12.6: Addition and Subtraction of Rational Expressions
Learning Objectives
- Add and subtract rational expressions with the same denominator.
- Find the least common denominator of rational expressions.
- Add and subtract rational expressions with different denominators.
- Solve real-world problems involving addition and subtraction of rational expressions.
Introduction
Like fractions, rational expressions represent a portion of a quantity. Remember that when we add or subtract fractions we must first make sure that they have the same denominator. Once the fractions have the same denominator, we combine the different portions by adding or subtracting the numerators and writing that answer over the common denominator.
Add and Subtract Rational Expressions with the Same Denominator
Fractions with common denominators combine in the following manner.
\begin{align*} \frac{a} {c} + \frac{b} {c} = \frac{a + b} {c} && \text{and} && \frac{a} {c} - \frac{b} {c} = \frac{a - b} {c}\end{align*}
Example 1
Simplify.
a) \begin{align*} \frac{8} {7} - \frac{2} {7} + \frac{4} {7}\end{align*}
b) \begin{align*} \frac{4x^2 - 3} {x + 5} + \frac{2x^2 - 1} {x + 5}\end{align*}
c) \begin{align*} \frac{x^2 - 2x + 1} {2x + 3} - \frac{3x^2 - 3x + 5} {2x + 3}\end{align*}
Solution
a) Since the denominators are the same we combine the numerators.
\begin{align*} \frac{8} {7} - \frac{2} {7} + \frac{4} {7} = \frac{8 - 2 + 4} {7} = \frac{10} {7}\ \text{Answer}\end{align*}
b) Since the denominators are the same we combine the numerators.
\begin{align*} \frac{4x^2 - 3 + 2x^2 - 1} {x + 5}\end{align*}
Simplify by collecting like terms.
\begin{align*} \frac{6x^2 - 4} {x + 5}\ \text{Answer}\end{align*}
c) Since the denominators are the same we combine the numerators. Make sure the subtraction sign is distributed to all terms in the second expression.
\begin{align*}& \frac{x^2 - 2x + 1 - (3x^2 - 3x + 5)} {2x + 3} \\ & = \frac{x^2 - 2x + 1 - 3x^2 + 3x - 5} {2x + 3} \\ & = \frac{-2x^2 + x - 4} {2x + 3}\ \text{Answer}\end{align*}
Find the Least common Denominator of Rational Expressions
To add and subtract fractions with different denominators, we must first rewrite all fractions so that they have the same denominator. In general, we want to find the least common denominator. To find the least common denominator, we find the least common multiple (LCM) of the expressions in the denominators of the different fractions. Remember that the least common multiple of two or more integers is the least positive integer having each as a factor.
Consider the integers 234, 126 and 273.
To find the least common multiple of these numbers we write each integer as a product of its prime factors.
Here we present a systematic way to find the prime factorization of a number.
- Try the prime numbers, in order, as factors.
- Use repeatedly until it is no longer a factor.
- Then try the next prime:
\begin{align*}234 & = 2 \cdot 117 \\ & = 2 \cdot 3 \cdot 39 \\ & = 2 \cdot 3 \cdot 3 \cdot 13 \\ 234 & = 2 \cdot 3^2 \cdot 13\end{align*}
\begin{align*}126 & = 2 \cdot 63 \\ & = 2 \cdot 3 \cdot 21 \\ & = 2 \cdot 3 \cdot 3 \cdot 7 \\ 126 & = 2 \cdot 3^2 \cdot 7\end{align*}
\begin{align*}273 & = 3 \cdot 91 \\ & = 3 \cdot 7 \cdot 13\end{align*}
Once we have the prime factorization of each number, the least common multiple of the numbers is the product of all the different factors taken to the highest power that they appear in any of the prime factorizations. In this case, the factor of two appears at most once, the factor of three appears at most twice, the factor of seven appears at most once, the factor of 13 appears at most once. Therefore,
\begin{align*}\text{LCM} = 2 \cdot 3^2 \cdot 7 \cdot 13 = 1638 \ \text{Answer}\end{align*}
If we have integers that have no common factors, the least common multiple is just the product of the integers. Consider the integers 12 and 25.
\begin{align*}12 = 2^2 \cdot 3 && \text{and} && 25 = 5^2\end{align*}
The \begin{align*}\;\mathrm{LCM} = 2^2 \cdot 3 \cdot 5^2 = 300\end{align*}, which is just the product of 12 and 25.
