8.5: Building Rational Expressions and the LCD
The Process
Recall, from Section 8.2 the equality property of fractions.
If \(\dfrac{a}{b} = \dfrac{c}{d}\), then \(ad = bc\).
Using the fact that \(1 = \dfrac{b}{b}, b \not = 0\), and that \(1\) is the multiplicative identity, it follows that if \(\dfrac{P}{Q}\) is a rational expression, then
\(\dfrac{P}{Q} \cdot \dfrac{b}{b} = \dfrac{Pb}{Qb}, b \not = 0\)
This equation asserts that a rational expression can be transformed into an equivalent rational expression by multiplying both the numerator and denominator by the same nonzero number.
This process is known a(s the process of building rational expressions and it is exactly the opposite of reducing rational expressions. The process is shown in these examples:
\(\dfrac{3}{4}\) can be built to \(\dfrac{12}{16}\) since:
\(\dfrac{3}{4} \cdot 1 = \dfrac{3}{4} \cdot \dfrac{4}{4} = \dfrac{3 \cdot 4}{4 \cdot 4} = \dfrac{12}{16}\)
\(\dfrac{-4}{5}\) can be built to \(\dfrac{-8}{10}\) since
\(\dfrac{-4}{5} \cdot 1 = \dfrac{-4}{5} \cdot \dfrac{2}{2} = \dfrac{-4 \cdot 2}{5 \cdot 2} = \dfrac{-8}{10}\)
\(\dfrac{3}{7}\) can be built to \(\dfrac{3xy}{7xy}\) since
\(\dfrac{3}{7} \cdot 1 = \dfrac{3}{7} \cdot \dfrac{xy}{xy} = \dfrac{3xy}{7xy}\)
\(\dfrac{4a}{3b}\) can be built to \(\dfrac{4a^2(a+1)}{3ab(a+1)}\) since:
\(\dfrac{4a}{3b} \cdot 1 = \dfrac{4a}{3b} \cdot \dfrac{a(a + 1)}{a(a+1)} = \dfrac{4a^2(a + 1)}{3ab(a + 1)}\).
Suppose we're given a rational expression \(\dfrac{P}{Q}\) and wish to build it into a rational expression with denominator \(Qb^2\), that is,
\(\dfrac{P}{Q} \rightarrow \dfrac{?}{Qb^2}\)
Since we changed the denominator, we must certainly change the numerator in the same way. To determine how to change the numerator we need to know how the denominator was changed. Since one rational expression is built into another equivalent expression by multiplication by 1, the first denominator must have been multiplied by some quantity. Observation of
\(\dfrac{P}{Q} \rightarrow \dfrac{?}{Qb^2}\)
tells us that \(Q\) was multiplied by \(b^2\). Hence, we must multiply the numerator \(P\) by \(b^2\). Thus
\(\dfrac{P}{Q} \rightarrow \dfrac{Pb^2}{Qb^2}\)
Quite often a simple comparison of the original denominator with the new denominator will tell us the factor being used. However, there will be times when the factor is unclear by simple observation. We need a method for finding the factor.
Observe the following examples; then try to speculate on the method.
\(\dfrac{3}{4} = \dfrac{?}{20}\)
The original denominator 4 was multiplied by 5 to yield 20. What arithmetic process will yield 5 using 4 and 20?
\(\dfrac{9}{10} = \dfrac{?}{10y}\)
The original denominator 10 was multiplied by \(y\) to yield \(10y\).
\(dfrac{-6xy}{2a^3b} = \dfrac{?}{16a^5b^3}\)
The original denominator \(2a^3b\) was multiplied by \(8a^2b^2\) to yield \(16a^5b^3\)
\(\dfrac{5ax}{(a+1)^2} = \dfrac{?}{4(a+1)^2(a-2)}\)
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.
\(\dfrac{8}{3} = \dfrac{N}{15}\)
The original denominator is \(3\) and the new denominator is \(15\). Divide the original denominator into the new denominator and multiply the numerator \(8\) by this result. \(15÷3=5\) Then, \(8 \cdot 5=40\). So,
\(\dfrac{8}{3} = \dfrac{40}{15}\) and \(N = 40\).
