2.1
Solutions of Linear Equations
in One Variable
2.1
OBJECTIVES
1. Identify a linear equation
2. Combine like terms to solve an equation
We begin this chapter by considering one of the most important tools of mathematics—the
equation. The ability to recognize and solve various types of equations and inequalities is
probably the most useful algebraic skill you will learn, and we will continue to build on the
methods developed here throughout the remainder of the text. To start, let’s describe what
we mean by an equation.
An equation is a mathematical statement in which two expressions represent the same
quantity. An equation has three parts:
2x 3
Left side
Equals
sign
5x 6
Right side
The equation simply says that the expression on the left and the expression on the right
represent the same quantity.
In this chapter, we will work with a particular kind of equation.
Definitions:
NOTE Linear equations are
also called first-degree
equations because the highest
power of the variable is the
first power, or first degree.
Linear Equation in One Variable
A linear equation in one variable is any equation that can be written in the
form
ax b 0
in which a and b are any real numbers
a0
NOTE We also say the solution
satisfies the equation.
The solution of an equation in one variable is any number that will make the equation a
true statement. The solution set for such an equation is simply the set consisting of all
solutions.
Example 1
Verify that 3 is a solution for the equation.
5x 6 2x 3
Replacing x with 3 gives
NOTE We use the question
mark over the symbol of
equality when we are checking
to see if the statement is true.
50
5(3) 6 2(3) 3
15 6 6 3
9 9
A true statement.
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Checking Solutions
SOLUTIONS TO LINEAR EQUATIONS IN ONE VARIABLE
SECTION 2.1
51
CHECK YOURSELF 1
Verify that 7 is a solution for this equation.
5x 15 2x 6
Solving linear equations in one variable will require using equivalent equations.
Definitions: Equivalent Equations
Two equations are equivalent if they have the same solution set.
For example, the three equations
5x 5 2x 4
NOTE You can easily verify this
by replacing x with 3 in each
equation.
3x 9
are all equivalent because they all have the same solution set, 3. Note that replacing x
with 3 will give a true statement in the third equation, but it is not as clear that 3 is a solution for the other two equations. This leads us to an equation-solving strategy of isolating
the variable, as is the case in the equation x 3.
To form equivalent equations that will lead to the solution of a linear equation, we need
two properties of equations: addition and multiplication. The addition property is defined
here.
Rules and Properties:
NOTE Adding the same
quantity to both sides of an
equation gives an equivalent
equation, which holds true
whether c is positive or
negative.
x 3
If
then
Addition Property of Equations
ab
acbc
Recall that subtraction can always be defined in terms of addition, so
a c a (c)
The addition property also allows us to subtract the same quantity from both sides of an
equation.
The multiplication property is defined here.
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Rules and Properties:
NOTE Multiplying both sides
of an equation by the same
nonzero quantity gives an
equivalent equation.
If
then
Multiplication Property of Equations
ab
ac bc
when c 0
It is also the case that division can be defined in terms of multiplication, so
a
1
a
c
c
c0
The multiplication property allows us to divide both sides of an equation by the same nonzero quantity.
52
CHAPTER 2
LINEAR EQUATIONS AND INEQUALITIES
Example 2
Applying the Properties of Equations
Solve for x.
3x 5 4
We start by using the addition property to add 5 to both sides of the equation.
NOTE Why did we add 5? We
NOTE We choose
3x 9
(2)
Now we want to get the x term alone on the left with a coefficient of 1 (we call this
isolating the x). To do this, we use the multiplication property and multiply both sides
1
by .
3
1
1
(3x) (9)
3
3
1
because
3
1
is the reciprocal of 3 and
3
1
31
3
3x 5 5 4 5
3 3(x) 3
1
So, x 3
(3)
In set notation, we write 3, which represents the set of all solutions. No other value
of x makes the original equation true. We could also use set-builder notation. We write
xx 3, which is read, “Every x such that x equals three.” We will use both notations
throughout the text.
Because any application of the addition or multiplication properties leads to an equivalent equation, equations (1), (2), and (3) in Example 2 all have the same solution, 3.
To check this result, we can replace x with 3 in the original equation:
3(3) 5 4
954
44
A true statement.
You may prefer a slightly different approach in the last step of the solution above. From
equation (2),
3x 9
The multiplication property can be used to divide both sides of the equation by 3. Then,
3x
9
3
3
x3
Of course, the result is the same.
CHECK YOURSELF 2
Solve for x.
4x 7 17
The steps involved in using the addition and multiplication properties to solve an equation are the same if more terms are involved in an equation.
© 2001 McGraw-Hill Companies
added 5 because it is the
opposite of 5, and the
resulting equation will have the
variable term on the left and
the constant term on the right.