The procedure for finding the lowest common multiple of polynomials is similar. We rewrite each polynomial in factored form and we form the LCM by taken each factor to the highest power it appears in any of the separate expressions.
Example 2
Find the LCM of \begin{align*}48x^2y\end{align*} and \begin{align*}60xy^3z\end{align*}.
Solution
First rewrite the integers in their prime factorization.
\begin{align*}48 & = 2^4 \cdot 3 \\ 60 & = 2^2 \cdot 3 \cdot 5\end{align*}
Therefore, the two expressions can be written as
\begin{align*}48x^2y & = 2^4 \cdot 3 \cdot x^2 \cdot y \\ 60xy^3z & = 2^2 \cdot 3 \cdot 5 \cdot x \cdot y^3 \cdot z\end{align*}
The LCM is found by taking each factor to the highest power that it appears in either expression.
\begin{align*}\text{LCM} = 2^4 \cdot 3 \cdot 5 \cdot x^2 \cdot y^3 \cdot z = 240 x^2 y^3z.\end{align*}
Example 3
Find the LCM of \begin{align*}2x^2 + 8x + 8\end{align*} and \begin{align*}x^3 - 4x^2 - 12x\end{align*}.
Solution
Factor the polynomials completely.
\begin{align*}2x^{2} + 8x + 8 & = 2(x^{2} + 4x + 4) = 2(x + 2)^{2} \\ x^{3} - 4x^{2} - 12x & = x(x^{2}-4x-12) = x(x-6)(x+2)\end{align*}
The LCM is found by taking each factor to the highest power that it appears in either expression.
\begin{align*}\text{LCM} = 2x (x + 2)^{2} (x - 6)\ \text{Answer}\end{align*}
It is customary to leave the LCM in factored form because this form is useful in simplifying rational expressions and finding any excluded values.
Example 4
Find the LCM of \begin{align*}x^{2}-25\end{align*} and \begin{align*}x^{2}+3x+2\end{align*}.
Solution
Factor the polynomials completely:
\begin{align*}x^{2}-25 & = (x + 5)(x - 5) \\ x^{2}+3x+2 & = (x+2)(x+1)\end{align*}
Since the two expressions have no common factors, the LCM is just the product of the two expressions.
\begin{align*}\text{LCM} = (x + 5)(x - 5)(x+2)(x+1)\ \text{Answer}\end{align*}
Add and Subtract Rational Expressions with Different Denominators
Now we are ready to add and subtract rational expressions. We use the following procedure.
- Find the least common denominator (LCD) of the fractions.
- Express each fraction as an equivalent fraction with the LCD as the denominator.
- Add or subtract and simplify the result.
Example 5
Add \begin{align*} \frac{4} {12} + \frac{5} {18}\end{align*}.
Solution
We can write the denominators in their prime factorization \begin{align*}12 = 2^2 \cdot 3\end{align*} and \begin{align*}18 = 2 \cdot 3^2\end{align*}. The least common denominator of the fractions is the LCM of the two numbers: \begin{align*}2^2 \cdot 3^2 = 36\end{align*}. Now we need to rewrite each fraction as an equivalent fraction with the LCD as the denominator.
