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Algebraic Expressions and Equations: Solving Equations of the Form x+a=b and x-a=b

Module by: Wade Ellis, Denny Burzynski. E-mail the authorsEdited By: Math Editors

Summary: This module is from Fundamentals of Mathematics by Denny Burzynski and Wade Ellis, Jr. This module discusses how to solve equations of the form x + a = b x+a=b and x - a = b x-a=b. By the end of the module students should understand the meaning and function of an equation, understand what is meant by the solution to an equation and be able to solve equations of the form x+a=bx+a=b size 12{x+a=b} {} and x−a=bx−a=b size 12{x - a=b} {}.

Section Overview

  • Equations
  • Solutions and Equivalent Equations
  • Solving Equations

Equations

Equation

An equation is a statement that two algebraic expressions are equal.

The following are examples of equations:

x+6 ︸ This expression = = 10 ︸ This expression x+6 ︸ This expression = = 10 ︸ This expression x-4 ︸ This expression = = - 11 ︸ This expression x-4 ︸ This expression = = - 11 ︸ This expression 3y-5 ︸ This expression = = - 2+2y ︸ This expression 3y-5 ︸ This expression = = - 2+2y ︸ This expression

Notice that x+6x+6 size 12{x+6} {}, x−4x−4 size 12{x - 4} {}, and 3y−53y−5 size 12{3y - 5} {} are not equations. They are expressions. They are not equations because there is no statement that each of these expressions is equal to another expression.

Solutions and Equivalent Equations

Conditional Equations

The truth of some equations is conditional upon the value chosen for the variable. Such equations are called conditional equations. There are two additional types of equations. They are examined in courses in algebra, so we will not consider them now.

Solutions and Solving an Equation

The set of values that, when substituted for the variables, make the equation true, are called the solutions of the equation.
An equation has been solved when all its solutions have been found.

Sample Set A

Example 1

Verify that 3 is a solution to x+7=10x+7=10 size 12{x+7 = "10"} {}.

When x=3x=3 size 12{x=3} {},

x+7 = 10 becomes 3+7 = 10 10 = 10 which is a  true  statement, verifying that 3   is a solution to   x+7=10 x+7 = 10 becomes 3+7 = 10 10 = 10 which is a  true  statement, verifying that 3   is a solution to   x+7=10

Example 2

Verify that - 6 - 6 is a solution to 5 y+8=−225 y+8=−22 size 12{5y+8= - "22"} {}

When y=-6y=-6 size 12{y=-6} {},

5y+8 = - 22 becomes 5(- 6)+8 = - 22 - 30+8 = - 22 - 22 = - 22 which is a  true  statement, verifying that - 6  is a solution to 5y+8=- 22 5y+8 = - 22 becomes 5(- 6)+8 = - 22 - 30+8 = - 22 - 22 = - 22 which is a  true  statement, verifying that - 6  is a solution to 5y+8=- 22

Example 3

Verify that 5 is not a solution to a−1=2 a+3a−1=2 a+3 size 12{a - 1=2a+3} {}.

When a=5a=5 size 12{a=5} {},

a-1 = 2a+3 becomes 5-1 = 2⋅5+3 5-1 = 10+3 4 = 13 a  false  statement, verifying that   5   is not a solution to  a-1=2a+3 a-1 = 2a+3 becomes 5-1 = 2⋅5+3 5-1 = 10+3 4 = 13 a  false  statement, verifying that   5   is not a solution to  a-1=2a+3

Example 4

Verify that -2 is a solution to 3m−2=−4m−163m−2=−4m−16 size 12{3m - 2= - 4m - "16"} {}.

When m=-2m=-2 size 12{x=3} {},

3m-2 = - 4m-16 becomes 3(- 2)-2 = - 4(- 2)-16 - 6-2 = 8-16 - 8 = - 8 which is a   true  statement, verifying that - 2   is a solution to  3m-2=- 4m-16 3m-2 = - 4m-16 becomes 3(- 2)-2 = - 4(- 2)-16 - 6-2 = 8-16 - 8 = - 8 which is a   true  statement, verifying that - 2   is a solution to  3m-2=- 4m-16

Practice Set A

Exercise 1

Verify that 5 is a solution to m+6=11m+6=11 size 12{m+6="11"} {}.

