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Learning Objectives

  • Simplify rational expressions.
  • Find excluded values of rational expressions.
  • Simplify rational models of real-world situations.

Introduction

A rational expression is reduced to lowest terms by factoring the numerator and denominator completely and canceling common factors. For example, the expression

\frac{x\cdot \cancel{z}}{y\cdot \cancel{z}}=\frac{x}{y}

simplifies to simplest form by canceling the common factor z.

Simplify Rational Expressions.

To simplify rational expressions means that the numerator and denominator of the rational expression have no common factors. In order to simplify to lowest terms, we factor the numerator and denominator as much as we can and cancel common factors from the numerator and the denominator of the fraction.

Example 1

Reduce each rational expression to simplest terms.

a)  \frac {4x-2}{2x^2+x-1}

b)  \frac {x^2-2x+1}{8x-8}

c) \frac {x^2-4}{x^2-5x+6}

Solution

a) Factor the numerator and denominator completely.  \frac {2(2x-1)}{(2x-1)(x+1)}

Cancel the common term (2x-1).  \frac {2}{x+1} Answer

b)Factor the numerator and denominator completely.  \frac {(x-1)(x-1)}{8(x-1)}

Cancel the common term (x-1).  \frac {x-1}{8} Answer

c) Factor the numerator and denominator completely.  \frac {(x-2)(x+2)}{(x-2)(x-3)}

Cancel the common term (x-2).  \frac {x+2}{x-3} Answer

Common mistakes in reducing fractions:

When reducing fractions, you are only allowed to cancel common factors from the denominator but NOT common terms. For example, in the expression

 \frac {(x+1).(x-3)}{(x+2).(x-3)}

we can cross out the (x - 3) factor because  \frac {(x-3)}{(x-3)}=1.

We write

\frac{(x+1)\cdot(\cancel{x-3})}{(x+2)\cdot(\cancel{x-3})}=\frac{(x+1)}{(x+2)}

However, don’t make the mistake of canceling out common terms in the numerator and denominator. For instance, in the expression.

 \frac {x^2+1}{x^2-5}

we cannot cross out the x^{2} terms.

\frac{x^2+1}{x^2-5}\neq \frac{\cancel{x^2}+1}{\cancel{x^2}-5}

When we cross out terms that are part of a sum or a difference we are violating the order of operations (PEMDAS). We must remember that the fraction sign means division. When we perform the operation

\frac{(x^2+1)}{(x^2-5)}

we are dividing the numerator by the denominator

(x^2+1)\div(x^2-5)

The order of operations says that we must perform the operations inside the parenthesis before we can perform the division.

Try this with numbers:

& \frac {9+1}{9-5}=\frac {10}{4}=2.5 &&&& \text{But if we cancel incorrectly we obtain the following}\ & \frac{\cancel{9}+1}{\cancel{9}-5}= -\frac{1}{5}= -0.2. \\& \text{CORRECT} &&&& \text{INCORRECT}

Find Excluded Values of Rational Expressions

Whenever a variable expression is present in the denominator of a fraction, we must be aware of the possibility that the denominator could be zero. Since division by zero is undefined, certain values of the variable must be excluded. These values are the vertical asymptotes (i.e. values that cannot exist for x). For example, in the expression \left ( \frac{2}{x-3} \right ), the value of x = 3 must be excluded.

To find the excluded values we simply set the denominator equal to zero and solve the resulting equation.

Example 2

Find the excluded values of the following expressions.

a)  \frac {x}{x+4}

b)  \frac {2x+1}{x^2-x-6}

c)  \frac {4}{x^2-5x}

Solution

a) When we set the denominator equal to zero we obtain.  x+4=0\Rightarrow x=-4 is the excluded value

b) When we set the denominator equal to zero we obtain.  x^2-x-6=0

Solve by factoring. (x-3)(x+2)=0

 \Rightarrow x=3 and x = -2 are the excluded values.

c) When we set the denominator equal to zero we obtain.  x^2-5x=0

Solve by factoring.  x(x-5)=0

 \Rightarrow x=0 and x = 5 are the excluded values.

Removable Zeros

Notice that in the expressions in Example 1, we removed a division by zero when we simplified the problem. For instance,

 \frac {4x-2}{2x^2+x-1}

was rewritten as

 \frac {2(2x-1)}{(2x-1)(x+1)}.

This expression experiences division by zero when x = \frac{1}{2} and x = -1.

