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3 x 8 = 6 size 12{ { {3x} over {8} } =6} {}

x = 16 size 12{x="16"} {}

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y = 3 size 12{-y=3} {}

y = - 3 size 12{y"=-"3} {}

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

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

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Combining techniques in equation solving

Having examined solving equations using the addition/subtraction and the multi­plication/division principles of equality, we can combine these techniques to solve more complicated equations.

When beginning to solve an equation such as 6 x - 4 = 16 size 12{6 ital "x-"4= - "16"} {} , it is helpful to know which property of equality to use first, addition/subtraction or multiplication/di­vision. Recalling that in equation solving we are trying to isolate the variable (disas­sociate numbers from it), it is helpful to note the following.

To associate numbers and letters, we use the order of operations.

  1. Multiply/divide
  2. Add/subtract

To undo an association between numbers and letters, we use the order of opera­tions in reverse.

  1. Add/subtract
  2. Multiply/divide

Sample set b

Solve each equation. (In these example problems, we will not show the checks.)

6 x 4 = - 16 size 12{6x-4"=-""16"} {}
-4 is associated with x by subtraction. Undo the association by adding 4 to both sides.

6 x 4 + 4 = - 16 + 4 size 12{6x-4+4"=-""16"+4} {}

6 x = - 12 size 12{6x"=-""12"} {}
6 is associated with x by multiplication. Undo the association by dividing both sides by 6

6 x 6 = 12 6 size 12{ { {6x} over {6} } = { {-"12"} over {6} } } {}

x = - 2 size 12{x"=-"2} {}

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8 k + 3 = -45 . size 12{-8k+3"=-""45" "." } {}
3 is associated with k by addition. Undo the association by subtracting 3 from both sides.

8 k + 3 3 = -45 3 size 12{-8k+3-3"=-""45"-3} {}

8 k = - 48 size 12{-8k"=-""48"} {}
-8 is associated with k by multiplication. Undo the association by dividing both sides by -8.

8 k 8 = 48 8 size 12{ { {-8k} over {-8} } = { {-"48"} over {-8} } } {}

k = 6 size 12{k=6} {}

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5 m 6 4 m = 4 m 8 + 3 m . size 12{5m-6-4m=4m-8+3m "." } {} Begin by solving this equation by combining like terms.

m 6 = 7 m 8 size 12{m-6=7m-8} {} Choose a side on which to isolate m . Since 7 is greater than 1, we'll isolate m on the right side.
Subtract m from both sides.

m 6 m = 7 m 8 m size 12{-m-6-m=7m-8-m} {}

6 = 6 m 8 size 12{-6=6m-8} {}
8 is associated with m by subtraction. Undo the association by adding 8 to both sides.

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

2 = 6 m size 12{2=6m} {}
6 is associated with m by multiplication. Undo the association by dividing both sides by 6.

2 6 = 6 m 6 size 12{ { {2} over {6} } = { {6m} over {6} } } {} Reduce.

1 3 = m size 12{ { {1} over {3} } =m} {}

Notice that if we had chosen to isolate m on the left side of the equation rather than the right side, we would have proceeded as follows:

m 6 = 7 m 8 size 12{m-6=7m-8} {}
Subtract 7 m from both sides.

m 6 7 m = 7 m 8 7 m size 12{m-6-7m=7m-8-7m} {}

6 m 6 = -8 size 12{-6m-6"=-"8} {}
Add 6 to both sides,

6 m 6 + 6 = -8 + 6 size 12{-6m-6+6"=-"8+6} {}

6 m = - 2 size 12{-6m"=-"2} {}
Divide both sides by -6.

6 m 6 = 2 6 size 12{ { {-6m} over {-6} } = { {-2} over {-6} } } {}

m = 1 3 size 12{m= { {1} over {3} } } {}

This is the same result as with the previous approach.

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8 x 7 = - 2 size 12{ { {8x} over {7} } "=-"2} {}
7 is associated with x by division. Undo the association by multiplying both sides by 7.

7 8 x 7 = 7 2 size 12{7 cdot { {8x} over {7} } =7 left (-2 right )} {}

7 8 x 7 = - 14 size 12{ { {7}} cdot { {8x} over { { {7}}} } "=-""14"} {}

8 x = - 14 size 12{8x"=-""14"} {}
8 is associated with x by multiplication. Undo the association by dividing both sides by 8.

8 x 8 = 7 4 size 12{ { { { {8}}x} over { { {8}}} } = { {-7} over {4} } } {}

x = 7 4 size 12{x= { {-7} over {4} } } {}

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Practice set b

Solve each equation. Be sure to check each solution.

5 m + 7 = - 13 size 12{5m+7"=-""13"} {}

m = - 4 size 12{m"=-"4} {}

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3 a 6 = 9 size 12{-3a-6=9} {}

a = - 5 size 12{a"=-"5} {}

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2 a + 10 3 a = 9 size 12{2a+"10"-3a=9} {}

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

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11 x 4 13 x = 4 x + 14 size 12{"11"x-4-"13"x=4x+"14"} {}

x = - 3 size 12{x"=-"3} {}

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3 m + 8 = - 5 m + 1 size 12{-3m+8"=-"5m+1} {}

m = - 7 2 size 12{m"=-" { {7} over {2} } } {}

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5 y + 8 y 11 = - 11 size 12{5y+8y-"11""=-""11"} {}

y = 0 size 12{y=0} {}

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Exercises

Solve each equation. Be sure to check each result.

