Aus Online Mathematik Brückenkurs 1

First, we move all the terms over to the left-hand side:

\displaystyle \frac{4x}{4x-7}-\frac{1}{2x-3}-1=0

Then, we multiply the top and bottom of all three terms by appropriate factors so that they have the same common denominator, in the following way,

\displaystyle \frac{4x}{4x-7}\centerdot \frac{2x-3}{2x-3}-\frac{1}{2x-3}\centerdot \frac{4x-7}{4x-7}-\frac{\left( 2x-3 \right)\left( 4x-7 \right)}{\left( 2x-3 \right)\left( 4x-7 \right)}=0

and so that we can rewrite the left-hand side giving

\displaystyle \frac{4x\left( 2x-3 \right)-\left( 4x-7 \right)-\left( 2x-3 \right)\left( 4x-7 \right)}{\left( 2x-3 \right)\left( 4x-7 \right)}=0

We expand the numerator

\displaystyle \frac{8x^{2}-12x-\left( 4x-7 \right)-\left( 8x^{2}-14x-12x+21 \right)}{\left( 2x-3 \right)\left( 4x-7 \right)}=0

and simplify

\displaystyle \frac{10x-14}{\left( 2x-3 \right)\left( 4x-7 \right)}=0

This equation is satisfied when the numerator is zero (provided the denominator is not also zero) and this happens when

\displaystyle 10x-14=0

which gives \displaystyle x={7}/{5}\;.

It can easily happen that we calculate incorrectly, so we must check that the answer \displaystyle x={7}/{5}\; satisfies the equation:

\displaystyle \begin{align} & \text{LHS }~~~=\text{ }~~~\frac{4\centerdot \frac{7}{5}}{4\centerdot \frac{7}{5}-7}-\frac{1}{2\centerdot \frac{7}{5}-3}\text{ }=\text{ }\left\{ \text{ multiply top and bottom by 5} \right\} \\ & \text{ }~~~ \\ & =~\frac{4\centerdot \frac{7}{5}}{4\centerdot \frac{7}{5}-7}\centerdot \frac{5}{5}-\frac{1}{2\centerdot \frac{7}{5}-3}\centerdot \frac{5}{5}=\frac{4\centerdot 7}{4\centerdot 7-7\centerdot 5}-\frac{5}{2\centerdot 7-3\centerdot 5} \\ & \\ & =\frac{4}{4-5}-\frac{5}{14-15}=-4-\left( -5 \right)=1\text{ }~~~=\text{ RHS}~~~ \\ \end{align}