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Elementary Algebra: Solving Linear Equations in One Variable

Module by: John Redden. E-mail the author

Summary: Elementary Algebra: An introduction to solving linear equations in one variable.

Module Overview

Learning how to solve various algebraic equations is one of our main goals in algebra. This module introduces the basic techniques for solving linear equations in one variable. (Prerequisites: Working knowledge of real numbers and their operations.)

Define Linear Equations in One Variable

We begin by establishing some definitions.

Definition 1: Equation
An equation is a statement indicating that two algebraic expressions are equal.
Definition 2: Linear Equation in One Variable
A linear equation in one variable xx size 12{x} {} is an equation that can be written in the form ax+b=0ax+b=0 size 12{ ital "ax"+b=0} {} where aa size 12{a} {} and bb size 12{b} {} are real numbers and a≠0a≠0 size 12{a <> 0} {}.

Following are some examples of linear equations in one variable, all of which will be solved in the course of this module.

x+3=−5 x+3=−5
(1)
x 3 + 1 2 = 2 3 x 3 + 1 2 = 2 3
(2)
5(3x+2)−2=−2(1−7x) 5(3x+2)−2=−2(1−7x)
(3)

Solutions to Linear Equations in One Variable

The variable in the linear equation 2x+3=132x+3=13 size 12{2x+3="13"} {} is xx size 12{x} {}. Values that can replace the variable to make a true statement compose the solution set. Linear equations have at most one solution. After some thought, you might deduce that x=5x=5 size 12{x=5} {} is a solution to 2x+3=132x+3=13 size 12{2x+3="13"} {}. To verify this we substitute the value 5 in for xx size 12{x} {} and see that we get a true statement, 25+3=10+3=1325+3=10+3=13 size 12{2 left (5 right )+3="10"+3="13"} {}.

Example 1

Is x=3x=3 size 12{x=3} {} a solution to −2x−3=−9−2x−3=−9 size 12{ - 2x - 3= - 9} {}?

Yes, because −23−3=−6−3=−9−23−3=−6−3=−9 size 12{ - 2 left (3 right ) - 3= - 6 - 3= - 9} {}

Example 2

Is a=−12a=−12 size 12{a= - { { size 8{1} } over { size 8{2} } } } {} a solution to −10a+5=25−10a+5=25 size 12{ - "10"a+5="25"} {}?

No, because −10−12+5=5+5=10≠25−10−12+5=5+5=10≠25 size 12{ - "10" left ( - { { size 8{1} } over { size 8{2} } } right )+5=5+5="10" <> "25"} {}

When evaluating expressions, it is a good practice to replace all variables with parenthesis first, then substitute in the appropriate values. By making use of parenthesis we could avoid some common errors using the order of operations.

Example 3

Is y=−3 y=−3 size 12{y= - 3} {}  a solution to 2y−5=−y−142y−5=−y−14 size 12{2y - 5= - y - "14"} {} ? 2(    )−5 = −(    )−14 Replace variables with parenthesis. 2( −3 )−5 = −( −3 )−14 Substitute the appropriate value. −6−5 = 3−14 Simplify. −11 = −11 True. 2(    )−5 = −(    )−14 Replace variables with parenthesis. 2( −3 )−5 = −( −3 )−14 Substitute the appropriate value. −6−5 = 3−14 Simplify. −11 = −11 True. Yes because y=−3y=−3 produces a true mathematical statement.

Solving Linear Equations in One Variable

When the coefficients of linear equations are numbers other than nice easy integers, guessing at solutions becomes an unreasonable prospect. We begin to develop an algebraic technique for solving by first looking at the properties of equality.

Properties of Equality

Given algebraic expressions A and B where c is a real number:

property 1: Addition Property of Equality

If A=B then A+c=B+c If A=B then A+c=B+c

property 2: Subtraction Property of Equality

If A=B then A-c=B-c If A=B then A-c=B-c

property 3: Multiplication Property of Equality

If A=B and c≠0 then cA=cB If A=B and c≠0 then cA=cB

property 4: Division Property of Equality

If A=B and c≠0 then  A c = B c If A=B and c≠0 then  A c = B c

Note:

Multiplying or dividing both sides of an equation by zero is carefully avoided. Dividing by zero is undefined and multiplying both sides by zero will result in an equation 0=0.

