Lösung 3.1:4a

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A general strategy when solving equations is to try to get the unknown variable by itself on one side.
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In this case, we start by subtracting z from both sides,
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<math>z+3i-z=2z-2-z.</math>
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Then we have a z left on the right-hand side,
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<math>3i=z-2</math>.
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We add <math>2</math> to both sides to remove the <math>-2</math> from the right hand side
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<math>3i+2=z-2+2</math>
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and after that we can just read off the solution:
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<math>2+3i=z</math>
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To check that we have calculated correctly, we substitute z=2+3i into the original equation and see that it is satisfied:
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<math>\begin{align}LHS &= z +3i = 2+3i+3i=2+6i,\\
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RHS &= 2z-2 = 2(2+3i)-2 = 4 + 6i -2 = 2+6i.\end{align}</math>
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Version vom 10:37, 23. Sep. 2008

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 z from both sides,


\displaystyle z+3i-z=2z-2-z.


Then we have a z left on the right-hand side,


\displaystyle 3i=z-2.


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


To check that we have calculated correctly, we substitute z=2+3i into the original equation and see that it is satisfied:


\displaystyle \begin{align}LHS &= z +3i = 2+3i+3i=2+6i,\\ RHS &= 2z-2 = 2(2+3i)-2 = 4 + 6i -2 = 2+6i.\end{align}


Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_3.1:4a“