Solution

Substitute 5 into m+6=11m+6=11 size 12{m+6="11"} {}. Does 5 plus 6 equal 11? Yes.Thus, 5 is a solution.

Exercise 2

Verify that - 5 - 5 is a solution to 2 m−4=−142 m−4=−14 size 12{2m - 4= - "14"} {}.

Solution

Substitute -5 into 2m−4=−142m−4=−14 size 12{2m - 4= - "14"} {}. does 2 time negative 5 minus 4 equal negative 14? Yes. Thus, -5 is a solution.

Exercise 3

Verify that 0 is a solution to 5 x+1=15 x+1=1 size 12{5x+1=1} {}.

Solution

Substitute 0 into 5 x+1=15 x+1=1 size 12{5x+1=1} {}. Does 5 times zero plus one equal 1? Yes. Thus, 0 is a solution.

Exercise 4

Verify that 3 is not a solution to −3 y+1=4 y+5−3 y+1=4 y+5 size 12{ - 3y+1=4y+5} {}.

Solution

Substitute 3 into −3 y+1=4 y+5−3 y+1=4 y+5 size 12{ - 3y+1=4y+5} {}. Does negative 3 times 3 plus 1 equal 4 times 3 plus 5? No. Thus, 3 is not a solution.

Exercise 5

Verify that -1 is a solution to 6 m−5+2m=7 m−66 m−5+2m=7 m−6 size 12{6m - 5+2m=7m - 6} {}.

Solution

Substitute -1 into 6m−5+2m=7 m−66m−5+2m=7 m−6 size 12{6m - 5+2m=7m - 6} {}. Does 6 times negative 1 minus 5 plus 2 times negative 1 equal 7 times negative 1 minus 6? Yes. Thus, -1 is a solution.

Equivalent Equations

Some equations have precisely the same collection of solutions. Such equations are called equivalent equations. For example, x-5=- 1x-5=- 1 size 12{"x - 5 "=" -1"} {}, x+7=11x+7=11 size 12{"x "+" 7 "=" 11"} {}, and x=4x=4 size 12{x=4} {} are all equivalent equations since the only solution to each is x=4x=4 size 12{x=4} {}. (Can you verify this?)

Solving Equations

We know that the equal sign of an equation indicates that the number represented by the expression on the left side is the same as the number represented by the expression on the right side.

Table 1
This number is the same as this number
↓ ↓ ↓
x x size 12{x} {} = 4
x + 7 x + 7 size 12{x+7} {} = 11
x − 5 x − 5 size 12{x - 5} {} = -1

Addition/Subtraction Property of Equality

From this, we can suggest the addition/subtraction property of equality.
 Given any equation,

  1. We can obtain an equivalent equation by adding the same number to both sides of the equation.
  2. We can obtain an equivalent equation by subtracting the same number from both sides of the equation.

The Idea Behind Equation Solving

The idea behind equation solving is to isolate the variable on one side of the equation. Signs of operation (+, -, ⋅,÷) are used to associate two numbers. For example, in the expression 5+35+3 size 12{5+3} {}, the numbers 5 and 3 are associated by addition. An association can be undone by performing the opposite operation. The addition/subtraction property of equality can be used to undo an association that is made by addition or subtraction.

Subtraction is used to undo an addition.

Addition is used to undo a subtraction.

The procedure is illustrated in the problems of (Reference).

Sample Set B

Use the addition/subtraction property of equality to solve each equation.

Example 5

x+4=6x+4=6 size 12{x+4=6} {}.
4 is associated with xx size 12{x} {} by addition. Undo the association by subtracting 4 from both sides.

x + 4 − 4 = 6 − 4 x + 0 = 2 x = 2 x + 4 − 4 = 6 − 4 x + 0 = 2 x = 2 alignl { stack { size 12{x+4 - 4=6 - 4} {} # size 12{x+0=2} {} # size 12{x=2} {} } } {}

Check: When x=2x=2 size 12{x=2} {}, x+4x+4 size 12{x+4} {} becomes

Does 2 + 4 = 6? Yes.