Check by reducing \(\dfrac{40}{15}\)
\(\dfrac{2x}{5b^2y} = \dfrac{N}{20b^5y^4}\)
The original denominator is \(5b^2y\) and the new denominator is \(20b^5y^4\). Divide the original denominator into the new denominator and multiply the numerator \(2x\) by this result.
\(\dfrac{20b^5y^4}{5b^2y} = 4b^3y^3\)
So, \(2x \cdot 4b^3y^3 = 8b^3xy^3\). Thus,
\(\dfrac{2x}{5b^2y} = \dfrac{8b^3xy^3}{20b^5y^4}\) and \(N = 8b^3xy^3\).
\(\dfrac{-6a}{a+2} = \dfrac{N}{(a+2)(a-7)}\)
The new denominator divided by the original denominator is
\(\dfrac{(a+2)(a-7)}{a+2} = a-7\)
Multiply \(-6a\) by \(a-7\).
\(-6a(a-7) = -6a^2 + 42a\)
\(\dfrac{-6a}{a+2} = \dfrac{-6a^2 + 42a}{(a+2)(a-7)}\) and \(N = -6a^2 + 42a\)
\(\dfrac{-3(a-1)}{a-4} = \dfrac{N}{a^2 - 16}\).
The new denominator divided by the original denominator is
\(\dfrac{a^2-16}{a-4} = \dfrac{(a+4)\cancel{(a-4)}}{\cancel{a-4}}\)
Multiply \(-3(a-1)\) by \(a + 4\)
\(-3(a-1)(a+4) = -3(a^2 + 3a - 4)\)
\( = -3a^2 - 9a + 12\)
\(\dfrac{-3(a-1)}{a-4} = \dfrac{-3a^2 - 9a + 12}{a^2 - 16}\) and \(N = -3a^2 - 9a + 12\)
\(\begin{array}{flushleft}
7x&=\dfrac{N}{x^2y^3}&\text{ Write } 7x \text{ as } \dfrac{7x}{1}\\
\dfrac{7x}{1}&=\dfrac{N}{x^2y^3}&\text{ Now we can see clearly that the original denominator }\\
&& 1 \text{ was multiplied by } x^2y^3. \text{ We need to multiply the }\\
&& \text{ numerator } 7x \text{ by } x^2y^3\\
7x&=\dfrac{7x \cdot x^2y^3}{x^2y^3}
\end{array}\)
\(7x = \dfrac{7x^3y^3}{x^2y^3} \text{ and } N = 7x^3y^3\)
\(\begin{array}{flushleft}
\dfrac{5x}{x+3}&=\dfrac{5x^2-20x}{N} & \text{ The same process works in this case. Divide the original }\\
&& \text{ numerator } 5x \text{ into the new numerator } 5x^2 - 20x\\
\dfrac{5x^2 - 20x}{5x} &= \dfrac{\cancel{5x}(x-4)}{\cancel{5x}}\\
&=x-4\\
\end{array}\)
\((x+3)(x-4) \text{ Multiply the denominator by } x-4\)
\(\dfrac{5x}{x+3} = \dfrac{5x^2-20}{(x+3)(x-4)} \text{ and } N = 5x^2 - 20\)
\(\begin{array}{flushleft}
\dfrac{4x}{3-x} &= \dfrac{N}{x-3} & \text{ The two denominators have nearly the same terms; each has }\\
&&\text{ The opposite sign. Factor } -1 \text{ from the original denominator}\\
3-x &= -1(-3+x)\\
&=-(x-3)\\
\end{array}\)
\(\dfrac{4x}{3-x} = \dfrac{4x}{-(x-3)} = \dfrac{-4x}{x-3} \text{ and } N = -4x\)
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 .
