Solution 2.2:1d
From Förberedande kurs i matematik 1
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- | {{ | + | Move ''x'' to the left-hand side by subtracting 2''x'' from both sides, |
- | < | + | |
- | {{ | + | {{Displayed math||<math>5x+7-2x=2x-6-2x</math>}} |
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+ | which gives | ||
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+ | {{Displayed math||<math>3x+7=-6\,\textrm{.}</math>}} | ||
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+ | Subtract 7 from both sides, | ||
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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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+ | {{Displayed math||<math>3x=-13\,\textrm{.}</math>}} | ||
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+ | 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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+ | to get x, | ||
+ | |||
+ | {{Displayed math||<math>x=-\frac{13}{3}\,\textrm{.}</math>}} |
Current revision
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{.} |