The solution to x+4=6x+4=6 size 12{x+4=6} {} is x=2x=2 size 12{x=2} {}.

Example 6

m−8=5m−8=5 size 12{m - 8=5} {}. 8 is associated with mm size 12{m} {} by subtraction. Undo the association by adding 8 to both sides.

m − 8 + 8 = 5 + 8 m + 0 = 13 m = 13 m − 8 + 8 = 5 + 8 m + 0 = 13 m = 13 alignl { stack { size 12{m - 8+8=5+8} {} # size 12{m+0="13"} {} # size 12{m="13"} {} } } {}

Check: When m=13m=13 size 12{m="13"} {},

becomes
m - 8 = 5. After substituting, does 13 - 8 = 5? Yes.
a true statement.

The solution to m−8=5m−8=5 size 12{m - 8=5} {} is m=13m=13 size 12{m="13"} {}.

Example 7

−3−5=y−2+8−3−5=y−2+8 size 12{ - 3 - 5=y - 2+8} {}. Before we use the addition/subtraction property, we should simplify as much as possible.

− 3 − 5 = y − 2 + 8 − 3 − 5 = y − 2 + 8 size 12{ - 3 - 5=y - 2+8} {}

−8=y+6−8=y+6 size 12{ - 8=y+6} {}
6 is associated with yy size 12{y} {} by addition. Undo the association by subtracting 6 from both sides.

−8−6=y+6−6−14=y+0−14=y−8−6=y+6−6−14=y+0−14=yalignl { stack { size 12{ - 8 - 6=y+6 - 6} {} # size 12{ - "14"=y+0} {} # size 12{ - "14"=y} {} } } {}
This is equivalent to y=−14y=−14 size 12{y= - "14"} {}.

Check: When y=−14y=−14 size 12{y= - "14"} {},

− 3 − 5 = y − 2 + 8 − 3 − 5 = y − 2 + 8 size 12{ - 3 - 5=y - 2+8} {}

becomes
Does negative 3 minus 5 equal negative 14 minus 2 plus 8? Yes.,
 a true statement.

The solution to −3−5=y−2+8−3−5=y−2+8 size 12{ - 3 - 5=y - 2+8} {} is y=−14y=−14 size 12{y= - "14"} {}.

Example 8

−5a+1+6a=−2−5a+1+6a=−2 size 12{ - 5a+1+6a= - 2} {}. Begin by simplifying the left side of the equation.

- 5a+1+6a ︸ - 5+6=1 = - 2 - 5a+1+6a ︸ - 5+6=1 = - 2

a+1=−2a+1=−2 size 12{a+1= - 2} {} 1 is associated with aa size 12{a} {} by addition. Undo the association by subtracting 1 from both sides.

a + 1 − 1 = − 2 − 1 a + 0 = − 3 a = − 3 a + 1 − 1 = − 2 − 1 a + 0 = − 3 a = − 3 alignl { stack { size 12{a+1 - 1= - 2 - 1} {} # size 12{a+0= - 3} {} # size 12{a= - 3} {} } } {}

Check: When a=−3a=−3 size 12{a= - 3} {},

− 5a + 1 + 6a = − 2 − 5a + 1 + 6a = − 2 size 12{ - 5a+1+6a= - 2} {}

becomes
Does negative 5 times negative 3 plus 1 plus 6 times negative 3 equal negative 2? Yes.,
a true statement.

The solution to −5a+1+6a=−2−5a+1+6a=−2 size 12{ - 5a+1+6a= - 2} {} is a=−3a=−3 size 12{a= - 3} {}.

Example 9

7k−4=6k+17k−4=6k+1 size 12{7k - 4=6k+1} {}. In this equation, the variable appears on both sides. We need to isolate it on one side. Although we can choose either side, it will be more convenient to choose the side with the larger coefficient. Since 8 is greater than 6, we’ll isolate kk size 12{k} {} on the left side.

7k−4=6k+17k−4=6k+1 size 12{7k - 4=6k+1} {} Since 6k6k size 12{6k} {} represents +6k+6k size 12{+6k} {}, subtract 6k6k size 12{6k} {} from each side.

