Aus Online Mathematik Brückenkurs 1

In this case, we see that the left-hand side contains the factor \displaystyle x+\text{3}, which we can take out to obtain

\displaystyle \begin{align} & \left( x+\text{3} \right)\left( x-1 \right)-\left( x+\text{3} \right)\left( 2x-9 \right)=\left( x+\text{3} \right)\left( \left( x-1 \right)-\left( 2x-9 \right) \right) \\ & =\left( x+\text{3} \right)\left( x-1-2x+9 \right)=\left( x+\text{3} \right)\left( -x+8 \right). \\ \end{align}

This rewriting of the equation results in the new equation

\displaystyle \left( x+\text{3} \right)\left( -x+8 \right)=0

which has the solutions \displaystyle x=-\text{3} and \displaystyle x=\text{8}.

We check the solution \displaystyle x=\text{8 } by substituting it into the equation:

LHS \displaystyle =\left( 8+3 \right)\centerdot \left( 8-1 \right)-\left( 8+3 \right)\centerdot \left( 2\centerdot 8-9 \right)=11\centerdot 7-11\centerdot 7=0= RHS