Lösung 2.2:1d

Aus Online Mathematik Brückenkurs 1

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Move
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Move ''x'' to the left-hand side by subtracting 2''x'' from both sides,
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<math>x</math>
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to the left-hand side by subtracting
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<math>2x</math>
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from both sides,
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<math>5x+7-2x=2x-6-2x</math>
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{{Displayed math||<math>5x+7-2x=2x-6-2x</math>}}
which gives
which gives
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{{Displayed math||<math>3x+7=-6\,\textrm{.}</math>}}
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<math>3x+7=-6</math>
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Subtract 7 from both sides,
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Subtract
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<math>7</math>
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from both sides,
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<math>3x+7-7=-6-7</math>
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so that the term
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<math>3x</math>
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alone remains on the left-hand side
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<math>3x=-13</math>
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Then, divide both sides by
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<math>3</math>
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{{Displayed math||<math>3x+7-7=-6-7</math>}}
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so that the term 3''x'' alone remains on the left-hand side
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<math>\frac{3x}{3}=-\frac{13}{3}</math>
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{{Displayed math||<math>3x=-13\,\textrm{.}</math>}}
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Then, divide both sides by 3,
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to get x:
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{{Displayed math||<math>\frac{3x}{3}=-\frac{13}{3}\,\textrm{,}</math>}}
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to get x,
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<math>x=-\frac{13}{3}</math>
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{{Displayed math||<math>x=-\frac{13}{3}\,\textrm{.}</math>}}

Version vom 13:23, 23. Sep. 2008

Move x to the left-hand side by subtracting 2x from both sides,

Vorlage:Displayed math

which gives

Vorlage:Displayed math

Subtract 7 from both sides,

Vorlage:Displayed math

so that the term 3x alone remains on the left-hand side

Vorlage:Displayed math

Then, divide both sides by 3,

Vorlage:Displayed math

to get x,

Vorlage:Displayed math

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_2.2:1d“