7k-4-6k ︸ 7-6=1 = 6k+1-6k ︸ 6-6=0 7k-4-6k ︸ 7-6=1 = 6k+1-6k ︸ 6-6=0

k−4=1k−4=1 size 12{k - 4=1} {} 4 is associated with kk size 12{k} {} by subtraction. Undo the association by adding 4 to both sides.

k − 4 + 4 = 1 + 4 k = 5 k − 4 + 4 = 1 + 4 k = 5 alignl { stack { size 12{k - 4+4=1+4} {} # size 12{k=5} {} } } {}

Check: When k=5k=5 size 12{k=5} {},

7k − 4 = 6k + 1 7k − 4 = 6k + 1 size 12{7k - 4=6k+1} {}

becomes
Does 7 times 5 minus 4 equal 6 times 5 plus 1? Yes.
 a true statement.

The solution to 7k−4=6k+17k−4=6k+1 size 12{7k - 4=6k+1} {} is k=5k=5 size 12{k=5} {}.

Example 10

−8+x=5−8+x=5 size 12{ - 8+x=5} {}. -8 is associated with xx size 12{x} {} by addition. Undo the by subtracting -8 from both sides. Subtracting -8 we get −−8=+8−−8=+8 size 12{ - left ( - 8 right )"=+"8} {}. We actually add 8 to both sides.

− 8 + x + 8 = 5 + 8 − 8 + x + 8 = 5 + 8 size 12{ - 8+x+8=5+8} {}

x = 13 x = 13 size 12{x="13"} {}

Check: When x=13x=13 size 12{x="13"} {}

− 8 + x = 5 − 8 + x = 5 size 12{ - 8+x=5} {}

becomes
Does negative 8 plus 13 equal 5? Yes.,
a true statement.

The solution to −8+x=5−8+x=5 size 12{ - 8+x=5} {} is x=13x=13 size 12{x="13"} {}.

Practice Set B

Exercise 6

y+9=4y+9=4 size 12{y+9=4} {}

Solution

y=−5y=−5 size 12{y= - 5} {}

Exercise 7

a−4=11a−4=11 size 12{a - 4="11"} {}

Solution

a=15a=15 size 12{a="15"} {}

Exercise 8

−1+7=x+3−1+7=x+3 size 12{ - 1+7=x+3} {}

Solution

x=3x=3 size 12{x=3} {}

Exercise 9

8m+4−7m=−2−38m+4−7m=−2−3 size 12{8m+4 - 7m= left ( - 2 right ) left ( - 3 right )} {}

Solution

m=2m=2 size 12{m=2} {}

Exercise 10

12k−4=9k−6+2k12k−4=9k−6+2k size 12{"12"k - 4=9k - 6+2k} {}

Solution

k=−2k=−2 size 12{k= - 2} {}

Exercise 11

−3+a=−4−3+a=−4 size 12{ - 3+a= - 4} {}

Solution

a=−1a=−1 size 12{a= - 1} {}

Exercises

For the following 10 problems, verify that each given value is a solution to the given equation.

Exercise 12

x−11=5x−11=5 size 12{x - "11"=5} {}, x=16x=16 size 12{x="16"} {}

Solution

Substitute x=4x=4 size 12{x=4} {} into the equation 4x−11=54x−11=5 size 12{4x - "11"=5} {}.
16−11=55=516−11=55=5alignl { stack { size 12{"16" - "11"=5} {} # size 12{5=5} {} } } {}
 x=4x=4 size 12{x=4} {} is a solution.

Exercise 13

y−4=−6y−4=−6 size 12{y - 4= - 6} {}, y=−2y=−2 size 12{y= - 2} {}

Exercise 14

2m−1=12m−1=1 size 12{2m - 1=1} {}, m=1m=1 size 12{m=1} {}

Solution

Substitute m=1m=1 size 12{m=1} {} into the equation 2m−1=12m−1=1 size 12{2m - 1=1} {}.
Does 2 minus 1 equal 1? Yes.
m=1m=1 size 12{m=1} {} is a solution.

