Applied Algebra: Modeling and functions

Factor out any common factors from the numerator and denominator. Then divide numerator and denominator by the common factors.

1. \(\dfrac{8x^3 y}{6x^2 y^3} = \dfrac{4x \cdot 2x^2 y}{3y^2\cdot 2x^2 y}= \dfrac{4x}{3y^2}\)

2. \(\dfrac{6x-3}{3} = \dfrac{\Ccancel[blue]{3}(2x+1) }{\Ccancel[blue]{3}}= 2x+1 \)

Factor the numerator and denominator. Then divide numerator and denominator by the common factors.

1. \(\dfrac{4x + 2}{4} = \dfrac{\Ccancel[blue]{2}(2x + 1)}{\Ccancel[blue]{2} (2)}\dfrac{2x+1}{2}\)

2. \(\dfrac{9x^2+3}{6x+3} = \dfrac{\Ccancel[blue]{3}(3x^2 + 1)}{\Ccancel[blue]{3} (2x+1)} = \dfrac{3x^2 + 1}{2x+1}\)

1. Factor numerator and denominator to obtain

   \begin{equation*} \frac{(x - 6) (x - 1)}{(6 - x) (6 + x)} \end{equation*}

   The factor \(x-6\) in the numerator is the opposite of the factor \(6-x\) in the denominator. That is, \(x - 6 = -1 (6 - x)\text{.}\) Thus,

   \begin{equation*} \frac{-1 \Ccancel[blue]{(6 - x )}(x - 1)}{\Ccancel[blue]{(6 - x )}(6 + x)}=\frac{-1 (x - 1)}{6 + x}=\frac{1-x}{6+x} \end{equation*}

2. The numerator of the fraction is a difference of two cubes, and the denominator is a difference of two squares. Factor each to obtain

   \begin{equation*} \frac{\Ccancel[blue]{(3x-1)}(9x^2+3x+1)}{\Ccancel[blue]{(3x-1)}(3x+1)}=\frac{9x^2+3x+1}{3x+1} \end{equation*}

1. The denominator of the first fraction factors into \((x + 1) (x - 1)\text{.}\) There are no common factors to divide out, so we multiply the numerators together and multiply the denominators together.

   \begin{equation*} \frac{5}{x^2-1}\cdot\frac{x+2}{x}=\frac{5(x+2)}{x(x^2-1)}=\frac{5x+10}{x^3-x} \end{equation*}

2. Factor each numerator and each denominator. Look for common factors.

   \begin{align*} \frac{4y^2-1}{4-y^2}\cdot\frac{y^2-2y}{4y+2} \amp = \frac{(2y-1)(\bcancel{2y+1})}{\cancel{(2-y)}(2+y)} \cdot \frac{y\cancelto{\alert{-1}}{(y-2)}}{2(\bcancel{2y+1})}\amp\amp\blert{\text{Divide out common factors.}}\\ \amp =\frac{-y(2y-1)}{2(y+2)} \amp\amp\blert{\text{Note: }~y-2=-(2-y)} \end{align*}

1. Multiply the first fraction by the reciprocal of the second fraction.

   \begin{align*} \frac{x^2-1}{x+3}\div\frac{x^2-x-2}{x^2+5x+6} \amp =\frac{x^2-1}{x+3}\cdot\frac{x^2+5x+6}{x^2-x-2} \amp\amp\blert{\text{Factor.}}\\ \amp = \frac{(x-1)(\Ccancel[blue]{x+1})}{\bcancel{x+3}}\cdot\frac{(\bcancel{x+3})(x+2)}{(\Ccancel[blue]{x+1})(x-2)}\\ \amp = \frac{(x-1)(x+2)}{x-2} \end{align*}

2. Multiply the first fraction by the reciprocal of the second fraction.

   \begin{align*} \frac{6ab}{2a+b}\div(4a^2b) \amp =\frac{\cancelto{\alert{3}}{6}\bcancel{ab}}{2a+b}\cdot\frac{1}{\cancelto{\alert{2}}{4}a\cdot\bcancel{ab}} \amp\amp\blert{\text{Divide out common factors.}}\\ \amp = \frac{3}{2a(2a+b)}=\frac{3}{4a^2+2ab} \end{align*}

1. Because these are like fractions, we add their numerators and keep the same denominator.

   \begin{equation*} \frac{2x}{9z^2}+\frac{5x}{9z^2}=\frac{2x+5x}{9z^2}=\frac{7x}{9z^2} \end{equation*}

2. Be careful to subtract the entire numerator of the second fraction: Use parentheses to show that the subtraction applies to both terms of \(5x - 3\text{.}\)

   \begin{align*} \frac{2x-1}{x+3}-\frac{5x-3}{x+3}\amp = \frac{2x-1\alert{-(5x-3)}}{x+3}\\ \amp =\frac{2x-1\alert{-5x+3}}{x+3}=\frac{-3x+2}{x+3} \end{align*}

The LCD is a polynomial that has as factors both \(x + 2\) and \(x-3\text{.}\) The simplest such polynomial is \((x + 2) (x - 3)\text{,}\) or \(x^2 - x - 6\text{.}\) For our purposes, it will be more convenient to leave the LCD in factored form, so the LCD is \((x + 2) (x - 3)\text{.}\)

Factor the denominators of each of the given fractions.

