Aus Online Mathematik Brückenkurs 1

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

\displaystyle \begin{align} & \rlap{/}2\cdot{}\rlap{/}3\cdot 3\cdot\frac{5x}{\rlap{/}6} - 2\cdot{}\rlap{/}3\cdot{}\rlap{/}3\cdot\frac{x+2}{\rlap{/}9} = \rlap{/}2\cdot 3\cdot 3\cdot \frac{1}{\rlap{/}2} \\[5pt] &\qquad\Leftrightarrow\quad 3\cdot 5x-2\cdot (x+2) = 3\cdot 3\,\textrm{.}\\ \end{align}

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

\displaystyle 13x-4=9\,\textrm{.}

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

1. Add 4 to both sides, \displaystyle \vphantom{x_2}13x-4+4=9+4\,, which gives \displaystyle \ 13x=13\,\textrm{.}
2. Divide both sides by 13, \displaystyle \frac{13x}{13}=\frac{13}{13}\,, which gives the answer \displaystyle \ x=1\,\textrm{.}

The equation has \displaystyle x=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)

\displaystyle \begin{align} \text{LHS} &= \frac{5\cdot 1}{6}-\frac{1+2}{9} = \frac{5}{6}-\frac{3}{9} = \frac{5}{6}-\frac{1}{3}\\[5pt] &= \frac{5}{6}-\frac{1\cdot 2}{3\cdot 2} = \frac{5-2}{6} = \frac{3}{6} = \frac{1}{2} = \text{RHS}\,\textrm{.} \end{align}