Exercise 15

5y+6=−145y+6=−14 size 12{5y+6= - "14"} {}, y=−4y=−4 size 12{y= - 4} {}

Exercise 16

3x+2−7x=−5x−63x+2−7x=−5x−6 size 12{3x+2 - 7x= - 5x - 6} {}, x=−8x=−8 size 12{x= - 8} {}

Solution

Substitute x=−8x=−8 size 12{x= - 8} {} into the equation 3x+2−7=−5x−63x+2−7=−5x−6 size 12{3x+2 - 7= - 5x - 6} {}.
Does negative 24 plus 2 minus 7 equal 40 minus 6? Yes.
x=−8x=−8 size 12{x= - 8} {} is a solution.

Exercise 17

−6a+3+3a=4a+7−3a−6a+3+3a=4a+7−3a size 12{ - 6a+3+3a=4a+7 - 3a} {}, a=−1a=−1 size 12{a= - 1} {}

Exercise 18

−8+x=−8−8+x=−8 size 12{ - 8+x= - 8} {}, x=0x=0 size 12{x=0} {}

Solution

Substitute x=0x=0 size 12{x=0} {} into the equation −8+x=−8−8+x=−8 size 12{ - 8+x= - 8} {}.
Does negative 8 plus 0 equal negative 8? Yes.
x=0x=0 size 12{x=0} {} is a solution.

Exercise 19

8b+6=6−5b8b+6=6−5b size 12{8b+6=6 - 5b} {}, b=0b=0 size 12{b=0} {}

Exercise 20

4x−5=6x−204x−5=6x−20 size 12{4x - 5=6x - "20"} {}, x=152x=152 size 12{x= { {"15"} over {2} } } {}

Solution

Substitute x=152x=152 size 12{x= { {"15"} over {2} } } {}into the equation 4x−5=6x−204x−5=6x−20 size 12{4x - 5=6x - "20"} {}.
Does 30 minus 5 equal 45 minus 20? Yes.
x=152x=152 size 12{x= { {"15"} over {2} } } {} is a solution.

Exercise 21

−3y+7=2y−15−3y+7=2y−15 size 12{ - 3y+7=2y - "15"} {}, y=225y=225 size 12{y= { {"22"} over {5} } } {}

Solve each equation. Be sure to check each result.

Exercise 22

y−6=5y−6=5 size 12{y - 6=5} {}

Solution

y=11y=11 size 12{y="11"} {}

Exercise 23

m+8=4m+8=4 size 12{m+8=4} {}

Exercise 24

k−1=4k−1=4 size 12{k - 1=4} {}

Solution

k=5k=5 size 12{k=5} {}

Exercise 25

h−9=1h−9=1 size 12{h - 9=1} {}

Exercise 26

a+5=−4a+5=−4 size 12{a+5= - 4} {}

Solution

a=−9a=−9 size 12{a= - 9} {}

Exercise 27

b−7=−1b−7=−1 size 12{b - 7= - 1} {}

Exercise 28

x+4−9=6x+4−9=6 size 12{x+4 - 9=6} {}

Solution

x=11x=11 size 12{x="11"} {}

Exercise 29

y−8+10=2y−8+10=2 size 12{y - 8+"10"=2} {}

Exercise 30

z+6=6z+6=6 size 12{z+6=6} {}

Solution

z=0z=0 size 12{z=0} {}

Exercise 31

w−4=−4w−4=−4 size 12{w - 4= - 4} {}

Exercise 32

x+7−9=6x+7−9=6 size 12{x+7 - 9=6} {}

Solution

x=8x=8 size 12{x=8} {}

Exercise 33

y−2+5=4y−2+5=4 size 12{y - 2+5=4} {}

Exercise 34

m+3−8=−6+2m+3−8=−6+2 size 12{m+3 - 8= - 6+2} {}

Solution

m=1m=1 size 12{m=1} {}

Exercise 35

z+10−8=−8+10z+10−8=−8+10 size 12{z+"10" - 8= - 8+"10"} {}

Exercise 36

2+9=k−82+9=k−8 size 12{2+9=k - 8} {}

Solution

k=19k=19 size 12{k="19"} {}

Exercise 37

−5+3=h−4−5+3=h−4 size 12{ - 5+3=h - 4} {}

Exercise 38

3m−4=2m+63m−4=2m+6 size 12{3m - 4=2m+6} {}

Solution

m=10m=10 size 12{m="10"} {}

Exercise 39

5a+6=4a−85a+6=4a−8 size 12{5a+6=4a - 8} {}

Exercise 40

8b+6+2b=3b−7+6b−88b+6+2b=3b−7+6b−8 size 12{8b+6+2b=3b - 7+6b - 8} {}

Solution

b=−21b=−21 size 12{b= - "21"} {}

Exercise 41

12h−1−3−5h=2h+5h+3(−4)12h−1−3−5h=2h+5h+3(−4) size 12{"12"h - 1 - 3 - 5h=2h+5h+3 \( - 4 \) } {}

