Solution 2.1:6a

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Current revision (11:24, 23 September 2008) (edit) (undo)
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Before we try dealing with the whole expression, we focus on simplifying the two factors individually by rewriting them using a common denominator:
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Before we try dealing with the whole expression, we focus on simplifying the two factors individually by rewriting them using a common denominator
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<math>\begin{align}
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& x-y+\frac{x^{2}}{y-x}=\frac{\left( x-y \right)\left( y-x \right)}{y-x}+\frac{x^{2}}{y-x} \\
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& \left\{ y-x=-\left( x-y \right) \right\} \\
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& =\frac{-\left( x-y \right)^{2}}{y-x}+\frac{x^{2}}{y-x}=\frac{-\left( x-y \right)^{2}+x^{2}}{y-x}=\frac{-\left( x^{2}-2xy+y^{2} \right)+x^{2}}{y-x} \\
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& =\frac{-x^{2}+2xy-y^{2}+x^{2}}{y-x}=\frac{2xy-y^{2}}{y-x}=\frac{y\left( 2x-y \right)}{y-x}, \\
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\end{align}</math>
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<math>\begin{align}
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& \frac{y}{2x-y}-1=\frac{y}{2x-y}-\frac{2x-y}{2x-y}=\frac{y-\left( 2x-y \right)}{2x-y}=\frac{y-2x+y}{2x-y} \\
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& =\frac{2y-2x}{2x-y}=\frac{2\left( y-x \right)}{2x-y} \\
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\\
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\end{align}</math>
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{{Displayed math||<math>\begin{align}
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x-y+\frac{x^{2}}{y-x} &= \frac{\left( x-y \right)\left( y-x \right)}{y-x}+\frac{x^{2}}{y-x} = \{\ y-x=-(x-y)\ \}\\[5pt]
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&= \frac{-(x-y)^{2}}{y-x}+\frac{x^{2}}{y-x} = \frac{-(x-y)^{2}+x^{2}}{y-x}\\[5pt]
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&= \frac{-(x^{2}-2xy+y^{2})+x^{2}}{y-x} = \frac{-x^{2}+2xy-y^{2}+x^{2}}{y-x}\\[5pt]
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&= \frac{2xy-y^{2}}{y-x} = \frac{y(2x-y)}{y-x},\\[15pt]
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\frac{y}{2x-y}-1
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&= \frac{y}{2x-y}-\frac{2x-y}{2x-y} = \frac{y-(2x-y)}{2x-y} = \frac{y-2x+y}{2x-y}\\[5pt]
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& =\frac{2y-2x}{2x-y} = \frac{2(y-x)}{2x-y}\,\textrm{.}
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\end{align}</math>}}
Then, we multiply the factors together and simplify by elimination:
Then, we multiply the factors together and simplify by elimination:
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{{Displayed math||<math>\biggl(x-y+\frac{x^{2}}{y-x}\biggr) \biggl(\frac{y}{2x-y}-1\biggr) = \frac{y(2x-y)}{y-x}\cdot\frac{2(y-x)}{2x-y}=2y\,\textrm{.}</math>}}
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<math>\left( x-y+\frac{x^{2}}{y-x} \right)\left( \frac{y}{2x-y}-1 \right)=\frac{y\left( 2x-y \right)}{y-x}\centerdot \frac{2\left( y-x \right)}{2x-y}=2y.</math>
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Current revision

Before we try dealing with the whole expression, we focus on simplifying the two factors individually by rewriting them using a common denominator

\displaystyle \begin{align}

x-y+\frac{x^{2}}{y-x} &= \frac{\left( x-y \right)\left( y-x \right)}{y-x}+\frac{x^{2}}{y-x} = \{\ y-x=-(x-y)\ \}\\[5pt] &= \frac{-(x-y)^{2}}{y-x}+\frac{x^{2}}{y-x} = \frac{-(x-y)^{2}+x^{2}}{y-x}\\[5pt] &= \frac{-(x^{2}-2xy+y^{2})+x^{2}}{y-x} = \frac{-x^{2}+2xy-y^{2}+x^{2}}{y-x}\\[5pt] &= \frac{2xy-y^{2}}{y-x} = \frac{y(2x-y)}{y-x},\\[15pt] \frac{y}{2x-y}-1 &= \frac{y}{2x-y}-\frac{2x-y}{2x-y} = \frac{y-(2x-y)}{2x-y} = \frac{y-2x+y}{2x-y}\\[5pt] & =\frac{2y-2x}{2x-y} = \frac{2(y-x)}{2x-y}\,\textrm{.} \end{align}

Then, we multiply the factors together and simplify by elimination:

\displaystyle \biggl(x-y+\frac{x^{2}}{y-x}\biggr) \biggl(\frac{y}{2x-y}-1\biggr) = \frac{y(2x-y)}{y-x}\cdot\frac{2(y-x)}{2x-y}=2y\,\textrm{.}
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