Lösung 2.2:3b
Aus Online Mathematik Brückenkurs 1
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Version vom 13:55, 22. Okt. 2008
First, we move all the terms over to the left-hand side,
| \displaystyle \frac{4x}{4x-7}-\frac{1}{2x-3}-1=0\,\textrm{.} |
Then, we multiply the top and bottom of all three terms by appropriate factors so that they have the same common denominator, in the following way,
| \displaystyle \frac{4x}{4x-7}\cdot\frac{2x-3}{2x-3} - \frac{1}{2x-3}\cdot\frac{4x-7}{4x-7} - \frac{(2x-3)(4x-7)}{(2x-3)(4x-7)} = 0 |
and so that we can rewrite the left-hand side giving
| \displaystyle \frac{4x(2x-3) - (4x-7) - (2x-3)(4x-7)}{(2x-3)(4x-7)}=0\,\textrm{.} |
We expand the numerator
| \displaystyle \frac{8x^{2}-12x-(4x-7)-(8x^{2}-14x-12x+21)}{(2x-3)(4x-7)} = 0 |
and simplify
| \displaystyle \frac{10x-14}{(2x-3)(4x-7)}=0\,\textrm{.} |
This equation is satisfied when the numerator is zero (provided the denominator is not also zero) and this happens when
| \displaystyle 10x-14=0\,, |
which gives \displaystyle x=7/5\,.
It can easily happen that we calculate incorrectly, so we check that the answer \displaystyle x=7/5 satisfies the equation,
| \displaystyle \begin{align}
\text{LHS } &= \frac{4\cdot\frac{7}{5}}{4\cdot\frac{7}{5}-7} - \frac{1}{2\cdot\frac{7}{5}-3} = \{\,\text{multiply top and bottom by 5}\,\}\\[5pt] &= \frac{4\cdot\frac{7}{5}}{4\cdot\frac{7}{5}-7}\cdot\frac{5}{5} - \frac{1}{2\cdot \frac{7}{5}-3}\cdot\frac{5}{5} = \frac{4\cdot 7}{4\cdot 7-7\cdot 5}-\frac{5}{2\cdot 7-3\cdot 5}\\[5pt] &= \frac{4}{4-5}-\frac{5}{14-15} = -4-(-5) = 1 = \text{RHS.} \end{align} |
