Processing Math: Done
To print higher-resolution math symbols, click the
Hi-Res Fonts for Printing button on the jsMath control panel.
jsMath

Lösung 3.1:4a

Aus Online Mathematik Brückenkurs 2

(Unterschied zwischen Versionen)
Wechseln zu: Navigation, Suche
K
K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel))
Zeile 3: Zeile 3:
In this case, we start by subtracting <math>z</math> from both sides,
In this case, we start by subtracting <math>z</math> from both sides,
-
{{Displayed math||<math>z+3i-z=2z-2-z\,\textrm{.}</math>}}
+
{{Abgesetzte Formel||<math>z+3i-z=2z-2-z\,\textrm{.}</math>}}
Then we have a <math>z</math> left on the right-hand side,
Then we have a <math>z</math> left on the right-hand side,
-
{{Displayed math||<math>3i=z-2\,\textrm{.}</math>}}
+
{{Abgesetzte Formel||<math>3i=z-2\,\textrm{.}</math>}}
We add <math>2</math> to both sides to remove the <math>-2</math> from the right hand side,
We add <math>2</math> to both sides to remove the <math>-2</math> from the right hand side,
-
{{Displayed math||<math>3i+2=z-2+2\,,</math>}}
+
{{Abgesetzte Formel||<math>3i+2=z-2+2\,,</math>}}
and after that we can just read off the solution,
and after that we can just read off the solution,
-
{{Displayed math||<math>2+3i=z\,\textrm{.}</math>}}
+
{{Abgesetzte Formel||<math>2+3i=z\,\textrm{.}</math>}}
To check that we have calculated correctly, we substitute <math>z=2+3i</math> into the original equation and see that it is satisfied,
To check that we have calculated correctly, we substitute <math>z=2+3i</math> into the original equation and see that it is satisfied,
-
{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
\text{LHS} &= z +3i = 2+3i+3i=2+6i\,,\\[5pt]
\text{LHS} &= z +3i = 2+3i+3i=2+6i\,,\\[5pt]
\text{RHS} &= 2z-2 = 2(2+3i)-2 = 4 + 6i -2 = 2+6i\,\textrm{.}
\text{RHS} &= 2z-2 = 2(2+3i)-2 = 4 + 6i -2 = 2+6i\,\textrm{.}
\end{align}</math>}}
\end{align}</math>}}

Version vom 13:06, 10. 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 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“