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Lösung 3.1:4a

Aus Online Mathematik Brückenkurs 2

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A general strategy when solving equations is to try to get the unknown variable by itself on one side.
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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In this case, we start by subtracting <math>z</math> from both sides,
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{{Displayed math||<math>z+3i-z=2z-2-z\,\textrm{.}</math>}}
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<math>z+3i-z=2z-2-z.</math>
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Then we have a <math>z</math> left on the right-hand side,
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{{Displayed math||<math>3i=z-2\,\textrm{.}</math>}}
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Then we have a z left on the right-hand side,
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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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{{Displayed math||<math>3i+2=z-2+2\,,</math>}}
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<math>3i=z-2</math>.
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and after that we can just read off the solution,
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{{Displayed math||<math>2+3i=z\,\textrm{.}</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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To check that we have calculated correctly, we substitute <math>z=2+3i</math> into the original equation and see that it is satisfied,
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{{Displayed math||<math>\begin{align}
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<math>3i+2=z-2+2</math>
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\text{LHS} &= z +3i = 2+3i+3i=2+6i\,,\\[5pt]
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\text{RHS} &= 2z-2 = 2(2+3i)-2 = 4 + 6i -2 = 2+6i\,\textrm{.}
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\end{align}</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 07:15, 30. Okt. 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,

z+3i−z=2z−2−z.

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

3i=z−2.

We add 2 to both sides to remove the −2 from the right hand side,

3i+2=z−2+2

and after that we can just read off the solution,

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,

LHSRHS=z+3i=2+3i+3i=2+6i=2z−2=2(2+3i)−2=4+6i−2=2+6i.
Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_3.1:4a“