Exercise 42

−4a+5−2a=−3a−11−2a−4a+5−2a=−3a−11−2a size 12{ - 4a+5 - 2a= - 3a - "11" - 2a} {}

Solution

a=16a=16 size 12{a="16"} {}

Exercise 43

−9n−2−6+5n=3n−2−5−6n−9n−2−6+5n=3n−2−5−6n size 12{ - 9n - 2 - 6+5n=3n - left (2 right ) left ( - 5 right ) - 6n} {}

Calculator Exercises

Exercise 44

y−2.161=5.063y−2.161=5.063 size 12{y - 2 "." "161"=5 "." "063"} {}

Solution

y=7.224y=7.224 size 12{y=7 "." "224"} {}

Exercise 45

a−44.0014=−21.1625a−44.0014=−21.1625 size 12{a - "44" "." "0014"= - "21" "." "1625"} {}

Exercise 46

−0.362−0.416=5.63m−4.63m−0.362−0.416=5.63m−4.63m size 12{ - 0 "." "362" - 0 "." "416"=5 "." "63"m - 4 "." "63"m} {}

Solution

m=−0.778m=−0.778 size 12{m= - 0 "." "778"} {}

Exercise 47

8.078−9.112=2.106y−1.106y8.078−9.112=2.106y−1.106y size 12{8 "." "078" - 9 "." "112"=2 "." "106"y - 1 "." "106"y} {}

Exercise 48

4.23k+3.18=3.23k−5.834.23k+3.18=3.23k−5.83 size 12{4 "." "23"k+3 "." "18"=3 "." "23"k - 5 "." "83"} {}

Solution

k=−9.01k=−9.01 size 12{k= - 9 "." "01"} {}

Exercise 49

6.1185x−4.0031=5.1185x−0.00586.1185x−4.0031=5.1185x−0.0058 size 12{6 "." "1185"x - 4 "." "0031"=5 "." "1185"x - 0 "." "0058"} {}

Exercise 50

21.63y+12.40−5.09y=6.11y−15.66+9.43y21.63y+12.40−5.09y=6.11y−15.66+9.43y size 12{"21" "." "63"y+"12" "." "40" - 5 "." "09"y=6 "." "11"y - "15" "." "66"+9 "." "43"y} {}

Solution

y=−28.06y=−28.06 size 12{y= - "28" "." "06"} {}

Exercise 51

0.029a−0.013−0.034−0.057=−0.038+0.56+1.01a0.029a−0.013−0.034−0.057=−0.038+0.56+1.01a size 12{0 "." "029"a - 0 "." "013" - 0 "." "034" - 0 "." "057"= - 0 "." "038"+0 "." "56"+1 "." "01"a} {}

Exercises for Review

Exercise 52

((Reference)) Is 7 calculators12 students7 calculators12 students size 12{ { {7" calculators"} over {"12"" students"} } } {} an example of a ratio or a rate?

Solution

rate

Exercise 53

((Reference)) Convert 3838 size 12{ {3} over {8} } {}% to a decimal.

Exercise 54

((Reference)) 0.4% of what number is 0.014?

Solution

3.5

Exercise 55

((Reference)) Use the clustering method to estimate the sum: 89+93+206+198+9189+93+206+198+91 size 12{"89"+"93"+"206"+"198"+"91"} {}

Exercise 56

((Reference)) Combine like terms: 4x+8y+12y+9x−2y4x+8y+12y+9x−2y size 12{4x+8y+"12"y+9x - 2y} {}.

Solution

13x+18y13x+18y size 12{"13"x+"18"y} {}

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