Aus Online Mathematik Brückenkurs 1

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

\displaystyle \begin{align} (x+3)(x-1) - (x+3)(2x-9) &= (x+3)\bigl((x-1)-(2x-9)\bigr)\\[5pt] &= (x+3)(x-1-2x+9)\\[5pt] &= (x+3)(-x+8)\,\textrm{.} \end{align}

This rewriting of the equation results in the new equation

\displaystyle (x+3)(-x+8)=0

which has the solutions \displaystyle x=-3 and \displaystyle x=8\,.

We check the solution \displaystyle x=8 by substituting it into the equation,

\displaystyle \text{LHS} = (8+3)\cdot (8-1) - (8+3)\cdot (2\cdot 8 - 9) = 11\cdot 7 - 11\cdot 7 = 0 = \textrm{RHS.}