\(\dfrac{3}{8} = \dfrac{N}{48}\)
- Answer
-
\(N=18\)
\(\dfrac{9a}{5b} = \dfrac{N}{35b^2x^3}\)
- Answer
-
\(N = 63abx^3\)
\(\dfrac{-2y}{y-1} = \dfrac{N}{y^2 - 1}\)
- Answer
-
\(N = -2y^2 - 2y\)
\(\dfrac{a+7}{a-5} = \dfrac{N}{a^2 - 3a - 10}\)
- Answer
-
\(N = a^2 + 9a + 14\)
\(4a = \frac{N}{6a^3(a-1)}\)
- Answer
-
\(N = 24a^4(a-1)\)
\(-2x = \dfrac{N}{8x^3y^3z^5}\)
- Answer
-
\(N = -16x^4y^3z^5\)
\(\dfrac{6ab}{b+3} = \dfrac{N}{b^2 + 6b + 9}\)
- Answer
-
\(N = 6ab^2 + 18ab\)
\(\dfrac{3m}{m+5} = \dfrac{3m^2 - 18m}{N}\)
- Answer
-
\(N = m^2 - m - 30\)
\(\dfrac{-2r^2}{r-3} = \dfrac{-2r^3 + 8r^2}{N}\)
- Answer
-
\(N = r^2 - 7r + 12\)
\(\dfrac{-8ab^2}{a-4} = \dfrac{N}{4-a}\)
- Answer
-
\(N = 8ab^2\)
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).
To add or subtract two or more rational expressions they must have the same denominator .
Building rational expressions allows us to transform fractions into fractions with the same denominators (which we can then add or subtract). The most convenient new denominator is the least common denominator (LCD) of the given fractions.
The Least Common Denominator (LCD)
In arithmetic, the least common denominator is the smallest (least) quantity that each of the given denominators will divide into without a remainder. For algebraic expressions, the LCD is the polynomial of least degree divisible by each denominator. Some examples are shown below.
\(\dfrac{3}{4}, \dfrac{1}{6}, \dfrac{5}{12}\)
The LCD is 12 since 12 is the smallest number that 4, 6, and 12 will divide into without a remainder.
\(\dfrac{1}{3}, \dfrac{5}{6}, \dfrac{5}{8}, \dfrac{7}{12}\)
The LCD is 24 since 24 is the smallest number that 3, 6, 8, and 12 will divide into without a remainder.
\(\dfrac{2}{x}, \dfrac{3}{x^2}\)
The LCD is \(x^2\) since \(x^2\) is the smallest quantity that \(x\) and \(x^2\) will divide into without a remainder.
\(\dfrac{5a}{6a^2b}, \dfrac{3a}{8ab^3}\)
The LCD is \(24a^2b^3\) since \(24a^2b^3\) is the smallest quantity that \(6a^2b\) and \(8ab^3\) will divide into without a remainder.
\(\dfrac{2y}{y-6}, \dfrac{4y^2}{(y-6)^3}, \dfrac{y}{y-1}\)
The LCD is \((y-6)^3(y-1)\) since \((y-6)^3 \cdot (y-1)\) is the smallest quantity that \(y-6, (y-6)^3\), and \(y-1\) will divide into without a remainder
We’ll now propose and demonstrate a method for obtaining the LCD.
- Factor each denominator. Use exponents for repeated factors. It is usually not necessary to factor numerical quantities.
- Write down each different factor that appears. If a factor appears more than once, use only the factor with the highest exponent.
- The LCD is the product of the factors written in step 2.
Sample Set B
Find the LCD
\(\dfrac{1}{x}, \dfrac{3}{x^3}, \dfrac{2}{4y}\)
1. The denominators are already factored.
2. Note that \(x\) appears as \(x\) and \(x^3\). Use only the \(x\) with the higher exponenent, \(x^3\). The term \(4y\) appears, so we must also use \(4y\).