To summarize, the equality is retained if we add, subtract, multiply and divide both sides of an equation by any nonzero real number. The central technique for solving linear equations involves applying these properties in order to isolate the variable on one side of the equation.

Example 4

Use the properties of equality to solve: x+3=−5 x+3=−5 x+3 = −5 x+3−3 = −5−3 Subtract 3 on both sides. x = −8 Simplify x+3 = −5 x+3−3 = −5−3 Subtract 3 on both sides. x = −8 Simplify The solution set is { −8 } { −8 } .

Example 5

Use the properties of equality to solve: −5x=−35 −5x=−35 −5x = −35 −5x −5 = −35 −5 Divide both sides by -5. x = 7 Simplify −5x = −35 −5x −5 = −35 −5 Divide both sides by -5. x = 7 Simplify The solution set is { 7 } { 7 } .

Two other important properties are:

property 5: Symmetric Property

If A=B then B=A. If A=B then B=A.

property 6: Transitive Property

If A=B and B=C then A=C. If A=B and B=C then A=C.

When solving, we often see 2=x 2=x but that is equivalent to x=2 x=2 .

Isolating the Variable

The idea behind solving in algebra is to isolate the variable. If given a linear equation of the form ax+b=c ax+b=c then we can solve it in two steps. First use the equality property of addition or subtraction to isolate the variable term. Next isolate the variable by using the equality property of multiplication or division. The property choice depends on the given operation, we choose to apply the opposite property of the given operation. For example, if given a term plus three we would first choose to subtract three on both sides of the equation. If given two times the variable then we would choose to divide both sides by two.

Example 6

Solve 2x+3=13 2x+3=13 . 2x+3 = 13 2x+3−3 = 13−3 Subtract 3 on both sides. 2x = 10 2x 2 = 10 2 Divide both sides by 2. x = 5 2x+3 = 13 2x+3−3 = 13−3 Subtract 3 on both sides. 2x = 10 2x 2 = 10 2 Divide both sides by 2. x = 5 The solution set is { 5 } { 5 } .

Example 7

Solve −3x−2=9 −3x−2=9 . −3x−2 = 9 −3x−2+2 = 9+2 Add 2 to both sides. −3x = 11 −3x −3 = 11 −3 Divide both sides by -3. x = − 11 3 −3x−2 = 9 −3x−2+2 = 9+2 Add 2 to both sides. −3x = 11 −3x −3 = 11 −3 Divide both sides by -3. x = − 11 3 The solution set is { − 11 3 } { − 11 3 }

Example 8

Solve x 3 + 1 2 = 2 3 x 3 + 1 2 = 2 3 . x 3 + 1 2 = 2 3 x 3 + 1 2 − 1 2 = 2 3 − 1 2 Subtract  1 2  on both sides. x 3 = 2 3 ( 2 2 )− 1 2 ( 3 3 ) x 3 = 4 6 − 3 6 x 3 = 1 6 3⋅ x 3 = 3⋅ 1 6 Multiply both sides by 3. x = 1 2 x 3 + 1 2 = 2 3 x 3 + 1 2 − 1 2 = 2 3 − 1 2 Subtract  1 2  on both sides. x 3 = 2 3 ( 2 2 )− 1 2 ( 3 3 ) x 3 = 4 6 − 3 6 x 3 = 1 6 3⋅ x 3 = 3⋅ 1 6 Multiply both sides by 3. x = 1 2 The solution set is { 1 2 } { 1 2 } .

In order to retain the equality, we must perform the same operation on both sides of the equation. To isolate the variable we want to remember to choose the opposite operation not the opposite number. For example, if we have -5x = 20 then we choose to divide both sides by -5, not 5.

Figure 1: Video Example 01

Multiplying by the Reciprocal

Recall that when multiplying reciprocals the result is 1, for example, ( 3 5 )( 5 3 )= 15 15 =1 ( 3 5 )( 5 3 )= 15 15 =1 . We can use this fact when the coefficient of the variable is a fraction.

Example 9

Solve − 4 5 x−5=15 − 4 5 x−5=15 . − 4 5 x−5 = 15 − 4 5 x−5+5 = 15+5 Add 5 on both sides. − 4 5 x = 20 − 5 4 ⋅( − 4 5 x ) = − 5 4 ⋅( 20 ) Multiply both sides by − 5 4 .  1x = −5⋅5 Simplify. x = −25 − 4 5 x−5 = 15 − 4 5 x−5+5 = 15+5 Add 5 on both sides. − 4 5 x = 20 − 5 4 ⋅( − 4 5 x ) = − 5 4 ⋅( 20 ) Multiply both sides by − 5 4 .  1x = −5⋅5 Simplify. x = −25 The solution set is { −25 } { −25 } .

