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Classify the equation as a conditional equation, an identity, or a contradiction and then state the solution:

8 ( 1 3 x ) + 15 ( 2 x + 7 ) = 2 ( x + 50 ) + 4 ( x + 3 ) + 1

identity; all real numbers

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Classify as a conditional equation, an identity, or a contradiction. Then state the solution.

10 + 4 ( p 5 ) = 0

Solution

.
Distribute. .
Combine like terms. .
Add 10 to both sides. .
Simplify. .
Divide. .
Simplify. .
The equation is true when p = 5 2 . This is a conditional equation.
The solution is p = 5 2 .
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Classify the equation as a conditional equation, an identity, or a contradiction and then state the solution: 11 ( q + 3 ) 5 = 19

conditional equation; q = 9 11

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Classify the equation as a conditional equation, an identity, or a contradiction and then state the solution: 6 + 14 ( k 8 ) = 95

conditional equation; k = 193 14

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Classify the equation as a conditional equation, an identity, or a contradiction. Then state the solution.

5 m + 3 ( 9 + 3 m ) = 2 ( 7 m 11 )

Solution

.
Distribute. .
Combine like terms. .
Subtract 14 m from both sides. .
Simplify. .
But 27 −22 . The equation is a contradiction.
It has no solution.
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Classify the equation as a conditional equation, an identity, or a contradiction and then state the solution:

12 c + 5 ( 5 + 3 c ) = 3 ( 9 c 4 )

contradiction; no solution

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Classify the equation as a conditional equation, an identity, or a contradiction and then state the solution:

4 ( 7 d + 18 ) = 13 ( 3 d 2 ) 11 d

contradiction; no solution

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Type of equation What happens when you solve it? Solution
Conditional Equation True for one or more values of the variables and false for all other values One or more values
Identity True for any value of the variable All real numbers
Contradiction False for all values of the variable No solution

Key concepts

  • General Strategy for Solving Linear Equations
    1. Simplify each side of the equation as much as possible.
      Use the Distributive Property to remove any parentheses.
      Combine like terms.
    2. Collect all the variable terms on one side of the equation.
      Use the Addition or Subtraction Property of Equality.
    3. Collect all the constant terms on the other side of the equation.
      Use the Addition or Subtraction Property of Equality.
    4. Make the coefficient of the variable term to equal to 1.
      Use the Multiplication or Division Property of Equality.
      State the solution to the equation.
    5. Check the solution.
      Substitute the solution into the original equation.

Practice makes perfect

Solve Equations Using the General Strategy for Solving Linear Equations

In the following exercises, solve each linear equation.

21 ( y 5 ) = −42

y = 3

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−16 ( 3 n + 4 ) = 32

n = −2

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5 ( 8 + 6 p ) = 0

p = 4 3

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( t 19 ) = 28

t = −9

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8 ( 9 b 4 ) 12 = 100

b = 2

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21 + 2 ( m 4 ) = 25

m = 6

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−6 + 6 ( 5 k ) = 15

k = 3 2

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2 ( 9 s 6 ) 62 = 16

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8 ( 6 t 5 ) 35 = −27

t = 1

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3 ( 10 2 x ) + 54 = 0

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−2 ( 11 7 x ) + 54 = 4

x = −2

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3 5 ( 10 x 5 ) = 27

x = 5

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1 5 ( 15 c + 10 ) = c + 7

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1 4 ( 20 d + 12 ) = d + 7

d = 1

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18 ( 9 r + 7 ) = −16

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15 ( 3 r + 8 ) = 28

r = −7

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−3 ( m 1 ) = 13

m = −15

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11 4 ( y 8 ) = 43

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18 2 ( y 3 ) = 32

y = −4

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24 8 ( 3 v + 6 ) = 0

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35 5 ( 2 w + 8 ) = −10

w = 1 2

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4 ( a 12 ) = 3 ( a + 5 )

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−2 ( a 6 ) = 4 ( a 3 )

a = 4

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2 ( 5 u ) = −3 ( 2 u + 6 )

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5 ( 8 r ) = −2 ( 2 r 16 )

r = 8

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3 ( 4 n 1 ) 2 = 8 n + 3

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9 ( 2 m 3 ) 8 = 4 m + 7

m = 3

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12 + 2 ( 5 3 y ) = −9 ( y 1 ) 2

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−15 + 4 ( 2 5 y ) = −7 ( y 4 ) + 4

y = −3

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8 ( x 4 ) 7 x = 14

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5 ( x 4 ) 4 x = 14

x = 34

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5 + 6 ( 3 s 5 ) = −3 + 2 ( 8 s 1 )

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−12 + 8 ( x 5 ) = −4 + 3 ( 5 x 2 )

x = −6

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4 ( u 1 ) 8 = 6 ( 3 u 2 ) 7

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7 ( 2 n 5 ) = 8 ( 4 n 1 ) 9

n = −1

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4 ( p 4 ) ( p + 7 ) = 5 ( p 3 )

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3 ( a 2 ) ( a + 6 ) = 4 ( a 1 )

a = −4

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( 9 y + 5 ) ( 3 y 7 )
= 16 ( 4 y 2 )