3. THe LCD is \(4x^3y\).
\(\dfrac{5}{(x-1)^2}, \dfrac{2x}{(x-1)(x-4)}, \dfrac{-5x}{x^2 -4x + 2}\)
1. Only the third denominator needs to be factored:
\(x^2 - 3x + 2 = (x-2)(x-1)\)
Now the three denoinators are \((x-1)^2, (x-1)(x-4)\), and \((x-2)(x-1)\).
2. Note that \(x-1\) appears as \((x-1)^2, x-1\), and \(x-1\). Use only the \(x-1\) with the highest exponent, \((x-1)^2\). Also appearing are \(x-4\) and \(x-2\).
3. The LCD is \((x-1)^2(x-4)(x-2)\).
\(\dfrac{-1}{6a^4}, \dfrac{3}{4a^3b}, \dfrac{1}{3a^3(b+5)}\)
1. The denominators are already factored.
2. We can see that the LCD of the numbers \(6, 4\), and \(3\) is \(12\). We also need \(a^4, b\), and \(b + 5\).
3. The LCD is \(12a^4b(b+5)\).
\(\dfrac{9}{x}, \dfrac{4}{8y}\)
1. The denominators are already factored.
2. \(x, 8y\).
3. The LCD is \(8xy\).
Practice Set B
Find the LCD.
\(\dfrac{3}{x^2}, \dfrac{4}{x^5}, \dfrac{-6}{xy}\)
- Answer
-
\(x^5y\)
\(\dfrac{x+1}{x-4}, \dfrac{x-7}{(x-4)^2}, \dfrac{-6}{x+1}\)
- Answer
-
\((x-4)^2(x+1)\)
\(\dfrac{2}{m-6}, \dfrac{-5n}{(m+1)^2(m-2)}, \dfrac{-3x}{x^2 - 6x + 9}\)
- Answer
-
\((m-6)(m+1)^2(m-2)^3\)
\(\dfrac{1}{x^2 - 1}, \dfrac{2}{x^2 - 2x - 3}, \dfrac{-3x}{x^2-6x+9}\)
- Answer
-
\((x+1)(x-1)(x-3)^2\)
\(\dfrac{3}{4y^2-8y}, \dfrac{8}{y^2-4y+4}, \dfrac{10y-1}{3y^3-6y^2}\)
- Answer
-
\(12y^2(y-2)^2\)
Sample Set C
Change the given rational expressions into rational expressions having the same denominator.
\(\begin{array}{flushleft}
\dfrac{3}{x^2}, \dfrac{4}{x} & \text{ The LCD, by inspection, is } x^2 \text{. Rewrite each expression with }\\
&x^2 \text{ as the new denominator. }\\
\dfrac{?}{x^2}, \dfrac{?}{x^2} & \text{ Determine the numerators. In } \dfrac{3}{x^2} \text{, the denominator was not }\\
& \text{ changed so we need not change the numerator}\\
& \text{ In the second fraction, the original denominator was } x\\
\dfrac{3}{x^2}, \dfrac{?}{x^2} & \text{ We can see that } x \text{ must be multiplied by } x \text{ to build it to } x^2\\
& \text{ So we must also multiply the numerator } 4 \text{ by } x \text{. Thus, } 4 \cdot x = 4x.\\
\dfrac{3}{x^2}, \dfrac{4x}{x^2}
\end{array}\)
\(\begin{array}{flushleft}
\dfrac{4b}{b-1}, \dfrac{-2b}{b+3} & \text{ By inspection, the LCD is } (b-1)(b+3)\\
