Lösung 2.2:1d

Aus Online Mathematik Brückenkurs 1

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

Version vom 08:26, 22. Okt. 2008

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

\displaystyle 5x+7-2x=2x-6-2x

which gives

\displaystyle 3x+7=-6\,\textrm{.}

Subtract 7 from both sides,

\displaystyle 3x+7-7=-6-7

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

\displaystyle 3x=-13\,\textrm{.}

Then, divide both sides by 3,

\displaystyle \frac{3x}{3}=-\frac{13}{3}\,\textrm{,}

to get x,

\displaystyle x=-\frac{13}{3}\,\textrm{.}
Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_2.2:1d“