Solution 2.1:7a
From Förberedande kurs i matematik 1
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- | {{ | + | If we multiply the top and bottom of the first fraction by <math>x+5</math> and the second by <math>x+3</math>, then they will both have the same denominator and we can work out the expression by subtracting the numerators |
- | + | ||
- | {{ | + | {{Displayed math||<math>\begin{align} |
+ | \frac{2}{x+3}-\frac{2}{x+5} &= \frac{2}{x+3}\cdot \frac{x+5}{x+5}-\frac{2}{x+5}\cdot \frac{x+3}{x+3}\\[7pt] | ||
+ | &= \frac{2(x+5)-2(x+3)}{(x+3)(x+5)}\\[5pt] | ||
+ | &= \frac{2x+10-2x-6}{(x+3)(x+5)}\\[5pt] | ||
+ | &= \frac{4}{(x+3)(x+5)}\,\textrm{.} | ||
+ | \end{align}</math>}} |
Current revision
If we multiply the top and bottom of the first fraction by \displaystyle x+5 and the second by \displaystyle x+3, then they will both have the same denominator and we can work out the expression by subtracting the numerators
\displaystyle \begin{align}
\frac{2}{x+3}-\frac{2}{x+5} &= \frac{2}{x+3}\cdot \frac{x+5}{x+5}-\frac{2}{x+5}\cdot \frac{x+3}{x+3}\\[7pt] &= \frac{2(x+5)-2(x+3)}{(x+3)(x+5)}\\[5pt] &= \frac{2x+10-2x-6}{(x+3)(x+5)}\\[5pt] &= \frac{4}{(x+3)(x+5)}\,\textrm{.} \end{align} |