Combining Like Terms and Simplifying

Linear equations typically will not be given in standard form and thus will require some additional preliminary steps. These additional steps are to first simplify the expressions on each side of the equal sign using the order of operations.

Opposite Side Like Terms

Given a linear equation in the form ax+b=cx+d ax+b=cx+d we must combine like terms on opposite sides of the equal sign. To do this we will use the addition or subtraction property of equality to combine like terms on either side of the equation.

Example 10

Solve for y: −2y+3=5y+17 −2y+3=5y+17 −2y+3 = 5y+17 −2y+3−5y = 5y+17−5y Subtract 5y on both sides. −7y+3 = 17 −7y+3−3 = 17−3 Subtract 3 on both sides. −7y = 14 −7y −7 = 14 −7 Divide both sides by -7. y = −2 −2y+3 = 5y+17 −2y+3−5y = 5y+17−5y Subtract 5y on both sides. −7y+3 = 17 −7y+3−3 = 17−3 Subtract 3 on both sides. −7y = 14 −7y −7 = 14 −7 Divide both sides by -7. y = −2 The solution set is { −2 } { −2 } .

Same Side Like Terms

We will often encounter linear equations where the expressions on each side of the equal sign could be simplified. If this is the case then it is usually best to simplify each side first. After which we then use the properties of equality to combine opposite side like terms.

Example 11

Solve for a: −4a+2−a=3+5a−2 −4a+2−a=3+5a−2 −4a+2−a = 3+5a−2 Add same side like terms first. −5a+2 = 5a+1 −5a+2−5a = 5a+1−5a Subtract 5a from both sides. −10a+2 = 1 −10a+2−2 = 1−2 Subtract 2 from both sides. −10a = −1 −10a −10 = −1 −10 Divide both sides by -10. a = 1 10 −4a+2−a = 3+5a−2 Add same side like terms first. −5a+2 = 5a+1 −5a+2−5a = 5a+1−5a Subtract 5a from both sides. −10a+2 = 1 −10a+2−2 = 1−2 Subtract 2 from both sides. −10a = −1 −10a −10 = −1 −10 Divide both sides by -10. a = 1 10 The solution set is { 1 10 } { 1 10 }

Simplifying Expressions First

When solving linear equations the goal is to determine what value, if any, will solve the equation. A general guideline is to use the order of operations to simplify the expressions on both sides first.

Example 12

Solve for x: 5(3x+2)−2=−2(1−7x) 5(3x+2)−2=−2(1−7x) 5(3x+2)−2 = −2(1−7x) Distribute. 15x+10−2 = −2+14x Add same side like terms. 15x+8 = −2+14x 15x+8−14x = −2+14x−14x Subtract 14x on both sides. x+8 = −2 x+8−8 = −2−8 Subtract 8 on both sides. x = −10 5(3x+2)−2 = −2(1−7x) Distribute. 15x+10−2 = −2+14x Add same side like terms. 15x+8 = −2+14x 15x+8−14x = −2+14x−14x Subtract 14x on both sides. x+8 = −2 x+8−8 = −2−8 Subtract 8 on both sides. x = −10 The solution set is { −10 } { −10 } .

Figure 2: Video Example 02

Conditional Equations, Identities, and Contradictions

There are three different kinds of equations defined as follows.

Definition 3: Conditional Equation
A conditional equation is true for particular values of the variable.
Definition 4: Identity
An identity is an equation that is true for all possible values of the variable. For example, x = x has a solution set consisting of all real numbers, ℜ ℜ .
Definition 5: Contradiction
A contradiction is an equation that is never true and thus has no solutions. For example, x + 1 = x has no solution. No solution can be expressed as the empty set {    }=∅ {    }=∅ .

So far we have seen only conditional linear equations which had one value in the solution set. If when solving an equation and the end result is an identity, like say 0 = 0, then any value will solve the equation. If when solving an equation the end result is a contradiction, like say 0 = 1, then there is no solution.