For the first fraction. 12 needs to be multiplied by a factor of 3 in order to change it into the LCD, so we multiply the numerator and the denominator by 3.
\begin{align*} \frac{4} {12} \cdot \frac{3} {3} = \frac{12} {36}\end{align*}
For the section fraction. 18 needs to be multiplied by a factor of 2 in order to change it into the LCD, so we multiply the numerator and the denominator by 2.
\begin{align*} \frac{5} {18} \cdot \frac{2} {2} = \frac{10} {36}\end{align*}
Once the denominators of the two fractions are the same we can add the numerators.
\begin{align*} \frac{12} {36} + \frac{10} {36} = \frac{22} {36}\end{align*}
The answer can be reduced by canceling a common factor of 2.
\begin{align*} \frac{12} {36} + \frac{10} {36} = \frac{22} {36} = \frac{11} {18}\ \text{Answer}\end{align*}
Example 6
Perform the following operation and simplify.
\begin{align*} \frac{2} {x + 2} - \frac{3} {2x - 5}\end{align*}
Solution
The denominators cannot be factored any further, so the LCD is just the product of the separate denominators.
\begin{align*}\text{LCD} = (x + 2) (2x - 5)\end{align*}
The first fraction needs to be multiplied by the factor \begin{align*}(2x - 5)\end{align*} and the second fraction needs to be multiplied by the factor \begin{align*}(x + 2)\end{align*}.
\begin{align*} \frac{2} {x + 2} \cdot \frac{(2x - 5)} {(2x - 5)} - \frac{3} {2x - 5} \cdot \frac{(x + 2)} {(x + 2)}\end{align*}
We combine the numerators and simplify.
\begin{align*} \frac{2(2x - 5) - 3(x + 2)} {(x + 2)(2x - 5)} = \frac{4x - 10 - 3x -6} {(x + 2)(2x - 5)}\end{align*}
Combine like terms in the numerator.
\begin{align*}\frac{x - 16} {(x + 2)(2x - 5)} \ \text{Answer}\end{align*}
Example 8
Perform the following operation and simplify.
\begin{align*} \frac{4x} {x -5} - \frac{3x} {5 -x}\end{align*}
Solution
Notice that the denominators are almost the same. They differ by a factor of -1.
Factor \begin{align*}(a - 1)\end{align*} from the second denominator.
\begin{align*} \frac{4x} {x - 5} - \frac{3x} {-(x - 5)}\end{align*}
The two negative signs in the second fraction cancel.
\begin{align*} \frac{4x} {x - 5} + \frac{3x} {(x - 5)}\end{align*}
Since the denominators are the same we combine the numerators.
\begin{align*} \frac{7x} {x - 5} \ \text{Answer}\end{align*}
Example 9
Perform the following operation and simplify.
\begin{align*} \frac{2x - 1} {x^2 - 6x + 9} - \frac{3x + 4} {x^2 - 9}\end{align*}
Solution
We factor the denominators.
\begin{align*} \frac{2x - 1} {(x - 3)^2} - \frac{3x + 4} {(x + 3)(x - 3)}\end{align*}
The LCD is the product of all the different factors taken to the highest power they have in either denominator. \begin{align*}\text{LCD} = (x - 3)^2 (x + 3)\end{align*}.
The first fraction needs to be multiplied by a factor of \begin{align*}(x + 3)\end{align*} and the second fraction needs to be multiplied by a factor of \begin{align*}(x - 3)\end{align*}.
\begin{align*} \frac{2x - 1} {(x -3)^2} \cdot \frac{(x + 3)} {(x + 3)} - \frac{3x + 4} {(x + 3)(x - 3)} \cdot \frac{(x -3)} {(x - 3)}\end{align*}
Combine the numerators.
\begin{align*} \frac{(2x -1)(x + 3) - (3x + 4)(x - 3)} {(x - 3)^2(x + 3)}\end{align*}
Eliminate all parentheses in the numerator.
\begin{align*} \frac{2x^2 + 5x - 3 - (3x^2 - 5x -12)} {(x - 3)^2 (x + 3)}\end{align*}
Distribute the negative sign in the second parenthesis.
\begin{align*} \frac{2x^2 + 5x - 3 - 3x^2 + 5x + 12} {(x -3)^2(x + 3)}\end{align*}
Combine like terms in the numerator.
\begin{align*} \frac{-x^2 + 10x + 9} {(x - 3)^2(x + 3)} \ \text{Answer}\end{align*}
Solve Real-World Problems Involving Addition and Subtraction of Rational Expressions
Example 9
In an electrical circuit with two resistors placed in parallel, the reciprocal of the total resistance is equal to the sum of the reciprocals of each resistance \begin{align*} \frac{1} {R_{tot}} = \frac{1} {R_1} + \frac{1} {R_2}\end{align*}. Find an expression for the total resistance in a circuit with two resistors wired in parallel.
