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Solve 3 ( m 6 ) 2 m = 4 + 1 for m .

3 ( m 6 ) 2 m = 4 + 1 3 m 18 2 m = 3 m 18 = 3 m = 15

C h e c k : 3 ( 15 6 ) 2 ( 15 ) = 4 + 1 Is this correct? 3 ( 9 ) 30 = 3 Is this correct? 27 30 = 3 Is this correct? 3 = 3 Yes, this is correct .

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Practice set b

Solve and check each equation.

16 x 3 15 x = 8 for x .

x = 11

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4 ( y 5 ) 3 y = 1 for y .

y = 19

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2 ( a 2 + 3 a 1 ) + 2 a 2 + 7 a = 0 for a .

a = 2

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5 m ( m 2 a 1 ) 5 m 2 + 2 a ( 5 m + 3 ) = 10 for a .

a = 10 + 5 m 6

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Often the variable we wish to solve for will appear on both sides of the equal sign. We can isolate the variable on either the left or right side of the equation by using the techniques of Sections [link] and [link] .

Sample set c

Solve 6 x 4 = 2 x + 8 for x .

6 x 4 = 2 x + 8 To isolate x on the left side, subtract 2 m from both sides . 6 x 4 2 x = 2 x + 8 2 x 4 x 4 = 8 Add 4 to both sides . 4 x 4 + 4 = 8 + 4 4 x = 12 Divide both sides by 4. 4 x 4 = 12 4 x = 3

C h e c k : 6 ( 3 ) 4 = 2 ( 3 ) + 8 Is this correct? 18 4 = 6 + 8 Is this correct? 14 = 14 Yes, this is correct .

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Solve 6 ( 1 3 x ) + 1 = 2 x [ 3 ( x 7 ) 20 ] for x .

On left side of an equation, arrows show that six is multiplied with each term inside the parentheses, and on right side, arrows show that three is multiplied with each term inside the parentheses.

6 18 x + 1 = 2 x [ 3 x 21 20 ] 18 x + 7 = 2 x [ 3 x 41 ] 18 x + 7 = 2 x 3 x + 41 18 x + 7 = x + 41 To isolate x on the right side, add 18 x to both sides . 18 x + 7 + 18 x = x + 41 + 18 x 7 = 17 x + 41 Subtract 41 from both sides . 7 41 = 17 x + 41 41 34 = 17 x Divide both sides by 17. 34 17 = 17 x 17 2 = x Since the equation 2 = x is equivalent to the equation x = 2 , we can write the answer as x = 2. x = 2

C h e c k : 6 ( 1 3 ( 2 ) ) + 1 = 2 ( 2 ) [ 3 ( 2 7 ) 20 ] Is this correct? 6 ( 1 + 6 ) + 1 = 4 [ 3 ( 9 ) 20 ] Is this correct? 6 ( 7 ) + 1 = 4 [ 27 20 ] Is this correct? 42 + 1 = 4 [ 47 ] Is this correct? 43 = 4 + 47 Is this correct? 43 = 43 Yes, this is correct .

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Practice set c

Solve 8 a + 5 = 3 a 5 for a .

a = 2

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Solve 9 y + 3 ( y + 6 ) = 15 y + 21 for y .

y = 1

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Solve 3 k + 2 [ 4 ( k 1 ) + 3 ] = 63 2 k for k .

k = 5

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Recognizing identities and contradictions

As we noted in Section [link] , some equations are identities and some are contradictions. As the problems of Sample Set D will suggest,

    Recognizing an identity

  1. If, when solving an equation, all the variables are eliminated and a true statement results, the equation is an identity.

    Recognizing a contradiction

  1. If, when solving an equation, all the variables are eliminated and a false statement results, the equation is a contradiction.

Sample set d

Solve 9 x + 3 ( 4 3 x ) = 12 for x .

On left side of an equation, arrows show that three is multiplied with each term inside the parentheses.

9 x + 12 9 x = 12 12 = 12

The variable has been eliminated and the result is a true statement. The original equation is an identity.

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Solve 2 ( 10 2 y ) 4 y + 1 = 18 for y .

On left side of an equation, arrows show that negative two is multiplied with each term inside the parentheses.