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( 7 m + 4 ) ( 2 m 5 )
= 14 ( 5 m 3 )

m = −4

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4 [ 5 8 ( 4 c 3 ) ]
= 12 ( 1 13 c ) 8

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5 [ 9 2 ( 6 d 1 ) ]
= 11 ( 4 10 d ) 139

d = −3

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3 [ −9 + 8 ( 4 h 3 ) ]
= 2 ( 5 12 h ) 19

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3 [ −14 + 2 ( 15 k 6 ) ]
= 8 ( 3 5 k ) 24

k = 3 5

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5 [ 2 ( m + 4 ) + 8 ( m 7 ) ]
= 2 [ 3 ( 5 + m ) ( 21 3 m ) ]

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10 [ 5 ( n + 1 ) + 4 ( n 1 ) ]
= 11 [ 7 ( 5 + n ) ( 25 3 n ) ]

n = −5

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5 ( 1.2 u 4.8 ) = −12

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4 ( 2.5 v 0.6 ) = 7.6

v = 1

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0.25 ( q 6 ) = 0.1 ( q + 18 )

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0.2 ( p 6 ) = 0.4 ( p + 14 )

p = −34

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0.5 ( 16 m + 34 ) = −15

m = −4

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Classify Equations

In the following exercises, classify each equation as a conditional equation, an identity, or a contradiction and then state the solution.

23 z + 19 = 3 ( 5 z 9 ) + 8 z + 46

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15 y + 32 = 2 ( 10 y 7 ) 5 y + 46

identity; all real numbers

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5 ( b 9 ) + 4 ( 3 b + 9 ) = 6 ( 4 b 5 ) 7 b + 21

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9 ( a 4 ) + 3 ( 2 a + 5 ) = 7 ( 3 a 4 ) 6 a + 7

identity; all real numbers

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18 ( 5 j 1 ) + 29 = 47

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24 ( 3 d 4 ) + 100 = 52

conditional equation; d = 2 3

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22 ( 3 m 4 ) = 8 ( 2 m + 9 )

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30 ( 2 n 1 ) = 5 ( 10 n + 8 )

conditional equation; n = 7

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7 v + 42 = 11 ( 3 v + 8 ) 2 ( 13 v 1 )

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18 u 51 = 9 ( 4 u + 5 ) 6 ( 3 u 10 )

contradiction; no solution

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3 ( 6 q 9 ) + 7 ( q + 4 ) = 5 ( 6 q + 8 ) 5 ( q + 1 )

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5 ( p + 4 ) + 8 ( 2 p 1 ) = 9 ( 3 p 5 ) 6 ( p 2 )

contradiction; no solution

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12 ( 6 h 1 ) = 8 ( 8 h + 5 ) 4

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9 ( 4 k 7 ) = 11 ( 3 k + 1 ) + 4

conditional equation; k = 26

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45 ( 3 y 2 ) = 9 ( 15 y 6 )

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60 ( 2 x 1 ) = 15 ( 8 x + 5 )

contradiction; no solution

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16 ( 6 n + 15 ) = 48 ( 2 n + 5 )

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36 ( 4 m + 5 ) = 12 ( 12 m + 15 )

identity; all real numbers

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9 ( 14 d + 9 ) + 4 d = 13 ( 10 d + 6 ) + 3

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11 ( 8 c + 5 ) 8 c = 2 ( 40 c + 25 ) + 5

identity; all real numbers

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Everyday math

Fencing Micah has 44 feet of fencing to make a dog run in his yard. He wants the length to be 2.5 feet more than the width. Find the length, L , by solving the equation 2 L + 2 ( L 2.5 ) = 44 .

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Coins Rhonda has $1.90 in nickels and dimes. The number of dimes is one less than twice the number of nickels. Find the number of nickels, n , by solving the equation 0.05 n + 0.10 ( 2 n 1 ) = 1.90 .

8 nickels

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Writing exercises

Using your own words, list the steps in the general strategy for solving linear equations.

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Explain why you should simplify both sides of an equation as much as possible before collecting the variable terms to one side and the constant terms to the other side.

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What is the first step you take when solving the equation 3 7 ( y 4 ) = 38 ? Why is this your first step?

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Solve the equation 1 4 ( 8 x + 20 ) = 3 x 4 explaining all the steps of your solution as in the examples in this section.

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Self check

After completing the exercises, use this checklist to evaluate your mastery of the objective of this section.

This is a table that has three rows and four columns. In the first row, which is a header row, the cells read from left to right: “I can…,” “confidently,” “with some help,” and “no-I don’t get it!” The first column below “I can…” reads: “solve equations using the general strategy for solving linear equations,” and “classify equations.” The rest of the cells are blank.

On a scale of 1-10, how would you rate your mastery of this section in light of your responses on the checklist? How can you improve this?

Practice Key Terms 3

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Source:  OpenStax, Elementary algebra. OpenStax CNX. Jan 18, 2017 Download for free at http://cnx.org/content/col12116/1.2
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