Solution
The expression for the relationship between total resistance and resistances placed in parallel says that the reciprocal of the total resistance is the sum of the reciprocals of the individual resistances.
Let’s simplify the expression \begin{align*} \frac{1} {R_1} + \frac{1} {R_2}\end{align*}.
The lowest common denominator is
\begin{align*} = R_1 R_2\end{align*}
Multiply the first fraction by \begin{align*} \frac{R_2} {R_2}\end{align*} and the second fraction by \begin{align*} \frac{R_1} {R_1}\end{align*}.
\begin{align*} \frac{R_2} {R_2} \cdot \frac{1} {R_1} + \frac{R_1} {R_1} \cdot \frac{1} {R_2}\end{align*}
Simplify.
\begin{align*} \frac{R_2 + R_1} {R_1R_2}\end{align*}
Therefore, the total resistance is the reciprocal of this expression.
\begin{align*} R_c = \frac{R_1R_2} {R_1 + R_2} \ \text{Answer}\end{align*}
Number Problems
These problems express the relationship between two numbers.
Example 11
The sum of a number and its reciprocal is \begin{align*} \frac{53} {14}\end{align*}. Find the numbers.
Solution
1. Define variables.
Let \begin{align*}x = a\end{align*} number
Then, \begin{align*} \frac{1} {x}\end{align*} is the reciprocal of the number
2. Set up an equation.
The equation that describes the relationship between the numbers is: \begin{align*} x + \frac{1} {x} = \frac{53} {14}\end{align*}
3. Solve the equation.
Find the lowest common denominator.
\begin{align*}\text{LCM} = 14x\end{align*}
Multiply all terms by \begin{align*}14x\end{align*}
\begin{align*} 14x \cdot x + 14x \cdot \frac{1} {x} = 14x \cdot \frac{53} {14}\end{align*}
Cancel common factors in each term.
\begin{align*}14x \cdot x+14 \cancel{x} \cdot \frac{1}{\cancel{x}}=\cancel{14}x \cdot \frac{53}{\cancel{14}}\end{align*}
Simplify.
\begin{align*} 14x^2 + 14 = 53x\end{align*}
Write all terms on one side of the equation.
\begin{align*} 14x^2 - 53x + 14 = 0\end{align*}
Factor.
\begin{align*} (7x - 2)(2x - 7) & = 0 \\ x = \frac{2} {7}\ \text{and}\ x & = \frac{7} {2}\end{align*}
Notice there are two answers for \begin{align*}x\end{align*}, but they are really the same. One answer represents the number and the other answer represents its reciprocal.
4. Check. \begin{align*} \frac{2} {7} + \frac{7} {2} = \frac{4 + 49} {14} = \frac{53} {14}\end{align*}. The answer checks out.
Work Problems
These are problems where two people or two machines work together to complete a job. Work problems often contain rational expressions. Typically we set up such problems by looking at the part of the task completed by each person or machine. The completed task is the sum of the parts of the tasks completed by each individual or each machine.
Part of task completed by first person + Part of task completed by second person = One completed task
To determine the part of the task completed by each person or machine we use the following fact.
Part of the task completed = rate of work × time spent on the task
In general, it is very useful to set up a table where we can list all the known and unknown variables for each person or machine and then combine the parts of the task completed by each person or machine at the end.
Example 12
Mary can paint a house by herself in 12 hours. John can paint a house by himself in 16 hours. How long would it take them to paint the house if they worked together?
Solution:
1. Define variables.
Let \begin{align*}t \ =\end{align*} the time it takes Mary and John to paint the house together.
2. Construct a table.
Since Mary takes 12 hours to paint the house by herself, in one hour she paints \begin{align*} \frac{1} {12}\end{align*} of the house.
