Aus Online Mathematik Brückenkurs 1

If we divide up the denominators that appear in the equation into small integer factors \displaystyle 6=2\centerdot 3, \displaystyle 9=3\centerdot 3 and 2, we see that the lowest common denominator is \displaystyle 2\centerdot 3\centerdot 3=18. Thus, we multiply both sides of the equation by \displaystyle 2\centerdot 3\centerdot 3 in order to avoid having denominators in the equation:

\displaystyle \begin{align} & 2\centerdot 3\centerdot 3\centerdot \frac{5x}{6}-2\centerdot 3\centerdot 3\centerdot \frac{x+2}{9}=2\centerdot 3\centerdot 3\centerdot \frac{1}{2} \\ & \Leftrightarrow 3\centerdot 5x-2\centerdot \left( x+2 \right)=3\centerdot 3 \\ \end{align}

We can rewrite the left-hand side as \displaystyle 3\centerdot 5x-2\centerdot \left( x+2 \right)=15x-2x-4=13x-4, so that we get the equation

\displaystyle 13x-4=9

We can now solve this first-degree equation by carrying out simple arithmetical calculations so as to get \displaystyle x by itself on one side:

1. Add 4 to both sides:

\displaystyle 13x-+4=9+4

which gives

\displaystyle 13x=13 .

2. Divide both sides by 13:

\displaystyle \frac{13x}{13}=\frac{13}{13}

which gives the answer \displaystyle x=1.

The equation has \displaystyle 1 as the solution.

When we have obtained an answer, it is important to go back to the original equation to check that \displaystyle x=1 really is the correct answer( i.e. that we haven't calculated incorrectly):

LHS = \displaystyle \frac{5\centerdot 1}{6}-\frac{1\centerdot 2}{9}=\frac{5}{6}-\frac{3}{9}=\frac{5}{6}-\frac{1}{3}=\frac{5}{6}-\frac{1\centerdot 2}{3\centerdot 2}=\frac{5-2}{6}=\frac{3}{6}=\frac{1}{2} = RHS