<< Chapter < Page Chapter >> Page >

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 + 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 [link] .

Sample set b

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

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

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 = 2 size 12{x=2} {} , x + 4 size 12{x+4} {} becomes

Does 2 + 4 = 6? Yes.

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

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

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 = 13 size 12{m="13"} {} ,

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

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

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

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

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

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

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

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

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

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

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 = 3 size 12{a= - 3} {} ,

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 5 a + 1 + 6 a = 2 size 12{ - 5a+1+6a= - 2} {} is a = 3 size 12{a= - 3} {} .

7 k 4 = 6 k + 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 k size 12{k} {} on the left side.

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

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

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

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

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

7 k 4 = 6 k + 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 7 k 4 = 6 k + 1 size 12{7k - 4=6k+1} {} is k = 5 size 12{k=5} {} .

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

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

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

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

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 size 12{ - 8+x=5} {} is x = 13 size 12{x="13"} {} .

Practice set b

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

y = 5 size 12{y= - 5} {}

a 4 = 11 size 12{a - 4="11"} {}

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

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

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

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

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

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

k = 2 size 12{k= - 2} {}

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

a = 1 size 12{a= - 1} {}

Exercises

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Solve each equation. Be sure to check each result.

y 6 = 5 size 12{y - 6=5} {}

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

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

k 1 = 4 size 12{k - 1=4} {}

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

h 9 = 1 size 12{h - 9=1} {}

a + 5 = 4 size 12{a+5= - 4} {}

a = 9 size 12{a= - 9} {}

b 7 = 1 size 12{b - 7= - 1} {}

x + 4 9 = 6 size 12{x+4 - 9=6} {}

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

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

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

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

w 4 = 4 size 12{w - 4= - 4} {}

x + 7 9 = 6 size 12{x+7 - 9=6} {}

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

y 2 + 5 = 4 size 12{y - 2+5=4} {}

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

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

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

2 + 9 = k 8 size 12{2+9=k - 8} {}

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

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

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

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

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

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

b = 21 size 12{b= - "21"} {}

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

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

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

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

Calculator exercises

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

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

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

0 . 362 0 . 416 = 5 . 63 m 4 . 63 m size 12{ - 0 "." "362" - 0 "." "416"=5 "." "63"m - 4 "." "63"m} {}

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

8 . 078 9 . 112 = 2 . 106 y 1 . 106 y size 12{8 "." "078" - 9 "." "112"=2 "." "106"y - 1 "." "106"y} {}

4 . 23 k + 3 . 18 = 3 . 23 k 5 . 83 size 12{4 "." "23"k+3 "." "18"=3 "." "23"k - 5 "." "83"} {}

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

6 . 1185 x 4 . 0031 = 5 . 1185 x 0 . 0058 size 12{6 "." "1185"x - 4 "." "0031"=5 "." "1185"x - 0 "." "0058"} {}

21 . 63 y + 12 . 40 5 . 09 y = 6 . 11 y 15 . 66 + 9 . 43 y size 12{"21" "." "63"y+"12" "." "40" - 5 "." "09"y=6 "." "11"y - "15" "." "66"+9 "." "43"y} {}

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

0 . 029 a 0 . 013 0 . 034 0 . 057 = 0 . 038 + 0 . 56 + 1 . 01 a size 12{0 "." "029"a - 0 "." "013" - 0 "." "034" - 0 "." "057"= - 0 "." "038"+0 "." "56"+1 "." "01"a} {}

Exercises for review

( [link] ) Is 7 calculators 12 students size 12{ { {7" calculators"} over {"12"" students"} } } {} an example of a ratio or a rate?

rate

( [link] ) Convert 3 8 size 12{ {3} over {8} } {} % to a decimal.

( [link] ) 0.4% of what number is 0.014?

3.5

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

( [link] ) Combine like terms: 4 x + 8 y + 12 y + 9 x 2 y size 12{4x+8y+"12"y+9x - 2y} {} .

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

Get Jobilize Job Search Mobile App in your pocket Now!

Get it on Google Play Download on the App Store Now




Source:  OpenStax, Algebra i for the community college. OpenStax CNX. Dec 19, 2014 Download for free at http://legacy.cnx.org/content/col11598/1.3
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Algebra i for the community college' conversation and receive update notifications?

Ask