• Mar 17th 2014, 01:31 PM
alanpaul
Hi
6 2
--------------------- + --------------
(x+5) (2x-3) x(2x-3)

Please help me to simplyfy this equation with simple steps

With Many Thanks
Alanpaul
• Mar 17th 2014, 01:35 PM
SlipEternal
Find a common denominator. The left fraction does not have a factor of $x$ in the denominator, so multiply top and bottom by $x$. The right fraction does not have $x+5$ as a factor in the denominator, so multiply top and bottom by $x+5$. Both have $2x-3$ as a factor in the denominator, so leave that alone.
• Mar 18th 2014, 02:08 PM
alanpaul
The left fraction does not have a factor of x in the denominator, so multiply top and bottom by x. Please can you give more simple way explain to understand that part. I understood all other steps.

With Many Thanks
Alan
• Mar 18th 2014, 02:30 PM
SlipEternal
When you multiply a value by 1, the result is the original value. So, we can multiply any value by 1 without changing it.

When you divide a value by itself, you get 1. For example, $\dfrac{5}{5} = 1$.

So, you can multiply any value by a different value divided by itself without changing the original value. For example,

$\dfrac{6}{(x+5)(2x-3)} = \dfrac{6}{(x+5)(2x-3)}\cdot 1 = \dfrac{6}{(x+5)(2x-3)}\cdot \dfrac{5}{5} = \dfrac{30}{5(x+5)(2x-3)}$

Since all I did was multiply the original value by 1, the expression has the same value. For any $x\neq 0$, the same is true when multiplying a value by $\dfrac{x}{x}$.

So, with all of that in mind, you have two fractions that you are trying to add, but they have different denominators. Break the denominators into factors. The factors are the expressions being multiplied together. The denominator of the fraction on the left has the factors $(x+5)$ and $(2x-3)$. The denominator of the fraction on the right has the factors $x$ and $(2x-3)$. So, in order for the two denominators to be equal, the fraction on the left needs a factor of $x$ in its denominator so that the list of its factors are $x$, $(x+5)$, and $(2x-3)$. The fraction on the right needs a factor of $(x+5)$ in its denominator so that the list of its factors are $x$, $(x+5)$, and $(2x-3)$. Those two changes will cause both denominators to be equal so that you can add the two fractions.

But, you don't want to change the value of either fraction. So, multiply each fraction by 1. For the fraction on the left, use the fact that $\dfrac{x}{x} = 1$. For the fraction on the right, use the fact that $\dfrac{x+5}{x+5} = 1$.

Here is what it looks like:

\begin{align*}\dfrac{6}{(x+5)(2x-3)} + \dfrac{2}{x(2x-3)} & = \dfrac{6}{(x+5)(2x-3)}\cdot 1 + \dfrac{2}{x(2x-3)}\cdot 1 \\ & = \dfrac{6}{(x+5)(2x-3)}\cdot \dfrac{x}{x} + \dfrac{2}{x(2x-3)}\cdot \dfrac{(x+5)}{(x+5)} \\ & = \dfrac{6x}{x(x+5)(2x-3)} + \dfrac{2(x+5)}{x(x+5)(2x-3)}\end{align*}

The two fractions at the end represent the same values as the two fractions at the beginning. This is because we multiplied them each by 1, which does not change their value. But, now, they have the same denominator, so you can add the two fractions.