Solution 2.1:6a
From Förberedande kurs i matematik 1
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{.} |