20 + 4 y 4 y + 1 = 18 19 = 18

The variable has been eliminated and the result is a false statement. The original equation is a contradiction.

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Practice set d

Classify each equation as an identity or a contradiction.

6 x + 3 ( 1 2 x ) = 3

identity, 3 = 3

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8 m + 4 ( 2 m 7 ) = 28

contradiction, 28 = 28

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3 ( 2 x 4 ) 2 ( 3 x + 1 ) + 14 = 0

identity, 0 = 0

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5 ( x + 6 ) + 8 = 3 [ 4 ( x + 2 ) ] 2 x

contradiction, 22 = 6

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Exercises

For the following problems, solve each conditional equation. If the equation is not conditional, identify it as an identity or a contradiction.

2 y + 7 = 3

y = 5

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5 x + 6 = 9

x = 3

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10 y 3 = 23

y = 2

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m 11 15 = 19

m = 44

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7 x 4 + 6 = 8

x = 8

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3 k 14 + 25 = 22

k = 14

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3 ( x 6 ) + 5 = 25

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16 ( y 1 ) + 11 = 85

y = 5

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23 y 19 = 22 y + 1

y = 20

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8 k + 7 = 2 k + 1

k = 1

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2 ( x 7 ) = 2 x + 5

contradiction

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4 ( 5 y + 3 ) + 5 ( 1 + 4 y ) = 0

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3 x + 7 = 3 ( x + 2 )

x = 3

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4 ( 4 y + 2 ) = 3 y + 2 [ 1 3 ( 1 2 y ) ]

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5 ( 3 x 8 ) + 11 = 2 2 x + 3 ( x 4 )

x = 19 14

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12 ( m 2 ) = 2 m + 3 m 2 m + 3 ( 5 3 m )

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

k = 3

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3 [ 4 2 ( y + 2 ) ] = 2 y 4 [ 1 + 2 ( 1 + y ) ]

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5 [ 2 m ( 3 m 1 ) ] = 4 m 3 m + 2 ( 5 2 m ) + 1

m = 2

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For the following problems, solve the literal equations for the indicated variable. When directed, find the value of that variable for the given values of the other variables.

Solve I = E R for R . Find the value of R when I = 0.005 and E = 0.0035.

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Solve P = R C for R . Find the value of R when P = 27 and C = 85.

R = 112

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Solve z = x x ¯ s for x . Find the value of x when z = 1.96 , s = 2.5 , and x ¯ = 15.

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Solve F = S x 2 S y 2 for S x 2 S x 2 represents a single quantity. Find the value of S x 2 when F = 2.21 and S y 2 = 3.24.

S x 2 = F · S y 2 ; S x 2 = 7.1604

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Solve p = n R T V for R .

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Solve x = 4 y + 7 for y .

y = x 7 4

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Solve y = 10 x + 16 for x .

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Solve 2 x + 5 y = 12 for y .

y = 2 x + 12 5

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Solve 9 x + 3 y + 15 = 0 for y .

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Solve m = 2 n h 5 for n .

n = 5 m + h 2

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Solve t = Q + 6 P 8 for P .

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Solve A star = + 9 j Δ for j .

j is equal to the product of star and triangle minus square over nine.

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Exercises for review

( [link] ) Simplify ( x + 3 ) 2 ( x 2 ) 3 ( x 2 ) 4 ( x + 3 ) .

( x + 3 ) 3 ( x 2 ) 7

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( [link] ) Find the product. ( x 7 ) ( x + 7 ) .

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( [link] ) Find the product. ( 2 x 1 ) 2 .

4 x 2 4 x + 1

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( [link] ) Solve the equation y 2 = 2.

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( [link] ) Solve the equation 4 x 5 = 3.

x = 15 4

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In biology, a pathogen (Greek: πάθος pathos "suffering", "passion" and -γενής -genēs "producer of") in the oldest and broadest sense, is anything that can produce disease. A pathogen may also be referred to as an infectious agent, or simply a germ. The term pathogen came into use in the 1880s.[1][2
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Source:  OpenStax, Elementary algebra. OpenStax CNX. May 08, 2009 Download for free at http://cnx.org/content/col10614/1.3
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