Example 13

Solve for x: 4(x+5)+6=2(2x+3) 4(x+5)+6=2(2x+3) 4(x+5)+6 = 2(2x+3) Distribute 4x+20+6 = 4x+6 Add same side like terms. 4x+26 = 4x+6 4x+26−4x = 4x+6−4x Subtract 4x on both sides. 26 = 6 False 4(x+5)+6 = 2(2x+3) Distribute 4x+20+6 = 4x+6 Add same side like terms. 4x+26 = 4x+6 4x+26−4x = 4x+6−4x Subtract 4x on both sides. 26 = 6 False There is no solution, ∅ ∅ .

Example 14

Solve for y: 3(3y+5)+5=10(y+2)−y 3(3y+5)+5=10(y+2)−y 3(3y+5)+5 = 10(y+2)−y Distribute 9y+15+5 = 10y+20−y Add same side like terms. 9y+20 = 9y+20 9y+20−20 = 9y+20−20 Subtract 20 on both sides. 9y = 9y 9y−9y = 9y−9y Subtract 9y on both sides. 0 = 0 True 3(3y+5)+5 = 10(y+2)−y Distribute 9y+15+5 = 10y+20−y Add same side like terms. 9y+20 = 9y+20 9y+20−20 = 9y+20−20 Subtract 20 on both sides. 9y = 9y 9y−9y = 9y−9y Subtract 9y on both sides. 0 = 0 True The equation is an identity, the solution set consists of all real numbers, ℜ ℜ .

Linear Literal Equations

Literal equations, or formulas, usually have more than one variable. Since the letters are placeholders for values, the steps for solving them are the same. Use the properties of equality to isolate the indicated variable.

Example 15

Solve for a: P=2a+b P=2a+b P = 2a+b P−b = 2a+b−b Subtract b on both sides. P−b = 2a P−b 2 = 2a 2 Divide both sides by 2. P−b 2 = a P = 2a+b P−b = 2a+b−b Subtract b on both sides. P−b = 2a P−b 2 = 2a 2 Divide both sides by 2. P−b 2 = a Solution: a= P−b 2 a= P−b 2

Example 16

Solve for x: z= x+y 2 z= x+y 2 z = x+y 2 2⋅z = 2⋅ x+y 2 Multiply both sides by 2. 2z = x+y 2z−y = x+y−y Subtract y on both sides. 2z−y = x z = x+y 2 2⋅z = 2⋅ x+y 2 Multiply both sides by 2. 2z = x+y 2z−y = x+y−y Subtract y on both sides. 2z−y = x Solution x=2z−y x=2z−y

Exercises

Checking Solutions

Exercise 1

Is  x =7 a solution to −3x+5=−16? Is  x =7 a solution to −3x+5=−16?

Solution

Yes

Exercise 2

Is x=2 a solution to −2x−7=28? Is x=2 a solution to −2x−7=28?

Solution

No

Exercise 3

Is x=−3 a solution to  1 3 x−4=−5? Is x=−3 a solution to  1 3 x−4=−5?

Solution

Yes

Exercise 4

Is x=−2 a solution to 3x−5=−2x−15? Is x=−2 a solution to 3x−5=−2x−15?

Solution

Yes

Exercise 5

Is x=− 1 2  a solution to 3(2x+1)=−4x−3? Is x=− 1 2  a solution to 3(2x+1)=−4x−3?