The factor \((x - 1)\) occurs once in the first denominator, the factor \(x\) occurs once in the second denominator, and the factor \((x + 1)\) occurs once in each denominator. Therefore, we include in our LCD one copy of each of these factors. The LCD is \(x (x + 1) (x - 1)\text{.}\)

Compare the denominator of the given fraction to the LCD. We see that the fraction \(\dfrac{3x}{x+2} \) needs a factor of \((x - 3)\) in its denominator, so \((x - 3)\) is the building factor for the first fraction. We multiply the numerator and denominator of the first fraction by \((x - 3)\) to obtain an equivalent fraction:

The fraction \(\dfrac{2x}{x-3} \) needs a factor of \((x + 2)\) in the denominator, so we multiply numerator and denominator by \((x + 2)\text{:}\)

1. The LCD for these fractions is \((x + 2) (x - 3)\text{.}\)

2. We build each fraction to an equivalent one with the LCD, as we did in Example A.76.

   \begin{equation*} \frac{3x}{x+2}=\frac{3x^2-9x}{x^2-x-6} ~~\text{ and }~~ \frac{2x}{x-3}=\frac{2x^2+4x}{x^2-x-6} \end{equation*}

3. Combine the numerators over the same denominator.

   \begin{align*} \frac{3x}{x+2}-\frac{2x}{x-3}\amp =\frac{3x^2-9x}{x^2-x-6}-\frac{2x^2+4x}{x^2-x-6} \amp\amp\blert{\text{Subtract the numerators.}}\\ \amp = \frac{(3x^2-9x)-(2x^2+4x)}{x^2-x-6}\\ \amp = \frac{x^2-13x}{x^2-x-6} \end{align*}

4. Reduce the result, if possible. If we factor both numerator and denominator, we find

   \begin{equation*} \frac{x(x-13)}{(x-3)(x+2)} \end{equation*}

   The fraction cannot be reduced.

1. To find the LCD, factor each denominator:

   \begin{align*} a\amp =a\\ a^2 + a\amp = a (a + 1) \end{align*}

   The LCD is \(a(a + 1)\text{.}\)

2. Build each term to an equivalent fraction with the LCD as denominator. (The building factors for each fraction are shown in color.) The third fraction already has the LCD for its denominator.

   \begin{align*} 1 = \frac{1 \cdot \alert{a (a + 1)}}{1 \cdot\alert{a (a + 1)}}\amp =\frac{a^2+a} {a (a + 1)}\\ \frac{2}{a}=\frac{2\cdot\alert{(a+1)} }{a\cdot\alert{(a+1)}}\amp =\frac{2a+2}{a(a+1)}\\ \frac{a^2 + 2}{a^2 + a} \amp = \frac{a^2 + 2}{a(a+1)} \end{align*}

3. Combine the numerators over the LCD.

   \begin{align*} 1 + \frac{2}{a}-\frac{a^2 + 2}{a^2 + a} \amp = \frac{a^2+a} {a (a + 1)} + \frac{2a+2}{a(a+1)}- \frac{a^2 + 2}{a(a+1)}\\ \amp = \frac{a^2+a +(2a+2)-(a^2+2) } {a (a + 1)}\\ \amp = \frac{3a} {a (a + 1)} \end{align*}

4. Reduce the fraction to find

   \begin{equation*} \frac{3\Ccancel[blue]{a}}{\Ccancel[blue]{a}(a+1)}=\frac{3}{a+1} \end{equation*}

Consider all of the simple fractions that appear in the complex fraction; in this example \(\dfrac{1}{2} \) and \(\dfrac{3}{4} \text{.}\) The LCD of these fractions is \(4\text{.}\) If we multiply the numerator and denominator of the complex fraction by \(4\text{,}\) we will eliminate the fractions within the fraction.

Be sure to multiply each term of the numerator and each term of the denominator by \(4\text{.}\)

Thus, the original complex fraction is equivalent to the simple fraction \(\displaystyle{\frac{4x+3}{4x-2}} \text{.}\)

1. \(\displaystyle{x^{-1}-y^{-1}=\frac{1}{x}-\frac{1}{y} ~~~\text{ or } ~~~\frac{y-x}{xy}}\)

2. \(\displaystyle{\left(x^{-2}+y^{-2}\right)^{-1}=\left(\frac{1}{x^{2}}+\frac{1}{y^{2}}\right)^{-1} =\left(\frac{y^{2}+x^{2}}{x^2y^2}\right)^{-1}=\frac{x^2y^2}{y^2+x^2}}\)