\dfrac{?}{(b-1)(b+3)}, \dfrac{?}{(b-1)(b+3)} & \text{ The denominator of the first rational expression has been multiplied }\\
& \text{ by } b + 3 \text{, so the numerator } 4b \text{ must be multiplied by } b + 3.\\
4b(b + 3) = 4b^2 + 12b\\
\dfrac{4b^2 + 12b}{(b-1)(b+3)}, \dfrac{?}{(b-1)(b+3)} & \text{ The denominator of the second rational expression has been multiplied }\\
& \text{ by } b-1 \text{, so the numerator } -2b \text{ must be multiplied by } b-1.\\
-2b(b-1) = -2b^2 + 2b\\
\dfrac{4b^2 + 12b}{(b-1)(b+3)}, \dfrac{-2b^2 + 2b}{(b-1)(b+3)}
\end{array}\)
\(\begin{array}{flushleft}
\dfrac{6x}{x^2 - 8x + 15}, \dfrac{-2x^2}{x^2 - 7x + 12}& \text{ We first find the LCD. Factor.}\\
\dfrac{6x}{(x-3)(x-5)}, \dfrac{-2x^2}{(x-3)(x-4)} & \text{The LCD is } (x-3)(x-5)(x-4) \text{. Rewrite each of these}\\
&\text{ fractions with new denominator } (x-3)(x-5)(x-4)\\
\dfrac{?}{(x-3)(x-5)(x-4)}, \dfrac{?}{(x-3)(x-5)(x-4)} & \text{ By comparing the denominator of the first fraction with the LCD }\\
& \text{ we see that we must multiply the numerator } 6x \text{ by } x-4\\
6x(x-4) = 6x^2-24x\\
\dfrac{6x^2 - 24x}{(x-3)(x-5)(x-4)}, \dfrac{?}{(x-3)(x-5)(x-4)} & \text{ By comparing the denominator of the second fraction with the LCD },
&\text{ we see that we must multiply the numerator } -2x^2 \text{ by } x - 5\\
-2x^2(x-5) = -2x^3 + 10x^2\\
\dfrac{6x^2 - 24x}{(x-3)(x-5)(x-4)}, \dfrac{-2x^3 + 10x^2}{(x-3)(x-5)(x-4)}
\end{array}\)
These examples have been done step-by-step and include explanations. This makes the process seem fairly long. In practice, however, the process is much quicker.
\(\begin{array}{flushleft}
\dfrac{6ab}{a^2-5a+4}, \dfrac{a+b}{a^2-8a+16}\\
\dfrac{6ab}{(a-1)(a-4)}, \dfrac{a+b}{(a-4)^2} & \text{ LCD } = (a-1)(a-4)^2\\
\dfrac{6ab(a-4)}{(a-1)(a-4)^2}, \dfrac{(a+b)(a-1)}{(a-1)(a-4)^2}
\end{array}\)
\(\begin{array}{flushleft}
\dfrac{x+1}{x^3 + 3x^2}, \dfrac{2x}{x^3 - 4x}, \dfrac{x-4}{x^2 - 4x + 4}\\
\dfrac{x+1}{x^2(x + 3)}, \dfrac{2x}{x(x+2)(x-2)}, \dfrac{x-4}{(x-2)^2} & \text{ LCD } = x^2(x+3)(x+2)(x-2)^2\\
\dfrac{(x+1)(x+2)(x-2)^2}{x^2(x+3)(x+2)(x-2)^2}, \dfrac{2x^2(x+3)(x-2)}{x^2(x+3)(x+2)(x-2)^2}, \dfrac{x^2(x+3)(x+2)(x-4)}{x^2(x+3)(x+2)(x-2)^2}
\end{array}\)
Practice Set C
Change the given rational expressions into rational expressions with the same denominators.