7 x = 42 size 12{7x="42"} {}

x = 6 size 12{x=6} {}

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8 x = 81 size 12{8x="81"} {}

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10 x = 120 size 12{"10"x="120"} {}

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

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11 x = 121 size 12{"11"x="121"} {}

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6 a = 48 size 12{-6a="48"} {}

a = - 8 size 12{a"=-"8} {}

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9 y = 54 size 12{-9y="54"} {}

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3 y = - 42 size 12{-3y"=-""42"} {}

y = 14 size 12{y="14"} {}

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5 a = - 105 size 12{-5a"=-""105"} {}

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2 m = - 62 size 12{2m"=-""62"} {}

m = - 31 size 12{m"=-""31"} {}

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3 m = - 54 size 12{3m"=-""54"} {}

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x 4 = 7 size 12{ { {x} over {4} } =7} {}

x = 28 size 12{x="28"} {}

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y 3 = 11 size 12{ { {y} over {3} } ="11"} {}

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z 6 = - 14 size 12{ { {-z} over {6} } "=-""14"} {}

z = 84 size 12{z="84"} {}

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w 5 = 1 size 12{ { {-w} over {5} } =1} {}

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3 m 1 = - 13 size 12{3m-1"=-""13"} {}

m = - 4 size 12{m"=-"4} {}

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4 x + 7 = - 17 size 12{4x+7"=-""17"} {}

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2 + 9 x = - 7 size 12{2+9x"=-"7} {}

x = - 1 size 12{x"=-"1} {}

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5 11 x = 27 size 12{5-"11"x="27"} {}

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32 = 4 y + 6 size 12{"32"=4y+6} {}

y = 13 2 size 12{y= { {"13"} over {2} } } {}

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5 + 4 = - 8 m + 1 size 12{-5+4"=-"8m+1} {}

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3 k + 6 = 5 k + 10 size 12{3k+6=5k+"10"} {}

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

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4 a + 16 = 6 a + 8 a + 6 size 12{4a+"16"=6a+8a+6} {}

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6 x + 5 + 2 x 1 = 9 x 3 x + 15 size 12{6x+5+2x-1=9x-3x+"15"} {}

x = 11 2 or 5 1 2 size 12{x= { {"11"} over {2} } " or 5" { {1} over {2} } } {}

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9 y 8 + 3 y + 7 = - 7 y + 8 y 5 y + 9 size 12{-9y-8+3y+7"=-"7y+8y-5y+9} {}

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3 a = a + 5 size 12{-3a=a+5} {}

a = - 5 4 size 12{a"=-" { {5} over {4} } } {}

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5 b = - 2 b + 8 b + 1 size 12{5b"=-"2b+8b+1} {}

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3 m + 2 8 m 4 = - 14 m + m 4 size 12{-3m+2-8m-4"=-""14"m+m-4} {}

m = - 1 size 12{m"=-"1} {}

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5 a + 3 = 3 size 12{5a+3=3} {}

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7 x + 3 x = 0 size 12{7x+3x=0} {}

x = 0 size 12{x=0} {}

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7 g + 4 11 g = - 4 g + 1 + g size 12{7g+4-"11"g"=-"4g+1+g} {}

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5 a 7 = 10 size 12{ { {5a} over {7} } ="10"} {}

a = 14 size 12{a="14"} {}

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2 m 9 = 4 size 12{ { {2m} over {9} } =4} {}

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3 x 4 = 9 2 size 12{ { {3x} over {4} } = { {9} over {2} } } {}

x = 6 size 12{x=6} {}

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8 k 3 = 32 size 12{ { {8k} over {3} } ="32"} {}

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3 a 8 3 2 = 0 size 12{ { {3a} over {8} } - { {3} over {2} } =0} {}

a = 4 size 12{a=4} {}

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5 m 6 25 3 = 0 size 12{ { {5m} over {6} } - { {"25"} over {3} } =0} {}

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Exercises for review

( [link] ) Use the distributive property to compute 40 28 size 12{"40" cdot "28"} {} .

40 30 2 = 1200 80 = 1120 size 12{"40" left ("30"-2 right )="1200"-"80"="1120"} {}

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( [link] ) Approximating π by 3.14, find the approximate circumference of the circle.

A circle with radius 8cm.

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( [link] ) Find the area of the parallelogram.

A parallelogram with base 20cm and height 11cm.

220 sq cm

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( [link] ) Find the value of 3 4 15 2 5 size 12{ { {-3 left (4-"15" right )-2} over {-5} } } {} .

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( [link] ) Solve the equation x 14 + 8 = - 2 . size 12{x-"14"+8"=-"2 "." } {}

x = 4 size 12{x=4} {}

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Source:  OpenStax, Fundamentals of mathematics. OpenStax CNX. Aug 18, 2010 Download for free at http://cnx.org/content/col10615/1.4
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