator. $\endgroup$
    – layman
    Oct 20, 2014 at 13:00
  • $\begingroup$ Now just multiply both sides of the equation by this common denominator and you will see that the fractions will cancel. You will not get any terms with $x$ that are too big or crazy to solve. $\endgroup$
    – layman
    Oct 20, 2014 at 13:01

3 Answers 3

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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.

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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:

$5(2x-1) = 1(x+1) \iff 10x-5=x+1 \iff 9x=6 \iff x=\frac23$

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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 $x$.

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