\(\dfrac{4}{x^3}, \dfrac{7}{x^5}\)
- Answer
-
\(\dfrac{4x^2}{x^5}, \dfrac{7}{x^5}\)
\(\dfrac{2x}{x+6}, \dfrac{x}{x-1}\)
- Answer
-
\(\dfrac{2x(x-1)}{(x+6)(x-1)}, \dfrac{x(x+6)}{(x+6)(x-1)}\)
\(\dfrac{-3}{b^2-b}, \dfrac{4b}{b^2-1}\)
- Answer
-
\(\dfrac{-3(b+1)}{b(b-1)(b+1)}, \dfrac{4b^2}{b(b-1)(b+1)}\)
\(\dfrac{8}{x^2-x-6}, \dfrac{-1}{x^2 + x - 2}\)
- Answer
-
\(\dfrac{8(x-1)}{(x-3)(x+2)(x-1)}, \dfrac{-1(x-3)}{(x-3)(x+2)(x-1)}\)
\(\dfrac{10x}{x^2 + 8x + 16}, \dfrac{5x}{x^2 - 16}\)
- Answer
-
\(\dfrac{10x(x-4)}{(x+4)^2(x-4)}, \dfrac{5x(x+4)}{(x+4)^2(x-4)}\)
\(\dfrac{-2ab^2}{a^3-6a^2}, \dfrac{6b}{a^4-2a^3}, \dfrac{-2a}{a^2 - 4a + 4}\)
- Answer
-
\(\dfrac{2^2b^2(a-2)^2}{a^3(a-6)(a-2)^2}, \dfrac{6b(a-6)(a-2)}{a^3(a-6)(a-2)^2}, \dfrac{-2a^4(a-6)}{a^3(a-6)(a-2)^2}\)
Exercises
For the following problems, replace N with the proper quantity.
\(\dfrac{3}{x} = \dfrac{N}{x^3}\)
- Answer
-
\(3x^2\)
\(\dfrac{4}{a} = \dfrac{N}{a^2}\)
\(\dfrac{-2}{x} = \dfrac{N}{xy}\)
- Answer
-
\(−2y\)
\(\dfrac{-7}{m} = \dfrac{N}{ms}\)
\(\dfrac{6a}{5} = \dfrac{N}{10b}\)
- Answer
-
\(12ab\)
\(\dfrac{a}{3z} = \dfrac{N}{12z}\)
\(\dfrac{x^2}{4y^2} = \dfrac{N}{20y^4}\)
- Answer
-
\(5x^2y^2\)
\(\dfrac{b^3}{6a} = \dfrac{N}{18a^5}\)
\(\dfrac{-4a}{5x^2y} = \dfrac{N}{15x^3y^3}\)
- Answer
-
\(-12axy^2\)
\(\dfrac{-10z}{7a^3b} = \dfrac{N}{21a^4b^5}\)
\(\dfrac{8x^2y}{5a^3} = \dfrac{N}{25a^3x^2}\)
- Answer
-
\(40x^4y\)
\(\dfrac{2}{a^2} = \dfrac{N}{a^2(a-1)}\)
\(\dfrac{5}{x^3} = \dfrac{N}{x^3(x-2)}\)
- Answer
-
\(5(x−2)\)
\(\dfrac{2a}{b^2} = \dfrac{N}{b^3-b}\)
\(\dfrac{4x}{a} = \dfrac{N}{a^4-4a^2}\)
- Answer
-
\(4ax(a+2)(a−2) \)
\(\dfrac{6b^3}{5a} = \dfrac{N}{10a^2-30a}\)
\(\dfrac{4x}{3b} = \dfrac{N}{3b^5 - 15b}\)
- Answer
-
\(4x(b^4 - 5)\)
\(\dfrac{2m}{m-1} = \dfrac{N}{(m-1)(m+2)}\)
\(\dfrac{3s}{s + 12} = \dfrac{N}{(s + 12)(s-7)}\)
- Answer
-
\(3s(s-7)\)
