Solution 2.1:5b
From Förberedande kurs i matematik 1
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| - | {{ | + | We can factorize the denominators as |
| - | < | + | |
| - | {{ | + | {{Displayed math||<math>\begin{align} |
| + | y^{2}-2y &= y(y-2)\\ | ||
| + | y^{2}-4 &= (y-2)(y+2)\quad\text{[difference of two squares]} | ||
| + | \end{align}</math>}} | ||
| + | |||
| + | and then we see that the terms' lowest common denominator is <math>y(y-2)(y+2)</math> because it is the product that contains the smallest number of factors which contain both <math>y(y-2)</math> and <math>(y-2)(y+2)</math>. | ||
| + | |||
| + | Now, we rewrite the fractions so that they have same denominators and start simplifying | ||
| + | |||
| + | {{Displayed math||<math>\begin{align} | ||
| + | \frac{1}{y^{2}-2y}-\frac{2}{y^{2}-4} | ||
| + | &= \frac{1}{y(y-2)}\cdot\frac{y+2}{y+2}-\frac{2}{(y-2)(y+2)}\cdot\frac{y}{y}\\[5pt] | ||
| + | &= \frac{y+2}{y(y-2)(y+2)} - \frac{2y}{(y-2)(y+2)y}\\[5pt] | ||
| + | &= \frac{y+2-2y}{y(y-2)(y+2)}\\[5pt] | ||
| + | &= \frac{-y+2}{y(y-2)(y+2)}\,\textrm{.} | ||
| + | \end{align}</math>}} | ||
| + | |||
| + | The numerator can be rewritten as <math>-y+2=-(y-2)</math> and we can eliminate the common factor <math>y-2</math>, | ||
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| + | {{Displayed math||<math>\frac{-y+2}{y(y-2)(y+2)} = \frac{-(y-2)}{y(y-2)(y+2)} = \frac{-1}{y(y+2)} = -\frac{1}{y(y+2)}\,\textrm{.}</math>}} | ||
Current revision
We can factorize the denominators as
| \displaystyle \begin{align}
y^{2}-2y &= y(y-2)\\ y^{2}-4 &= (y-2)(y+2)\quad\text{[difference of two squares]} \end{align} |
and then we see that the terms' lowest common denominator is \displaystyle y(y-2)(y+2) because it is the product that contains the smallest number of factors which contain both \displaystyle y(y-2) and \displaystyle (y-2)(y+2).
Now, we rewrite the fractions so that they have same denominators and start simplifying
| \displaystyle \begin{align}
\frac{1}{y^{2}-2y}-\frac{2}{y^{2}-4} &= \frac{1}{y(y-2)}\cdot\frac{y+2}{y+2}-\frac{2}{(y-2)(y+2)}\cdot\frac{y}{y}\\[5pt] &= \frac{y+2}{y(y-2)(y+2)} - \frac{2y}{(y-2)(y+2)y}\\[5pt] &= \frac{y+2-2y}{y(y-2)(y+2)}\\[5pt] &= \frac{-y+2}{y(y-2)(y+2)}\,\textrm{.} \end{align} |
The numerator can be rewritten as \displaystyle -y+2=-(y-2) and we can eliminate the common factor \displaystyle y-2,
| \displaystyle \frac{-y+2}{y(y-2)(y+2)} = \frac{-(y-2)}{y(y-2)(y+2)} = \frac{-1}{y(y+2)} = -\frac{1}{y(y+2)}\,\textrm{.} |
