Solution 2.1:6c
From Förberedande kurs i matematik 1
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| - | { | + | Because the denominators are <math>a^{2}-ab = a(a-b)</math> and <math>a-b</math>, both terms will have a common denominator <math>a(a-b)</math> if the top and bottom of the second term are multiplied by <math>a</math>, |
| - | < | + | |
| - | {{ | + | {{Displayed math||<math>\begin{align} |
| + | \frac{2a+b}{a^{2}-b}-\frac{2}{a-b} &= \frac{2a+b}{a(a-b)}-\frac{2}{a-b}\cdot\frac{a}{a}\\[5pt] | ||
| + | &= \frac{2a+b-2a}{a(a-b)}\\[5pt] | ||
| + | &= \frac{b}{a(a-b)}\,\textrm{.} | ||
| + | \end{align}</math>}} | ||
Current revision
Because the denominators are \displaystyle a^{2}-ab = a(a-b) and \displaystyle a-b, both terms will have a common denominator \displaystyle a(a-b) if the top and bottom of the second term are multiplied by \displaystyle a,
| \displaystyle \begin{align}
\frac{2a+b}{a^{2}-b}-\frac{2}{a-b} &= \frac{2a+b}{a(a-b)}-\frac{2}{a-b}\cdot\frac{a}{a}\\[5pt] &= \frac{2a+b-2a}{a(a-b)}\\[5pt] &= \frac{b}{a(a-b)}\,\textrm{.} \end{align} |
