Aus Online Mathematik Brückenkurs 1

Because both terms, \displaystyle x\left( x+3 \right) and \displaystyle x\left( 2x-9 \right) contain the factor \displaystyle x, we can take out \displaystyle x from the left-hand side and collect together the remaining expression:

\displaystyle \begin{align} & x\left( x+3 \right)-x\left( 2x-9 \right)=x\left( \left( x+3 \right)-\left( 2x-9 \right) \right) \\ & =x\left( x+3-2x+9 \right)=x\left( -x+12 \right) \\ \end{align}

The equation is thus

\displaystyle x\left( -x+12 \right)=0

and we obtain directly that the equation is satisfied if either \displaystyle x or \displaystyle -x+\text{12} is zero. The solutions to the equation are therefore \displaystyle x=0\text{ } and \displaystyle x=\text{12}.

Here, it can be worth checking that \displaystyle x=\text{12 } is a solution (the case \displaystyle x=0 is obvious):

LHS \displaystyle =12\centerdot \left( 12+3 \right)-12\centerdot \left( 2\centerdot 12-9 \right)=2\centerdot 15-12\centerdot 15=0= RHS