Lösung 2.3:3e
Aus Online Mathematik Brückenkurs 1
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Version vom 14:09, 22. Okt. 2008
In this case, we see that the left-hand side contains the factor \displaystyle x+3, which we can take out to obtain
| \displaystyle \begin{align}
(x+3)(x-1) - (x+3)(2x-9) &= (x+3)\bigl((x-1)-(2x-9)\bigr)\\[5pt] &= (x+3)(x-1-2x+9)\\[5pt] &= (x+3)(-x+8)\,\textrm{.} \end{align} |
This rewriting of the equation results in the new equation
| \displaystyle (x+3)(-x+8)=0 |
which has the solutions \displaystyle x=-3 and \displaystyle x=8\,.
We check the solution \displaystyle x=8 by substituting it into the equation,
| \displaystyle \text{LHS} = (8+3)\cdot (8-1) - (8+3)\cdot (2\cdot 8 - 9) = 11\cdot 7 - 11\cdot 7 = 0 = \textrm{RHS.} |