(1)
SOLUTIONS TO LINEAR EQUATIONS IN ONE VARIABLE
SECTION 2.1
53
Example 3
Applying the Properties of Equations
Solve for x.
5x 11 2x 7
Our objective is to use the properties of equations to isolate x on one side of an equivalent equation. We begin by adding 11 to both sides.
NOTE Adding 11 puts the
constant term on the right.
5x 11 11 2x 7 11
5x 2x 4
We continue by adding 2x to (or subtracting 2x from) both sides. We can do this because
of our addition property of equations.
NOTE If you prefer, write
5x 2x 2x 2x 4
Again:
3x 4
5x (2x) 2x (2x) 4
We have now isolated the variable term on the left side of the equation.
3x 4
1
In the final step, we multiply both sides by .
3
NOTE This is the same as
dividing both sides by 3. So
3x
4
3
3
x
4
3
1
1
(3x) (4)
3
3
x
4
3
In set notation, we write
4
. We leave it to you to check this result by substitution.
3
CHECK YOURSELF 3
Solve for x.
7x 12 2x 9
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Both sides of an equation should be simplified as much as possible before the addition
and multiplication properties are applied. If like terms are involved on one side (or on both
sides) of an equation, they should be combined before an attempt is made to isolate the
variable. Example 4 illustrates this approach.
Example 4
Applying the Properties of Equations with Like Terms
Solve for x.
NOTE Notice the like terms on
the left and right sides of the
equation.
8x 2 3x 8 3x 2
Here we combine the like terms 8x and 3x on the left and the like terms 8 and 2 on the
right as our first step. We then have
5x 2 3x 10
54
CHAPTER 2
LINEAR EQUATIONS AND INEQUALITIES
We can now solve as before.
5x 2 2 3x 10 2
Subtract 2 from both sides.
5x 3x 8
Then,
5x 3x 3x 3x 8
Subtract 3x from both sides.
2x 8
2x 8
2
2
Divide both sides by 2.
x4
4
or
The solution is 4, which can be checked by returning to the original equation.
CHECK YOURSELF 4
Solve for x.
7x 3 5x 10 4x 3
If parentheses are involved on one or both sides of an equation, the parentheses should
be removed by applying the distributive property as the first step. Like terms should then be
combined before an attempt is made to isolate the variable. Consider Example 5.
Example 5
Applying the Properties of Equations with Parentheses
Solve for x.
x 3(3x 1) 4(x 2) 4
First, apply the distributive property to remove the parentheses on the left and right sides.
x 9x 3 4x 8 4
Combine like terms on each side of the equation.
10x 3 4x 12
NOTE Recall that to isolate the
Now, isolate variable x on the left side.
x, we must get x alone on the
left side with a coefficient of 1.
10x 3 3 4x 12 3
Add 3 to both sides.
10x 4x 4x 4x 15
Subtract 4x from both sides.
6x 15
6x 15
6
6
x
5
2
Divide both sides by 6.
or
5
2
5
The solution is . Again, this can be checked by returning to the original equation.
2
© 2001 McGraw-Hill Companies
10x 4x 15
SOLUTIONS TO LINEAR EQUATIONS IN ONE VARIABLE
SECTION 2.1
55
CHECK YOURSELF 5
Solve for x.
x 5(x 2) 3(3x 2) 18
NOTE The LCM of a set of
denominators is also called the
lowest common denominator
(LCD).
To solve an equation involving fractions, the first step is to multiply both sides of the
equation by the least common multiple (LCM) of all denominators in the equation. This
will clear the equation of fractions, and we can proceed as before.
Example 6
Applying the Properties of Equations with Fractions
Solve for x.
x
2
5
2
3
6
First, multiply each side by 6, the least common multiple of 2, 3, and 6.
6
6
3
6
2 3 66
x
2
5
2 63 66
Apply the distributive property.
2 63 66
Simplify.
x
x
2
2
1
2
5
1
5
1
1
Next, isolate the variable x on the left side.
NOTE The equation is now
cleared of fractions.
3x 4 5
3x 9
x3
or
3
The solution, 3, can be checked as before by returning to the original equation.
© 2001 McGraw-Hill Companies
CHECK YOURSELF 6
Solve for x.
x
4
19
4
5
20
Be sure that the distributive property is applied properly so that every term of the equation is multiplied by the LCM.
CHAPTER 2
LINEAR EQUATIONS AND INEQUALITIES
Example 7
Applying the Properties of Equations with Fractions
Solve for x.
2x 1
x
1
5
2
First, multiply each side by 10, the LCM of 5 and 2.
