Lösung 3.1:4a
Aus Online Mathematik Brückenkurs 2
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Version vom 10:32, 11. Mär. 2009
A general strategy when solving equations is to try to get the unknown variable by itself on one side.
In this case, we start by subtracting \displaystyle z from both sides,
| \displaystyle z+3i-z=2z-2-z\,\textrm{.} |
Then we have a \displaystyle z left on the right-hand side,
| \displaystyle 3i=z-2\,\textrm{.} |
We add \displaystyle 2 to both sides to remove the \displaystyle -2 from the right hand side,
| \displaystyle 3i+2=z-2+2\,, |
and after that we can just read off the solution,
| \displaystyle 2+3i=z\,\textrm{.} |
To check that we have calculated correctly, we substitute \displaystyle z=2+3i into the original equation and see that it is satisfied,
| \displaystyle \begin{align}
\text{LHS} &= z +3i = 2+3i+3i=2+6i\,,\\[5pt] \text{RHS} &= 2z-2 = 2(2+3i)-2 = 4 + 6i -2 = 2+6i\,\textrm{.} \end{align} |