Solution

No

Solving in One Step

Exercise 6

Solve for x:  x−5=−8 Solve for x:  x−5=−8

Solution

x=−3 x=−3

Exercise 7

Solve for y:  −4+y=−9 Solve for y:  −4+y=−9

Solution

y=−5 y=−5

Exercise 8

Solve for x:  x− 1 2 = 1 3 Solve for x:  x− 1 2 = 1 3

Solution

x= 5 6 x= 5 6

Exercise 9

Solve for x:  x+2 1 2 =3 1 3 Solve for x:  x+2 1 2 =3 1 3

Solution

x= 5 6 x= 5 6

Exercise 10

Solve for x:  4x=−44 Solve for x:  4x=−44

Solution

x=−11 x=−11

Exercise 11

Solve for a: −3a=−30 Solve for a: −3a=−30

Solution

a=10 a=10

Exercise 12

Solve for y:  27=9y Solve for y:  27=9y

Solution

y=3 y=3

Exercise 13

Solve for x:   x 3 =− 1 2 Solve for x:   x 3 =− 1 2

Solution

x=− 3 2 x=− 3 2

Exercise 14

Solve for t:  − t 12 = 1 4 Solve for t:  − t 12 = 1 4

Solution

t=−3 t=−3

Exercise 15

Solve for x:   7 3 x=− 1 2 Solve for x:   7 3 x=− 1 2

Solution

x=− 3 14 x=− 3 14

Solve in Two Steps

Exercise 16

Solve for a:  3a−7=23 Solve for a:  3a−7=23

Solution

a=10 a=10

Exercise 17

Solve for y:  −3y+2=−13 Solve for y:  −3y+2=−13

Solution

y=5 y=5

Exercise 18

Solve for x:  −5x+8=8 Solve for x:  −5x+8=8

Solution

x=0 x=0

Exercise 19

Solve for x:   1 2 x+ 1 3 = 2 5 Solve for x:   1 2 x+ 1 3 = 2 5

Solution

x= 2 15 x= 2 15

Exercise 20

Solve for y:  3−2y=−11 Solve for y:  3−2y=−11

Solution

y=7 y=7

Exercise 21

Solve for x:  −10=2x−5 Solve for x:  −10=2x−5

Solution

x=− 5 2 x=− 5 2

Exercise 22

Solve for a:  4a− 2 3 =− 1 6 Solve for a:  4a− 2 3 =− 1 6

Solution

a= 1 8 a= 1 8

Exercise 23

Solve for x:   3 5 x− 1 2 = 1 10 Solve for x:   3 5 x− 1 2 = 1 10

Solution

x=1 x=1

Exercise 24

Solve for y:  − 4 5 y+ 1 3 = 1 15 Solve for y:  − 4 5 y+ 1 3 = 1 15

Solution

y= 1 3 y= 1 3

Exercise 25

Solve for x:  −x−5=−2 Solve for x:  −x−5=−2

Solution

x=−3 x=−3

Solve in Multiple Steps

Exercise 26

Solve for x:  3x−5=2x−17 Solve for x:  3x−5=2x−17

Solution

x=−12 x=−12

Exercise 27

Solve for y:  −2y−7=3y+13 Solve for y:  −2y−7=3y+13

Solution

y=−4 y=−4

Exercise 28

Solve for a:   1 2 a− 2 3 =a+ 1 5 Solve for a:   1 2 a− 2 3 =a+ 1 5

Solution

a=− 26 15 a=− 26 15

Exercise 29

Solve for x:  −2+4x+9=7x+8−2x Solve for x:  −2+4x+9=7x+8−2x

Solution

x=−1 x=−1

Exercise 30

Solve for a:  3a+5−x=2a+7 Solve for a:  3a+5−x=2a+7

Solution

No Solution, ∅ ∅

Exercise 31

Solve for b:  −7b+3=2−5b+1−2b Solve for b:  −7b+3=2−5b+1−2b

Solution

All Reals, ℜ ℜ

Exercise 32

Solve for y:  −5(2y−3)+2=12 Solve for y:  −5(2y−3)+2=12

Solution

y= 1 2 y= 1 2

Exercise 33

Solve for x:  3−2(x+4)=−3(4x−5) Solve for x:  3−2(x+4)=−3(4x−5)

Solution

x=2 x=2

Exercise 34

Solve for a:  −3(2a−3)+2=3(a+7) Solve for a:  −3(2a−3)+2=3(a+7)

Solution

a=− 10 9 a=− 10 9

Exercise 35

Solve for x:  10(3x+5)−5(4x+2)=2(5x+20) Solve for x:  10(3x+5)−5(4x+2)=2(5x+20)

Solution

All Reals, ℜ ℜ

Literal Equations

Exercise 36

Solve for w:  P=2l+2w Solve for w:  P=2l+2w

Solution

w= P−2l 2 w= P−2l 2

Exercise 37

Solve for b:  P=a+b+c Solve for b:  P=a+b+c

Solution

b=P−a−c b=P−a−c

Exercise 38

Solve for C:  F= 9 5 C+32 Solve for C:  F= 9 5 C+32

Solution

C= 5F−160 9 C= 5F−160 9

Exercise 39

Solve for r:  C=2πr Solve for r:  C=2πr

Solution

r= C 2π r= C 2π

Exercise 40

Solve for y:  z= x−y 5 Solve for y:  z= x−y 5

Solution

y=−5z+x y=−5z+x

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