Solution 2.1:5a

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Current revision (10:35, 23 September 2008) (edit) (undo)
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In the same way that we calculated fractions, we can subtract the terms' numerators if we first
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In the same way that we calculated fractions, we can subtract the terms' numerators if we first expand the fractions so that they have the same denominator. Because the denominators are <math>x-x^{2}=x(1-x)</math> and <math>x</math>, the lowest common denominator is <math>x(1-x)</math>,
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expand the fractions so that they have the same denominator. Because the denominators are
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<math>x-x^{2}=x\left( 1-x \right)</math>
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and
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<math>x</math>, the lowest common denominator is
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<math>x\left( 1-x \right)</math>:
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{{Displayed math||<math>\begin{align}
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\frac{1}{x-x^{2}}-\frac{1}{x\vphantom{x^2}}
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&= \frac{1}{x-x^{2}}-\frac{1}{x\vphantom{x^2}}\cdot \frac{1-x}{1-x\vphantom{x^2}}\\[5pt]
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&= \frac{1}{x-x^{2}}-\frac{1-x}{x-x^{2}}\\[5pt]
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&= \frac{1-(1-x)}{x-x^{2}}\\[5pt]
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&= \frac{1-1+x}{x-x^{2}}\\[5pt]
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&= \frac{x}{x-x^{2}}\,\textrm{.}
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\end{align}</math>}}
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<math>\begin{align}
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This fraction can be simplified by eliminating the factor ''x'' from the numerator and denominator
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& \frac{1}{x-x^{2}}-\frac{1}{x}=\frac{1}{x-x^{2}}-\frac{1}{x}\centerdot \frac{1-x}{1-x}=\frac{1}{x-x^{2}}-\frac{1-x}{x-x^{2}} \\
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& \\
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& =\frac{1-\left( 1-x \right)}{x-x^{2}}=\frac{1-1+x}{x-x^{2}}=\frac{x}{x-x^{2}} \\
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& \\
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\end{align}</math>
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{{Displayed math||<math>\frac{x}{x-x^{2}} = \frac{x}{x(1-x)} = \frac{1}{1-x}\,\textrm{.}</math>}}
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This fraction can be simplified by eliminating the factor
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<math>x</math>
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from the numerator and denominator.
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<math>\begin{align}
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& \frac{x}{x-x^{2}}=\frac{x}{x\left( 1-x \right)}=\frac{1}{1-x}. \\
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& \\
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\end{align}</math>
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Current revision

In the same way that we calculated fractions, we can subtract the terms' numerators if we first expand the fractions so that they have the same denominator. Because the denominators are \displaystyle x-x^{2}=x(1-x) and \displaystyle x, the lowest common denominator is \displaystyle x(1-x),

\displaystyle \begin{align}

\frac{1}{x-x^{2}}-\frac{1}{x\vphantom{x^2}} &= \frac{1}{x-x^{2}}-\frac{1}{x\vphantom{x^2}}\cdot \frac{1-x}{1-x\vphantom{x^2}}\\[5pt] &= \frac{1}{x-x^{2}}-\frac{1-x}{x-x^{2}}\\[5pt] &= \frac{1-(1-x)}{x-x^{2}}\\[5pt] &= \frac{1-1+x}{x-x^{2}}\\[5pt] &= \frac{x}{x-x^{2}}\,\textrm{.} \end{align}

This fraction can be simplified by eliminating the factor x from the numerator and denominator

\displaystyle \frac{x}{x-x^{2}} = \frac{x}{x(1-x)} = \frac{1}{1-x}\,\textrm{.}
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