\left. \begin{array} { l } { \frac { 14 } { x + y } + \frac { 3 } { x - y } = 5 } \\ { \frac { 21 } { x + y } - \frac { 1 } { x - y } = 2 } \end{array} \right.
Answers
(x - 2\right)} + \frac{{x}^{2}}{x - 2} - \frac{{x}^{3} + x - 4}{\left(x + 2\right)\left(x - 2\right)}\]
Make all denominators the same so that we can add or subtract the fractions
The lowest common denominator is \(\left(x-2\right)\left(x+2\right)\).
\[\frac{x - 2}{\left(x + 2\right)\left(x - 2\right)} + \frac{\left({x}^{2}\right)\left(x + 2\right)}{\left(x + 2\right)\left(x - 2\right)} - \frac{{x}^{3} + x - 4}{\left(x + 2\right)\left(x - 2\right)}\]
Write as one fraction\[\frac{x - 2 + \left({x}^{2}\right)\left(x + 2\right) - \left({x}^{3} + x - 4\right)}{\left(x + 2\right)\left(x - 2\right)}\]
Simplify\[\frac{x - 2 + {x}^{3} + 2{x}^{2} - {x}^{3} - x + 4}{\left(x + 2\right)\left(x - 2\right)} = \frac{2{x}^{2} + 2}{\left(x + 2\right)\left(x - 2\right)}\]
Take out the common factor and write the final answer\[\frac{2\left({x}^{2} + 1\right)}{\left(x + 2\right)\left(x - 2\right)}\]
WORKED EXAMPLE 21: SIMPLIFYING FRACTIONS
Simplify: \[\frac{2}{{x}^{2} - x} + \frac{{x}^{2} + x + 1}{{x}^{3} - 1} - \frac{x}{{x}^{2} - 1}, \quad \left(x\ne 0;x\ne ±1\right)\]
Factorise the numerator and denominator\[\frac{2}{x\left(x - 1\right)} + \frac{\left({x}^{2} + x + 1\right)}{\left(x - 1\right)\left({x}^{2} + x + 1\right)} - \frac{x}{\left(x - 1\right)\left(x + 1\right)}\]
Simplify and find the common denominator\[\frac{2\left(x + 1\right) + x\left(x + 1\right) - {x}^{2}}{x\left(x - 1\right)\left(x + 1\right)}\]
Write the final answer\[\frac{2x + 2 + {x}^{2} + x - {x}^{2}}{x\left(x - 1\right)\left(x + 1\right)} = \frac{3x + 2}{x\left(x - 1\right)\left(x + 1\right)}\]
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EXERCISE 1.10
Simplify (assume all denominators are non-zero):
\(\dfrac{3a}{15}\)
\[\frac{3a}{15} = \frac{a}{5}\]
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\(\dfrac{2a + 10}{4}\)
\begin{align*} \frac{2a + 10}{4}& = \frac{2(a+5)}{4}\\ &=\frac{a + 5}{2} \end{align*}
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\(\dfrac{5a + 20}{a + 4}\)
\begin{align*} \frac{5a + 20}{a + 4}& = \frac{5(a + 4)}{a + 4}\\ &=5 \end{align*}
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\(\dfrac{a^{2} - 4a}{a - 4}\)
\begin{align*} \frac{a^{2} - 4a}{a - 4}& = \frac{a(a - 4)}{a - 4}\\ &=a \end{align*}
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\(\dfrac{3a^{2} - 9a}{2a - 6}\)
\begin{align*} \frac{3a^{2} - 9a}{2a - 6}& = \frac{3a(a - 3)}{2(a - 3)}\\ &=\frac{3a}{2} \end{align*}
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\(\dfrac{9a + 27}{9a + 18}\)
\begin{align*} \frac{9a + 27}{9a + 18}& = \frac{9(a + 3)}{9(a + 2)}\\ &=\frac{a + 3}{a + 2} \end{align*}
Note restriction: \(a \ne -2\).
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\(\dfrac{6ab + 2a}{2b}\)
\begin{align*} \frac{6ab + 2a}{2b}& = \frac{2a(3b + 1)}{2b}\\ &=\frac{a(3b + 1)}{b} \end{align*}
Note restriction: \(b \ne 0\).
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\(\dfrac{16x^{2}y - 8xy}{12x - 6}\)
\begin{align*} \frac{16x^{2}y - 8xy}{12x - 6}& = \frac{8xy(2x - 1)}{6(2x - 1)}\\ & = \frac{8xy}{6}\\ &=\frac{4xy}{3} \end{align*}
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\(\dfrac{4xyp - 8xp}{12xy}\)
\begin{align*} \frac{4xyp - 8xp}{12xy} & = \frac{4xp(y - 2)}{12xy}\\ &=\frac{p(y - 2)}{3y} \end{align*}
Note restriction: \(y \ne 0\).
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\(\dfrac{9x^2 - 16}{6x - 8}\)
\begin{align*} \frac{9x^2 - 16}{6x - 8} &= \frac{(3x-4)(3x+4)}{2(3x-4)} \\ &= \frac{3x+4}{2} \end{align*}
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\(\dfrac{b^2 - 81a^2}{18a-2b}\)
\begin{align*} \frac{b^2 - 81a^2}{18a-2b} &= \frac{(b-9)(b+9)}{2(9-b)} \\ &= -\frac{b+9}{2} \end{align*}
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\(\dfrac{t^2 - s^2}{s^2 - 2st + t^2}\)
\begin{align*} \frac{t^2 - s^2}{s^2 - 2st + t^2} &= \frac{(t-s)(t+s)}{(s-t)^2} \\ &= \frac{t+s}{t-s} \end{align*}
Note restriction: \(s \ne t\)
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\(\dfrac{x^2 - 2x - 15}{5x - 25}\)
\begin{align*} \frac{x^2 - 2x - 15}{5x - 25} &= \frac{(x-5)(x+3)}{5(x - 5)} \\ &= \frac{x+3}{5} \end{align*}
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\(\dfrac{x^2 + 2x - 15}{x^2 + 8x + 15}\)
\begin{align*} \frac{x^2 + 2x - 15}{x^2 + 8x + 15} &= \frac{(x+5)(x-3)}{(x+3)(x+5)} \\ &= \frac{x-3}{x+3} \end{align*}
Note restriction: \(x \ne -3\).
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\(\dfrac{x^2 - x -6}{x^3 - 27}\)
\begin{align*} \frac{x^2 - x -6}{x^3 - 27} &= \frac{(x-3)(x+2)}{(x-3)(x^2 + 3x + 9)} \\ &= \frac{x+2}{x^2 + 3x + 9} \end{align*}
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\(\dfrac{a^2 + 6a - 16}{a^3 - 8}\)
\
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\(\dfrac{3x+2}{x^2 - 6x + 8} \times \dfrac{x-2}{3x^2 + 8x + 4}\)
\begin{align*} \frac{3x+2}{x^2 - 6x + 8} \times \frac{x-2}{3x^2 + 8x + 4} &= \frac{3x+2}{(x-4)(x-2)} \times \frac{x-2}{(3x+2)(x+2)} \\ &= \frac{1}{(x-4)(x+2)} \end{align*}
Note restrictions: \(x \ne 4\) and \(x \ne -2\).
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\(\dfrac{a^2 - 2a + 8}{a^2 + 6a + 8} \times \dfrac{a^2 + a - 12}{3} - \dfrac{3}{2}\)
\begin{align*} \frac{a^2 - 2a + 8}{a^2 + 6a + 8} \times \frac{a^2 + a - 12}{3} - \frac{3}{2} &= \frac{( & = \frac{(