\(\dfrac{a+1}{a-3} = \dfrac{N}{(a-3)(a-4)}\)
\(\dfrac{a+2}{a-2} = \dfrac{N}{(a-2)(a-4)}\)
- Answer
-
\((a+2)(a−4)\)
\(\dfrac{b+7}{b-6} = \dfrac{N}{(b-6)(b+6)}\)
\(\dfrac{5m}{2m + 1} = \dfrac{N}{(2m+1)(m-2)}\)
- Answer
-
\(5m(m−2)\)
\(\dfrac{4}{a+6} = \dfrac{N}{a^2 + 5a - 6}\)
\(\dfrac{9}{b-2} = \dfrac{N}{b^2 - 6b + 8}\)
- Answer
-
\(9(b−4)\)
\(\dfrac{3b}{b-3} = \dfrac{N}{b^2 - 11b + 24}\)
\(\dfrac{-2x}{x-7} = \dfrac{N}{x^2 - 4x - 21}\)
- Answer
-
\(−2x(x+3)\)
\(\dfrac{-6m}{m+6} = \dfrac{N}{m^2 + 10m + 24}\)
\(\dfrac{4y}{y+1} = \dfrac{N}{y^2 + 9y + 8}\)
- Answer
-
\(4y(y+8)\)
\(\dfrac{x + 2}{x - 2} = \dfrac{N}{x^2 - 4}\)
\(\dfrac{y-3}{y + 3} = \dfrac{N}{y^2 - 9}\)
- Answer
-
\((y-3)^2\)
\(\dfrac{a+5}{a-5} = \dfrac{N}{a^2 - 25}\)
- Answer
-
\((z - 4)^2\)
\(\dfrac{4}{2a + 1} = \dfrac{N}{2a^2 - 5a - 3}\)
\(\dfrac{1}{3b - 1} = \dfrac{N}{3b^2 + 11b - 4}\)
- Answer
-
\(b+4\)
\(\dfrac{a+2}{2a - 1} = \dfrac{N}{2a^2 + 9a - 5}\)
\(\dfrac{-3}{4x + 3} = \dfrac{N}{4x^2 - 13x - 12}\)
- Answer
-
\(−3(x−4)\)
\(\dfrac{b+2}{3b - 1} = \dfrac{N}{6b^2 + 7b - 3}\)
\(\dfrac{x - 1}{4x - 5} = \dfrac{N}{12x^2 - 11x - 5}\)
- Answer
-
\((x−1)(3x+1)\)
\(\dfrac{3}{x + 2} = \dfrac{3x - 21}{N}\)
\(\dfrac{4}{y + 6} = \dfrac{4y + 8}{N}\)
- Answer
-
\((y+6)(y+2)\)
\(\dfrac{-6}{a - 1} = \dfrac{-6a - 18}{N}\)
\(\dfrac{-8a}{a+3} = \dfrac{-8a^2 - 40a}{N}\)
- Answer
-
\((a+3)(a+5)\)
\(\dfrac{y+1}{y-8} = \dfrac{y^2 - 2y - 3}{N}\)
\(\dfrac{x - 4}{x + 9} = \dfrac{x^2 + x - 20}{N}\)
- Answer
-
\((x+9)(x+5)\)
\(\dfrac{3x}{2-x} = \dfrac{N}{x-2}\)
\(\dfrac{7a}{5-a} = \dfrac{N}{a-5}\)
- Answer
-
\(-7a\)
\(\dfrac{-m + 1}{3 - m} = \dfrac{N}{m-3}\)
\(\dfrac{k + 6}{10 - k} = \dfrac{N}{k- 10}\)
- Answer
-
\(−k−6\)
\(\dfrac{2}{a^2} = \dfrac{N}{a^2(a-1)}\)
For the following problems, convert the given rational expressions to rational expressions having the same denominators.
