Lösung 2.2:2b
Aus Online Mathematik Brückenkurs 1
First, we multiply both sides in the equation by \displaystyle 4\cdot 7=28, so that we get rid of the denominators in the equation,
| \displaystyle \begin{align}
& 4\cdot{}\rlap{/}7\cdot\frac{8x+3}{\rlap{/}7} - \rlap{/}4\cdot 7\cdot\frac{5x-7}{\rlap{/}4} = 4\cdot 7\cdot 2\\[5pt] &\qquad\Leftrightarrow\quad 4\cdot (8x+3) - 7\cdot (5x-7) = 56\,\textrm{.} \end{align} |
We can simplify the left-hand side to \displaystyle 4\cdot (8x+3) - 7\cdot (5x-7) = 32x+12-35x+49 = -3x+61\,. Hence, the equation is
| \displaystyle -3x+61=56\,\textrm{.} |
We solve this equation by subtracting 61 from both sides and then dividing by -3,
| \displaystyle \begin{align}
-3x+61-61&=56-61\,,\\[5pt] -3x&=-5\,,\\[5pt] \frac{-3x}{-3}&=\frac{-5}{-3}\,,\\[5pt] x&=\frac{5}{3}\,\textrm{.} \end{align} |
The answer is \displaystyle x={5}/{3}\,.
As the final part of the solution, check the answer by substituting \displaystyle x={5}/{3} into the original equation
| \displaystyle \begin{align}
\text{LHS} &= \frac{8\cdot\frac{5}{3}+3}{7}-\frac{5\cdot\frac{5}{3}-7}{4} = \frac{\bigl(8\cdot\frac{5}{3}+3\bigr)\cdot 3}{7\cdot 3} - \frac{\bigl( 5\cdot \frac{5}{3}-7\bigr)\cdot 3}{4\cdot 3}\\[5pt] &= \frac{8\cdot 5+3\cdot 3}{7\cdot 3}-\frac{5\cdot 5-7\cdot 3}{4\cdot 3} = \frac{40+9}{21}-\frac{25-21}{12}\\[5pt] &= \frac{49}{21}-\frac{4}{12} = \frac{7\cdot 7}{3\cdot 7} - \frac{2\cdot 2}{2\cdot 2\cdot 3} = \frac{7}{3}-\frac{1}{3} = \frac{7-1}{3} = \frac{6}{3} = 2 = \text{RHS.} \end{align} |