Since John takes 16 hours to pain the house by himself, in one hour he paints \begin{align*} \frac{1} {16}\end{align*} of the house.
Mary and John work together for t hours to paint the house together. Using,
Part of the task completed = rate of work ∙ time spent on the task
We can write that Mary completed \begin{align*} \frac{t} {12}\end{align*} of the house and John completed \begin{align*} \frac{t} {16}\end{align*} of the house in this time.
This information is nicely summarized in the table below:
Painter | Rate of work (per hour) | Time worked | Part of Task |
---|---|---|---|
Mary | \begin{align*} \frac{1} {12}\end{align*} | \begin{align*}t\end{align*} | \begin{align*} \frac{t} {12}\end{align*} |
John | \begin{align*} \frac{1} {16}\end{align*} | \begin{align*}t\end{align*} | \begin{align*} \frac{1} {16}\end{align*} |
3. Set up an equation.
Since Mary completed \begin{align*} \frac{t} {12}\end{align*} of the house and John completed \begin{align*} \frac{t} {16}\end{align*} and together they paint the whole house in \begin{align*}t\end{align*} hours, we can write the equation.
\begin{align*} \frac{t} {12} + \frac{t} {16} = 1.\end{align*}
4. Solve the equation.
Find the lowest common denominator.
\begin{align*}\text{LCM} = 48\end{align*}
Multiply all terms in the equation by the LCM.
\begin{align*} 48 \cdot \frac{t} {12} + 48 \cdot \frac{t} {16} = 48 \cdot 1\end{align*}
Cancel common factors in each term.
\begin{align*}\cancel{48}^4.\frac{t}{\cancel{12}}+\cancel{48}.^3.\frac{t}{\cancel{16}}=48.1\end{align*}
Simplify.
\begin{align*} 4t + 3t = 48\end{align*}
\begin{align*} 7t = 48 \Rightarrow t = \frac{48} {7} = 6.86 \ hours \ \text{Answer}\end{align*}
Check
The answer is reasonable. We expect the job to take more than half the time Mary takes but less than half the time John takes since Mary works faster than John.
Example 12
Suzie and Mike take two hours to mow a lawn when they work together. It takes Suzie 3.5 hours to mow the same lawn if she works by herself. How long would it take Mike to mow the same lawn if he worked alone?
Solution
1. Define variables.
Let \begin{align*}t \ =\end{align*} the time it takes Mike to mow the lawn by himself.
2. Construct a table.
Painter | Rate of Work (per Hour) | Time Worked | Part of Task |
---|---|---|---|
Suzie | \begin{align*} \frac{1} {3.5} = \frac{2} {7}\end{align*} | 2 | \begin{align*} \frac{4} {7}\end{align*} |
Mike | \begin{align*} \frac{1} {t}\end{align*} | 2 | \begin{align*} \frac{2} {t}\end{align*} |
3. Set up an equation.
Since Suzie completed \begin{align*} \frac{4} {7}\end{align*} of the lawn and Mike completed \begin{align*} \frac{2} {t}\end{align*} of the lawn and together they mow the lawn in 2 hours, we can write the equation: \begin{align*} \frac{4} {7} + \frac{2} {t} = 1\end{align*}.
4. Solve the equation.
Find the lowest common denominator.
\begin{align*}\text{LCM} = 7t\end{align*}
Multiply all terms in the equation by the LCM.
\begin{align*} 7t \cdot \frac{4} {7} + 7t \cdot \frac{2} {t} = 7t \cdot 1\end{align*}
Cancel common factors in each term.
\begin{align*}\cancel{7}t.\frac{4}{\cancel{7}}+ \cancel{7}t.\frac{2}{\cancel{t}}=7t.1\end{align*}
Simplify.
\begin{align*} 4t + 14 & = 7t \\ 3t & = 14 \Rightarrow t = \frac{14} {3} = 4\frac{2} {3} \ hours \ \text{Answer}\end{align*}
Check.
The answer is reasonable. We expect Mike to work slower than Suzie because working by herself it takes her less than twice the time it takes them to work together.