However, when we cancel common factors, we simplify the expression to \frac{2}{x+1}. The reduced form allows the value x = \frac{1}{2}. We thus removed a division by zero and the reduced expression has only x = -1 as the excluded value. Technically the original expression and the simplified expression are not the same. When we simplify to simplest form we should specify the removed excluded value. Thus,

 \frac {4x-2}{2x^2+x-1}=\frac {2}{x+1} , x \neq \frac {1}{2}

The expression from Example 1, part b reduces to

 \frac {x^2-2x+1}{8x-8}=\frac {x-1}{8} , x \neq 1

The expression from Example 1, part c reduces to

 \frac {x^2-4}{x^2-5x+6}=\frac {x+2}{x-3}, x \neq 2

Simplify Rational Models of Real-World Situations

Many real world situations involve expressions that contain rational coefficients or expressions where the variable appears in the denominator.

Example 3

The gravitational force between two objects in given by the formula F= G \frac{(m_1m_2)}{(d^2)}. if the gravitation constant is given by G = 6.67\times10^{-11} \ (N \cdot m^2/kg^2). The force of attraction between the Earth and the Moon is F = 2.0 \times 10^{20} \ N (with masses of m_1 = 5.97 \times 10^{24} \ kg for the Earth and m_2 = 7.36 \times 10^{22} \ kg for the Moon).

What is the distance between the Earth and the Moon?

Solution

\text{Let’s start with the Law of Gravitation formula}. && F & =G\frac {m_1m_2}{d^2} \\\text{Now plug in the known values}. && 2.0\times 10^{20} N & =6.67\times 10^{-11} \frac {N \cdot m^2}{kg^2}.\frac {(5.97\times 10^{24}kg)(7.36\times 10^{22}kg)}{d^2} \\\text{Multiply the masses together}. && 2.0\times 10^{20} N & =6.67\times 10^{-11} \frac {N \cdot m^2}{kg^2}.\frac {4.39\times 10^{47}kg^2}{d^2} \\\text{Cancel the}\ kg^2\ \text{units}. && 2 \cdot 0 \times 10^{20} N & = 6.67 \times 10^{-11} \frac{N \cdot m^2} {\cancel{kg^2}} \cdot \frac{4.39 \times 10^{47} \cancel{kg^2}} {d^2} \\\text{Multiply the numbers in the numerator}. && 2.0\times 10^{20} N & = \frac {2.93\times 10^{37}}{d^2}N \cdot m^2 \\\text{Multiply both sides by}\ d^2. && 2.0\times 10^{20} N \cdot d^2 & =\frac {2.93\times 10^{37}}{d^2} \cdot d^2 \cdot N \cdot m^2 \\\text{Cancel common factors}. && 2 \cdot 0 \times 10^{20} N \cdot d^{2} & = \frac{2.93 \times 10^{37}} {\cancel{d^2}} \cdot \cancel{d^2} \cdot N \cdot m^2 \\\text{Simplify}. && 2.0\times 10^{20} N \cdot d^2 & =2.93\times 10^{37}N \cdot m^2 \\\text{Divide both sides by}\ 2.0 \times 10^{20}\ N. && d^2 & =\frac {2.93 \times 10^{37}}{2.0\times 10^{20}}\frac {N \cdot m^2}{N} \\\text{Simplify}. && d^2 & =1.465\times 10^{17} m^2 \\\text{Take the square root of both sides}. && d & =3.84 \times 10^8m\ \text{Answer}

This is indeed the distance between the Earth and the Moon.

Example 4

The area of a circle is given by A=\pi r^{2} and the circumference of a circle is given by C=2 \pi r. Find the ratio of the circumference and area of the circle.

Solution

The ratio of the circumference and area of the circle is: \frac {2\pi r}{\pi r^2}

We cancel common factors from the numerator and denominator. \frac{2 \cancel{\pi} \cancel{R}}{\cancel{\pi} \cancel{R^2}_R}

Simplify. \frac {2}{r} Answer

Example 5

The height of a cylinder is 2 units more than its radius. Find the ratio of the surface area of the cylinder to its volume.

Solution

Define variables.

Let R \ = the radius of the base of the cylinder.

Then, R + 2 \ = the height of the cylinder

To find the surface area of a cylinder, we need to add the areas of the top and bottom circle and the area of the curved surface.

