Aus Online Mathematik Brückenkurs 1

First, we multiply both sides in the equation by \displaystyle 4\centerdot 7=28, so that we get rid of the denominators in the equation,

\displaystyle \begin{align} & 4\centerdot 7\centerdot \frac{8x+3}{7}-4\centerdot 7\centerdot \frac{5x-7}{4}=4\centerdot 7\centerdot 2 \\ & \Leftrightarrow 4\centerdot \left( 8x+3 \right)-7\centerdot \left( 5x-7 \right)=56 \\ \end{align}

We can simplify the left-hand side to ,

\displaystyle 4\centerdot \left( 8x+3 \right)-7\centerdot \left( 5x-7 \right)=32x+12-35x+49=-3x+61

Hence, the equation is

\displaystyle -3x+61=56

We solve this equation by subtracting \displaystyle 61 from both sides and then dividing by \displaystyle -3,

\displaystyle \begin{align} & -3x+61-61=56-61 \\ & -3x=-5 \\ & \frac{-3x}{-3}=\frac{-5}{-3} \\ & x=\frac{5}{3} \\ \end{align}

The answer is \displaystyle x={5}/{3}\;.

As the final part of the solution, check the answer by substituting \displaystyle x={5}/{3}\; into the original equation

\displaystyle \begin{align} & \text{LHS}\quad =\quad \frac{8\centerdot \frac{5}{3}+3}{7}-\frac{5\centerdot \frac{5}{3}-7}{4}=\frac{\left( 8\centerdot \frac{5}{3}+3 \right)\centerdot 3}{7\centerdot 3}-\frac{\left( 5\centerdot \frac{5}{3}-7 \right)\centerdot 3}{4\centerdot 3} \\ & \\ & =\frac{8\centerdot 5+3\centerdot 3}{7\centerdot 3}-\frac{5\centerdot 5-7\centerdot 3}{4\centerdot 3}=\frac{40+9}{21}-\frac{25-21}{12} \\ & \\ & =\frac{49}{21}-\frac{4}{12}=\frac{7\centerdot 7}{3\centerdot 7}-\frac{2\centerdot 2}{2\centerdot 2\centerdot 2}=\frac{7}{3}-\frac{1}{3}=\frac{7-1}{3} \\ & \\ & =\frac{6}{3}=2\quad =\quad \text{RHS} \\ \end{align}