Review Questions
Perform the indicated operation and simplify. Leave the denominator in factored form.
- \begin{align*} \frac{5} {24} - \frac{7} {24}\end{align*}
- \begin{align*} \frac{10} {21} + \frac{9} {35}\end{align*}
- \begin{align*} \frac{5} {2x + 3} + \frac{3} {2x + 3}\end{align*}
- \begin{align*} \frac{3x - 1} {x + 9} - \frac{4x + 3} {x + 9}\end{align*}
- \begin{align*} \frac{4x + 7} {2x^2} - \frac{3x - 4} {2x^2}\end{align*}
- \begin{align*} \frac{x^2} {x + 5} - \frac{25} {x + 5}\end{align*}
- \begin{align*} \frac{2x} {x - 4} + \frac{x} {4 - x}\end{align*}
- \begin{align*} \frac{10} {3x - 1} - \frac{7} {1 - 3x}\end{align*}
- \begin{align*} \frac{5} {2x + 3} - 3\end{align*}
- \begin{align*} \frac{5x + 1} {x + 4} + 2\end{align*}
- \begin{align*} \frac{1} {x} + \frac{2} {3x}\end{align*}
- \begin{align*} \frac{4} {5x^2} - \frac{2} {7x^3}\end{align*}
- \begin{align*} \frac{4x} {x + 1} - \frac{2} {2(x + 1)}\end{align*}
- \begin{align*} \frac{10} {x + 5} + \frac{2} {x + 2}\end{align*}
- \begin{align*} \frac{2x} {x - 3} - \frac{3x} {x + 4}\end{align*}
- \begin{align*} \frac{4x - 3} {2x + 1} + \frac{x + 2} {x - 9}\end{align*}
- \begin{align*} \frac{x^2} {x + 4} - \frac{3x^2} {4x - 1}\end{align*}
- \begin{align*} \frac{2} {5x + 2} - \frac{x + 1} {x^2}\end{align*}
- \begin{align*} \frac{x + 4} {2x} + \frac{2} {9x}\end{align*}
- \begin{align*} \frac{5x + 3} {x^2 + x} + \frac{2x + 1} {x}\end{align*}
- \begin{align*} \frac{4} {(x + 1) (x - 1)} - \frac{5} {(x + 1)(x + 2)}\end{align*}
- \begin{align*} \frac{2x} {(x + 2)(3x - 4)} + \frac{7x} {(3x - 4)^2}\end{align*}
- \begin{align*} \frac{3x + 5} {x(x - 1)} - \frac{9x - 1} {(x - 1)^2}\end{align*}
- \begin{align*} \frac{1} {(x - 2) (x - 3)} + \frac{4} {(2x + 5) (x - 6)}\end{align*}
- \begin{align*} \frac{3x - 2} {x - 2} + \frac{1} {x^2 - 4x + 4}\end{align*}
- \begin{align*} \frac{-x^2} {x^2 - 7x + 6} + x - 4\end{align*}
- \begin{align*} \frac{2x} {x^2 + 10x + 25} - \frac{3x} {2x^2 + 7x - 15}\end{align*}
- \begin{align*} \frac{1} {x^2 - 9} + \frac{2} {x^2 + 5x + 6}\end{align*}
- \begin{align*} \frac{-x + 4} {2x^2 - x - 15} + \frac{x} {4x^2 + 8x - 5}\end{align*}
- \begin{align*} \frac{4} {9x^2 - 49} - \frac{1} {3x^2 + 5x - 28}\end{align*}
- One number is 5 less than another. The sum of their reciprocals is \begin{align*} \frac{13} {36}\end{align*}. Find the two numbers.
- One number is 8 times more than another. The difference in their reciprocals is \begin{align*} \frac{21} {20}\end{align*}. Find the two numbers.
- A pipe can fill a tank full of oil in 4 hours and another pipe can empty the tank in 8 hours. If the valves to both pipes are open, how long would it take to fill the tank?