10
2
2x 1
x
1 10
5
2
5
2x 1
x
10
10(1) 10
5
2
1
Apply the distributive property on the left. Reduce.
1
2(2x 1) 10 5x
4x 2 10 5x
4x 8 5x
8x
Next, isolate x. Here we isolate x on the right side.
or
8
The solution for the original equation is 8.
CHECK YOURSELF 7
Solve for x.
3x 1
x1
2
4
3
Thus far, we have considered only equations of the form ax b 0, in which a 0.
If we allow the possibility that a 0, two additional equation forms arise. The resulting
equations can be classified into three types depending on the nature of their solutions.
1. An equation that is true for only particular values of the variable is called a
conditional equation. Here the equation can be written in the form
ax b 0
in which a 0. This case was illustrated in all our previous examples and exercises.
2. An equation that is true for all possible values of the variable is called an identity. In
this case, both a and b are 0, so we get the equation 0 0. This will be the case if
both sides of the equation reduce to the same expression (a true statement).
3. An equation that is never true, no matter what the value of the variable, is called a
contradiction. For example, if a is 0 but b is nonzero, we end up with something like
4 0. This will be the case if both sides of the equation reduce to a false statement.
Example 8 illustrates the second and third cases.
© 2001 McGraw-Hill Companies
56
SOLUTIONS TO LINEAR EQUATIONS IN ONE VARIABLE
SECTION 2.1
57
Example 8
Identities and Contradictions
(a) Solve for x.
2(x 3) 2x 6
Apply the distributive property to remove the parentheses.
2x 6 2x 6
NOTE See the definition of an
identity, above. By adding 6 to
both sides of this equation, we
have 0 0.
6 6
A true statement.
Because the two sides reduce to the true statement 6 6, the original equation is an
identity, and the solution set is the set of all real numbers.
(b) Solve for x.
3(x 1) 2x x 4
Again, apply the distributive property.
3x 3 2x x 4
x3x4
NOTE See the earlier definition
of a contradiction. Subtracting
3 from both sides, we have
0 1.
34
A false statement.
Because the two sides reduce to the false statement 3 4, the original equation is a contradiction. There are no values of the variable that can satisfy the equation. The solution set
has nothing in it. We call this the empty set and write or .
CHECK YOURSELF 8
Determine whether each of the following equations is a conditional equation, an
identity, or a contradiction.
(a) 2(x 1) 3 x
NOTE An algorithm is a stepby-step process for problem
solving.
(b) 2(x 1) 3 2x 1
(c) 2(x 1) 3 2x 1
An organized step-by-step procedure is the key to an effective equation-solving strategy.
The following algorithm summarizes our work in this section and gives you guidance in
approaching the problems that follow.
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Step by Step: Solving Linear Equations in One Variable
NOTE If the equation derived
in step 5 is always true, the
original equation was an
identity. If the equation is
always false, the original
equation was a contradiction.
Step 1 Remove any grouping symbols by applying the distributive property.
Step 2 Multiply both sides of the equation by the LCM of any denominators,
to clear the equation of fractions.
Step 3 Combine any like terms that appear on either side of the equation.
Step 4 Apply the addition property of equations to write an equivalent
equation with the variable term on one side of the equation and the
constant term on the other side.
Step 5 Apply the multiplication property of equations to write an equivalent
equation with the variable isolated on one side of the equation.
Step 6 Check the solution in the original equation.
CHAPTER 2
LINEAR EQUATIONS AND INEQUALITIES
When you are solving an equation for which a calculator is recommended, it is often
easiest to do all calculations as the last step.
Example 9
Evaluating Expressions Using a Calculator
Solve the following equation for x.
185(x 3.25) 1650
159.44
500
Following the steps of the algorithm, we get
185x 185 3.25 1650
159.44
500
Remove parentheses.
185x 185 3.25 1650 159.44 500
Multiply by the LCM.
185x 159.44 500 185 3.25 1650
Apply the addition property.
x
159.44 500 185 3.25 1650
185
Isolate the variable.
Now, remembering to insert parentheses around the numerator, we use a calculator to
simplify the expression on the right.
x 425.25
or
425.25
CHECK YOURSELF 9
Solve the following equation for x.
2200(x 17.5) 1550
2326
75
CHECK YOURSELF ANSWERS
3
2
1. 5(7) 15 2(7) 6
2. 6
3.
4. 8
5. 5
3
35 15 14 6
20 20
A true statement.
6. 7
7. 5
8. (a) Conditional; (b) contradiction; (c) identity
9. 62.5
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58
Name
2.1
Exercises
Section
Date
In exercises 1 to 14, solve each equation, and check your results. Express each answer in
set notation.