\(\dfrac{2}{a}, \dfrac{3}{a^4}\)
\(\dfrac{5}{b^2}, \dfrac{4}{b^3}\)
- Answer
-
\(\dfrac{5b}{b^3}, \dfrac{4}{b^3}\)
\(\dfrac{8}{z}, \dfrac{3}{4z^3}\)
\(\dfrac{9}{x^2}, \dfrac{1}{4x}\)
- Answer
-
\(\dfrac{36}{4x^2}, \dfrac{x}{4x^2}\)
\(\dfrac{2}{a+3}, \dfrac{4}{a+1}\)
\(\dfrac{2}{x + 5}, \dfrac{4}{x-5}\)
- Answer
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\(\dfrac{2(x-5)}{(x+5)(x-5)}, \dfrac{4(x+5)}{(x+5)(x-5)}\)
\(\dfrac{1}{x-7}, \dfrac{4}{x-1}\)
\(\dfrac{10}{y+2}, \dfrac{1}{y+8}\)
- Answer
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\(\dfrac{10(y+8)}{(y+2)(y+8)}, \dfrac{y+2}{(y+2)(y+8)}\)
\(\dfrac{4}{a^2}, \dfrac{a}{a+4}\)
\(\dfrac{-3}{b^2}, \dfrac{b^2}{b+5}\)
- Answer
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\(\dfrac{-3(b+5)}{b^2(b+5)}, \dfrac{b^4}{b^2(b+5)}\)
\(\dfrac{-6}{b-1}, \dfrac{5b}{4b}\)
\(\dfrac{10a}{a-6}, \dfrac{2}{a^2 - 6a}\)
- Answer
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\(\dfrac{10a^2}{a(a-6)}, \dfrac{2}{a(a-6)}\)
\(\dfrac{4}{x^2 + 2x}, \dfrac{1}{x^2 - 4}\)
\(\dfrac{x+1}{x^2 - x - 6}, \dfrac{x+4}{x^2 + x - 2}\)
- Answer
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\(\dfrac{(x+1)(x-1)}{(x-1)(x+2)(x-3)}, \dfrac{(x+4)(x-3)}{(x-1)(x+2)(x-3)}\)
\(\dfrac{x-5}{x^2 - 9x + 20}, \dfrac{4}{x^2 - 3x - 10}\)
\(\dfrac{-4}{b^2 + 5b - 6}, \dfrac{b+6}{b^2 - 1}\)
- Answer
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\(\dfrac{-4(b +1)}{(b+1)(b-1)(b + 6)}, \dfrac{(b+6)^2}{(b+1)(b-1)(b+6)}\)
\(\dfrac{b+2}{b^2 + 6b + 8}, \dfrac{b-1}{b^2 + 8b + 12}\)
\(\dfrac{x+7}{x^2 - 2x - 3}, \dfrac{x+3}{x^2 - 6x - 7}\)
- Answer
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\(\dfrac{(x+7)(x-7)}{(x+1)(x-3)(x-7)}, \dfrac{(x+3)(x-3)}{(x+1)(x-3)(x-7)}\)
\(\dfrac{2}{a^2 + a}, \dfrac{a+3}{a^2 - 1}\)
\(\dfrac{x-2}{x^2 + 7x + 6}, \dfrac{2x}{x^2 + 4x - 12}\)
- Answer
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\(\dfrac{(x-2)^2}{(x+1)(x-2)(x+6)}, \dfrac{2x(x+1)}{(x+1)(x-2)(x+6)}\)
\(\dfrac{x-2}{2x^2 + 5x - 3}, \dfrac{x-1}{5x^2 + 16x + 3}\)
\(\dfrac{2}{x-5}, \dfrac{-3}{5-x}\)
- Answer
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\(\dfrac{2}{x-5}, \dfrac{3}{x-5}\)
\(\dfrac{4}{a-6}, \dfrac{-5}{6-a}\)
\(\dfrac{6}{2-x}, \dfrac{5}{x-2}\)
- Answer
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\(\dfrac{-6}{x-2}, \dfrac{5}{x-2}\)
\(\dfrac{k}{5-k}, \dfrac{3k}{k-5}\)
\(\dfrac{2m}{m-8}, \dfrac{7}{8-m}\)
- Answer
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\(\dfrac{2m}{m-8}, \dfrac{-7}{m-8}\)
Exercises For Review
Factor \(m^2x^3 + mx^2 + mx\)
Factor \(y^2 - 10y + 21\)
- Answer
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\((y−7)(y−3)\)
Write the equation of the line that passes through the points (1, 1) and (4, −2). Express the equation in slope-intercept form.
Reduce \(\dfrac{y^2 - y - 6}{y-3}\)
- Answer
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\(y+2\)
Find the quotient. \(\dfrac{x^2 - 6x + 9}{x^2 - x - 6} \div \dfrac{x^2 + 2x - 15}{x^2 + 2x}\)