- Stefan could wash the cars by himself in 6 hours and Misha could wash the cars by himself in 5 hours. Stefan starts washing the cars by himself, but he needs to go to his football game after 2.5 hours. Misha continues the task. How long did it take Misha to finish washing the cars?
- Amanda and her sister Chyna are shoveling snow to clear their driveway. Amanda can clear the snow by herself in three hours and Chyna can clear the snow by herself in four hours. After Amanda has been working by herself for one hour, Chyna joins her and they finish the job together. How long does it take to clear the snow from the driveway?
- At a soda bottling plant one bottling machine can fulfill the daily quota in 10 hours, and a second machine can fill the daily quota in 14 hours. The two machines start working together but after four hours the slower machine broke and the faster machine had to complete the job by itself. How many hours does the fast machine works by itself?
Review Answers
- \begin{align*} -\frac{1} {12}\end{align*}
- \begin{align*} \frac{11} {15}\end{align*}
- \begin{align*} \frac{8} {2x + 3}\end{align*}
- \begin{align*} \frac{-x-4} {x + 9}\end{align*}
- \begin{align*} \frac{x = 11} {2x^2}\end{align*}
- \begin{align*}x - 5\end{align*}
- \begin{align*} \frac{x} {x - 4}\end{align*}
- \begin{align*} \frac{17} {3x - 1}\end{align*}
- \begin{align*} \frac{-6x -4} {2x + 3}\end{align*}
- \begin{align*} \frac{7x + 9} {x + 4}\end{align*}
- \begin{align*} \frac{5} {3x}\end{align*}
- \begin{align*} \frac{28x - 10} {35x^3}\end{align*}
- \begin{align*} \frac{4x - 1} {x + 1}\end{align*}
- \begin{align*} \frac{12x + 30} {(x + 5) (x + 2)}\end{align*}
- \begin{align*} \frac{-x^2 + 17x} {(x - 3) (x + 4)}\end{align*}
- \begin{align*} \frac{6x^2 - 34x + 19} {(2x + 1)(x - 9)}\end{align*}
- \begin{align*} \frac{x^3 - 13x^2} {(x + 4) (4x - 1)}\end{align*}
- \begin{align*} -\frac{3x^2 + 7x + 2} {x^2(5x + 2)}\end{align*}
- \begin{align*} \frac{9x + 40} {18x}\end{align*}
- \begin{align*} \frac{2x^2 + 8x + 4} {x(x + 1)}\end{align*}
- \begin{align*} \frac{-x + 13} {(x + 1) (x - 1) (x + 2)}\end{align*}
- \begin{align*} \frac{13x^2 + 6x} {(x + 2) (3x - 4)^2}\end{align*}
- \begin{align*} \frac{-6x^2 + 3x - 5} {x(x - 1)^2}\end{align*}
- \begin{align*} \frac{6x^2 - 17x - 6} {(x - 2)(x - 3)(2x + 5)(x - 6)}\end{align*}
- \begin{align*} \frac{3x^2 - 8x + 5} {(x - 2)^2}\end{align*}
- \begin{align*} \frac{-11x^2 = 34x - 24} {(x - 6)(x - 1)}\end{align*}
- \begin{align*} \frac{x^2 - 21x} {(2x - 3) (x + 5)^2}\end{align*}
- \begin{align*} \frac{3x - 4} {(x - 3) (x + 3) (x + 2)}\end{align*}
- \begin{align*} \frac{-x^2 + 4x = 4} {(2x + 5) (x - 3) (2x - 1)}\end{align*}
- \begin{align*} \frac{x + 9} {(3x + 7) (3x - 7) (x + 4)}\end{align*}
- \begin{align*}x = 4, x + 5 = 4\end{align*} or \begin{align*} x = - \frac{45} {13}, x + 5 = \frac{20} {13}\end{align*}
- \begin{align*} x = \frac{5} {6}, 8x = \frac{20} {3}\end{align*}
- 8 hours
- 2 hours and 55 minutes
- \begin{align*} 1 \frac{1} {7} \ hours\end{align*}, or 1 hour 9 minutes
- \begin{align*} 3 \frac{1} {7} \ hours\end{align*}