ANSWERS
1. 5x 8 17
3. 8 7x 41
2. 4x 9 11
4. 7 4x 21
5. 7x 5 6x 6
6. 9x 4 8x 3
7. 8x 4 3x 24
8. 5x 2 2x 5
9. 7x 4 2x 26
10. 11x 3 4x 31
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
11. 4x 3 1 2x
12. 8x 5 19 4x
13. 2x 8 7x 37
14. 3x 5 9x 22
15.
16.
17.
In exercises 15 to 32, simplify and then solve each equation. Express your answer in set
notation.
18.
15. 5x 2 x 9 3x 10
19.
16. 5x 5 x 7 x 2
20.
17. 7x 3 4x 5 5x 13
18. 8x 3 6x 7 5x 17
19. 5x 3(x 6)
20. 2(x 15) 7x
21.
22.
© 2001 McGraw-Hill Companies
23.
21. 5(8 x) 3x
22. 7x 7(6 x)
23. 2(2x 1) 3(x 1)
24. 3(3x 1) 4(3x 1)
24.
25.
26.
25. 8x 3(2x 4) 17
26. 7x 4(3x 4) 9
27.
28.
27. 7(3x 4) 8(2x 5) 13
28. 4(2x 1) 3(3x 1) 9
59
ANSWERS
29. 9 4(3x 1) 3(6 3x) 9
29.
30.
30. 13 4(5x 1) 3(7 5x) 15
31.
31. 5 2[x 2(x 1)] 55 4[x 3(x 2)]
32.
33.
32. 7 5[x 3(x 2)] 25 2[x 2(x 3)]
34.
35.
In exercises 33 to 46, clear fractions and then solve each equation. Express your answer in
set notation.
36.
37.
33.
2x
5
3
3
3
34.
3x
1
4
4
4
35.
x
x
11
6
5
36.
x
x
1
6
8
37.
2x
x
5
3
4
2
38.
5x
2x
5
6
3
6
39.
x
x7
1
5
3
3
40.
x
3
x1
6
4
4
41.
5x 3
x
2
4
3
42.
6x 1
2x
3
5
3
43.
2x 3
2x 1
8
5
3
15
44.
3x
3x 1
11
5
2
10
38.
39.
40.
41.
42.
43.
44.
45.
46.
47.
48.
45. 0.5x 6 0.2x
50.
46. 0.7x 7 0.3x 5
In exercises 47 to 56, classify each equation as a conditional equation, an identity, or a
contradiction.
60
47. 3(x 1) 2x 3
48. 2(x 3) 2x 6
49. 3(x 1) 3x 3
50. 2(x 3) x 5
© 2001 McGraw-Hill Companies
49.
ANSWERS
51. 3(x 1) 3x 3
52. 2(x 3) 3x 5
51.
52.
53.
53. 3x (x 3) 2(x 1) 2
54. 5x (x 4) 4(x 2) 4
x
x
x
55.
2
3
6
3x
2x
x
56.
4
3
6
54.
55.
56.
57.
In exercises 57 to 60, use a calculator to solve the given equations for x. Round your
answer to two decimal places and use set notation.
57.
59.
63(x 2.45) 325
3
200
58.
23x 14(x 9.75)
15.75
23.46
60.
58.
47(x 3.15) 263
11
315
59.
15.25x 12(2x 11.23)
8.4
15.6
61.
61. What is the common characteristic of equivalent equations?
60.
62.
62. What is meant by a solution to a linear equation?
63. Define (a) identity and (b) contradiction.
63.
64. Why does the multiplication property of equation not include multiplying both sides
of the equation by 0?
64.
65.
Label exercises 65 to 70 true or false.
66.
© 2001 McGraw-Hill Companies
67.
65. Adding the same value to both sides of an equation creates an equivalent equation.
68.
66. Multiplying both sides of an equation by 0 creates an equivalent equation.
67. To clear an equation of fractions, we multiply both sides by the GCF of the
denominator.
68. The multiplication property of equations allows us to divide both sides by the same
nonzero quantity.
61
ANSWERS
69.
69. Some equations have more than one solution.
70.
70. No matter what value is substituted for x, the expressions on either side of the equals
sign have the same value.
Answers
1. 5
3. 7
15. 7
17.
27. 5
29.
5. 11
5
2
4
3
7. 4
19. 9
31. 13
3
2
49. Contradiction
51. Identity
57. 6.82
59. 6.30
61.
67. False
69. True
41. 3
43.
21. 5
33. 7
45. 20
11.
23. 5
2
3
25.
35. 30
13. 9
5
2
37. 6
47. Conditional
53. Contradiction
63.
55. Identity
65. True
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39. 15
9. 6
62