4.2 Rational expressions: building rational expressions and the (Page 2/4)

Algebra ii for the community Page 2 / 4

Observe the following examples; then try to speculate on the method.

$\frac{3}{4} = \frac{?}{20} .$

The original denominator 4 was multiplied by 5 to yield 20. What arithmetic process will yield 5 using 4 and 20?

$\frac{9}{10} = \frac{?}{10 y} .$

The original denominator 10 was multiplied by $y$ to yield $10 y$ .

$\frac{- 6 x y}{2 a^{3} b} = \frac{?}{16 a^{5} b^{3}} .$

The original denominator $2 a^{3} b$ was multiplied by $8 a^{2} b^{2}$ to yield $16 a^{5} b^{3} .$

$\frac{5 a x}{{(a + 1)}^{2}} = \frac{?}{4 {(a + 1)}^{2} (a - 2)} .$

The original denominator ${(a + 1)}^{2}$ was multiplied by $4 (a - 2)$ to yield $4 {(a + 1)}^{2} (a - 2)$ .

To determine the quantity that the original denominator was multiplied by to yield the new denominator, we ask, "What did I multiply the original denominator by to get the new denominator?" We find this factor by dividing the original denominator into the new denominator.

It is precisely this quantity that we multiply the numerator by to build the rational expression.

Sample set a

Determine N in each of the following problems.

$\begin{array}{l} \begin{array}{l} \frac{8}{3} = \frac{N}{15} . & \begin{array}{l} The original denominator is 3 and the new \\ denominato is 15 . Divide the original \\ denominator into the new denominator and \\ multiply the numerator 8 by this result . \end{array} \\ 15 \div 3 = 5 \\ Then, 8 \cdot 5 = 40. So, \\ \frac{8}{3} = \frac{40}{15} and N = 40. \end{array} \\ Check by reducing \frac{40}{15} . \end{array}$

$\begin{array}{l} \frac{2 x}{5 b^{2} y} = \frac{N}{20 b^{5} y^{4}} . & \begin{array}{l} The original denominator is 5 b^{2} y and the new \\ denominator is 20 b^{5} y^{4} . Divide the original \\ denominator into the new denominator and \\ multiply the numerator 2 x by this result . \end{array} \\ \frac{20 b^{5} y^{4}}{5 b^{2} y} = 4 b^{3} y^{3} \\ So, 2 x \cdot 4 b^{3} y^{3} = 8 b^{3} x y^{3} . Thus, \\ \frac{2 x}{5 b^{2} y} = \frac{8 b^{3} x y^{3}}{20 b^{5} y^{4}} and N = 8 b^{3} x y^{3} . \end{array}$

$\begin{array}{l} \begin{array}{l} \frac{- 6 a}{a + 2} = \frac{N}{(a + 2) (a - 7)} . & The new denominator divided by the original denominator is \\ \frac{(a + 2) (a - 7)}{a + 2} = a - 7 \\ Multiply - 6 a by a - 7. \\ - 6 a (a - 7) = - 6 a^{2} + 42 a \end{array} \\ \frac{- 6 a}{a + 2} = \frac{- 6 a^{2} + 42 a}{(a + 2) (a - 7)} and N = - 6 a^{2} + 42 a . \end{array}$

$\begin{array}{l} \begin{array}{l} \frac{- 3 (a - 1)}{a - 4} = \frac{N}{a^{2} - 16} . & The new denominator divided by the original denominator is \\ \frac{a^{2} - 16}{a - 4} = \frac{(a + 4) (a - 4)}{a - 4} \\ = a + 4 \\ Multiply - 3 (a - 1) by a + 4. \\ \begin{matrix} - 3 (a - 1) (a + 4) & = & - 3 (a^{2} + 3 a - 4) \end{matrix} \\ \begin{matrix} = & - 3 a^{2} - 9 a + 12 \end{matrix} \end{array} \\ \frac{- 3 (a - 1)}{a - 4} = \frac{- 3 a^{2} - 9 a + 12}{a^{2} - 16} and N = - 3 a^{2} - 9 a + 12 \end{array}$