\text{The volume of the cylinder is} &&& V = \pi R^2 (R + 2) \\\text{The ratio of the surface area of the cylinder to its volume is} &&& \frac{2 \pi R^2 + 2 \pi R(R + 2)} {\pi R^2 (R + 2)} \\\text{Eliminate the parentheses in the numerator}. &&& \frac{2 \pi R^2 + 2 \pi R^2 + 4 \pi R} {\pi R^2 (R + 2)} \\\text{Combine like terms in the numerator}. &&& \frac{4 \pi R^2 + 4 \pi R} {\pi R^2 (R + 2)} \\\text{Factor common terms in the numerator}. &&& \frac{4 \pi R (R + 1)} {\pi R^2 (R + 2)} \\ \text{Cancel common terms in the numerator and denominator}. &&& \frac{4 \cancel{\pi} \cancel{R}(R+1)}{\cancel{\pi}\cancel{R^2}_R(R+2)} \\\text{Simplify}. &&& \frac{4(R + 1)} {R(R + 2)}\ \text{Answer}

Review Questions

Reduce each fraction to lowest terms.

  1.  \frac{4} {2x - 8}
  2.  \frac{x^2 + 2x} {x}
  3.  \frac{9x + 3} {12x + 4}
  4.  \frac{6x^2 + 2x} {4x}
  5.  \frac{x - 2} {x^2 - 4x + 4}
  6.  \frac{x^2 - 9} {5x + 15}
  7.  \frac{x^2 + 6x + 8} {x^2 + 4x}
  8.  \frac{2x^2 + 10x} {x^2 + 10x + 25}
  9.  \frac{x^2 + 6x + 5} {x^2 - x - 2}
  10.  \frac{x^2 - 16} {x^2 + 2x - 8}
  11.  \frac{3x^2 + 3x - 18} {2x^2 + 5x - 3}
  12.  \frac{x^3 + x^2 - 20x} {6x^2 + 6x - 120}

Find the excluded values for each rational expression.

  1.  \frac{2} {x}
  2.  \frac{4} {x + 2}
  3.  \frac{2x - 1} {(x - 1)^2}
  4.  \frac{3x + 1} {x^2 - 4}
  5.  \frac{x^2} {x^2 + 9}
  6.  \frac{2x^2 + 3x - 1} {x^2 - 3x - 28}
  7.  \frac{5x^3 - 4} {x^2 + 3x}
  8.  \frac{9} {x^3 + 11x^2 + 30x}
  9.  \frac{4x - 1} {x^2 + 3x - 5}
  10.  \frac{5x + 11} {3x^2 - 2x - 4}
  11.  \frac{x^2 - 1} {2x^2 + x + 3}
  12.  \frac{12} {x^2 + 6x + 1}
  13. In an electrical circuit with resistors placed in parallel, the reciprocal of the total resistance is equal to the sum of the reciprocals of each resistance. \frac{1}{R_{c}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}. If R_1 = 25 \ \Omega and the total resistance is R_c = 10 \ \Omega, what is the resistance R_2?
  14. Suppose that two objects attract each other with a gravitational force of 20 Newtons. If the distance between the two objects is doubled, what is the new force of attraction between the two objects?
  15. Suppose that two objects attract each other with a gravitational force of 36 Newtons. If the mass of both objects was doubled, and if the distance between the objects was doubled, then what would be the new force of attraction between the two objects?
  16. A sphere with radius r has a volume of \frac{4}{3} \pi r^{3} and a surface area of 4 \pi r^{2}. Find the ratio the surface area to the volume of a sphere.
  17. The side of a cube is increased by a factor of two. Find the ratio of the old volume to the new volume.
  18. The radius of a sphere is decreased by four units. Find the ratio of the old volume to the new volume.

Review Answers

  1.  \frac{2} {x - 4}
  2. x + 2, x \neq 0
  3.  \frac{3} {4}, x \neq - \frac{1} {3}
  4.  \frac{3x+1} {2}, x \neq 0
  5.  \frac{1} {x - 2}
  6.  \frac{x - 3} {5}, x \neq - 3
  7.  \frac{x + 2} {x}, x \neq - 4
  8.  \frac{2x} {x + 5}
  9.  \frac{x + 5} {x - 2}, x \neq -1
  10.  \frac{x - 4} {x - 2}, x \neq - 4
  11.  \frac{3x - 6} {2x - 1}, x \neq -3
  12. .
  13. x = 0
  14. x = -2
  15. x = 1
  16. x = 2, x = -2
  17. none
  18. x = -4, x = 7
  19. x = 0, x = -3
  20. x = 0, x = -5, x = -6
  21. x = 1.19, x = -4.19
  22. x = 1.54, x = -0.87
  23. none
  24. x = -0.17, x =-5.83
  25. R_c = 16\frac{2}{3} \ \Omega
  26. 5 Newtons
  27. 36 Newtons
  28.  \frac{3} {R}
  29.  \frac{1} {8}
  30.  \frac{R^3} {(R - 4)^3}

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CK.MAT.ENG.SE.1.Algebra-I.12.4

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