$\begin{array}{l} 7 x & = & \frac{N}{x^{2} y^{3}} . & Write 7 x as \frac{7 x}{1} . \\ \frac{7 x}{1} & = & \frac{N}{x^{2} y^{3}} & \begin{array}{l} Now we can see clearly that the original denominator \\ 1 was multiplied by x^{2} y^{3} . We need to multiply the \\ numerator 7 x by x^{2} y^{3} . \end{array} \\ 7 x & = & \frac{7 x \cdot x^{2} y^{3}}{x^{2} y^{3}} \\ 7 x & = & \frac{7 x^{3} y^{3}}{x^{2} y^{3}} and N = 7 x^{3} y^{3} . \end{array}$

$\begin{array}{l} \begin{array}{l} \frac{5 x}{x + 3} = \frac{5 x^{2} - 20 x}{N} . & \begin{array}{l} The same process works in this case . Divide the original \\ numerator 5 x into the new numerator 5 x^{2} - 20 x . \end{array} \\ \frac{5 x^{2} - 20 x}{5 x} = \frac{5 x (x - 4)}{5 x} \\ = x - 4 \\ Multiply the denominator by x - 4. \\ (x + 3) (x - 4) \end{array} \\ \frac{5 x}{x + 3} = \frac{5 x^{2} - 20}{(x + 3) (x - 4)} and N = 5 x^{2} - 20. \end{array}$

$\begin{array}{l} \begin{array}{l} \frac{4 x}{3 - x} = \frac{N}{x - 3} . & \begin{array}{l} The two denominators have nearly the same terms; each has \\ the opposite sign . Factor - 1 from the original denominator . \end{array} \\ 3 - x = - 1 (- 3 + x) \\ = - (x - 3) \end{array} \\ \frac{4 x}{3 - x} = \frac{4 x}{- (x - 3)} = \frac{- 4 x}{x - 3} and N = - 4 x . \end{array}$

It is important to note that we factored $- 1$ from the original denominator. We did not multiply it by $- 1$ . Had we multiplied only the denominator by $- 1$ we would have had to multiply the numerator by $- 1$ also.

Practice set a

Determine N .

$\frac{3}{8} = \frac{N}{48}$

$N = 18$

$\frac{9 a}{5 b} = \frac{N}{35 b^{2} x^{3}}$

$N = 63 a b x^{3}$

$\frac{- 2 y}{y - 1} = \frac{N}{y^{2} - 1}$

$N = - 2 y^{2} - 2 y$

$\frac{a + 7}{a - 5} = \frac{N}{a^{2} - 3 a - 10}$

$N = a^{2} + 9 a + 14$

$4 a = \frac{N}{6 a^{3} (a - 1)}$

$N = 24 a^{4} (a - 1)$

$- 2 x = \frac{N}{8 x^{3} y^{3} z^{5}}$

$N = - 16 x^{4} y^{3} z^{5}$

$\frac{6 a b}{b + 3} = \frac{N}{b^{2} + 6 b + 9}$

$N = 6 a b^{2} + 18 a b$

$\frac{3 m}{m + 5} = \frac{3 m^{2} - 18 m}{N}$

$N = m^{2} - m - 30$

$\frac{- 2 r^{2}}{r - 3} = \frac{- 2 r^{3} + 8 r^{2}}{N}$

$N = r^{2} - 7 r + 12$

$\frac{- 8 a b^{2}}{a - 4} = \frac{N}{4 - a}$

$N = 8 a b^{2}$

The reason for building rational expressions

Building rational expressions

Normally, when we write a rational expression, we write it in reduced form. The reason for building rational expressions is to make addition and subtraction of rational expressions convenient (simpler).

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Source: OpenStax, Algebra ii for the community college. OpenStax CNX. Jul 03, 2014 Download for free at http://cnx.org/content/col11671/1.1

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