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5.3 Further techniques in equation solving  (Page 2/2)

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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 .

Practice set b

Solve and check each equation.

16 x − 3 − 15 x = 8 for x .

x = 11

4 ( y − 5 ) − 3 y = − 1 for y .

y = 19

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

a = − 2

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

a = 10 + 5 m 6

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 .

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 .

Practice set c

Solve 8 a + 5 = 3 a − 5 for a .

a = − 2

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

y = − 1

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

k = 5

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.

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.

Practice set d

Classify each equation as an identity or a contradiction.

6 x + 3 ( 1 − 2 x ) = 3

identity, 3 = 3

− 8 m + 4 ( 2 m − 7 ) = 28

contradiction, − 28 = 28

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

identity, 0 = 0

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

contradiction, − 22 = 6

Exercises

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

3 x + 1 = 16

x = 5

6 y − 4 = 20

4 a − 1 = 27

a = 7

3 x + 4 = 40

2 y + 7 = − 3

y = − 5

8 k − 7 = − 23

5 x + 6 = − 9

x = − 3

7 a + 2 = − 26

10 y − 3 = − 23

y = − 2

14 x + 1 = − 55

x 9 + 2 = 6

x = 36

m 7 − 8 = − 11

y 4 + 6 = 12

y = 24

x 8 − 2 = 5

m 11 − 15 = − 19

m = − 44

k 15 + 20 = 10

6 + k 5 = 5

k = − 5

1 − n 2 = 6

7 x 4 + 6 = − 8

x = − 8

− 6 m 5 + 11 = − 13

3 k 14 + 25 = 22

k = − 14

3 ( x − 6 ) + 5 = − 25

16 ( y − 1 ) + 11 = − 85

y = − 5

6 x + 14 = 5 x − 12

23 y − 19 = 22 y + 1

y = 20

− 3 m + 1 = 3 m − 5

8 k + 7 = 2 k + 1

k = − 1

12 n + 5 = 5 n − 16

2 ( x − 7 ) = 2 x + 5

contradiction

− 4 ( 5 y + 3 ) + 5 ( 1 + 4 y ) = 0

3 x + 7 = − 3 − ( x + 2 )

x = − 3

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

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

x = 19 14

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

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

k = 3

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

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

m = 2

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.

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

R = 112

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

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

Solve p = n R T V for R .

Solve x = 4 y + 7 for y .

y = x − 7 4

Solve y = 10 x + 16 for x .

Solve 2 x + 5 y = 12 for y .

y = − 2 x + 12 5

Solve − 9 x + 3 y + 15 = 0 for y .

Solve m = 2 n − h 5 for n .

n = 5 m + h 2

Solve t = Q + 6 P 8 for P .

Solve A star = □   + 9 j Δ for j .

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

Solve A rhombus is equal to the sum of triangle and the product of star and square over twice delta. for A star .

Exercises for review

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

( x + 3 ) 3 ( x − 2 ) 7

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

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

4 x 2 − 4 x + 1

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

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

x = − 15 4

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Source:  OpenStax, Basic mathematics review. OpenStax CNX. Jun 06, 2012 Download for free at http://cnx.org/content/col11427/1.2
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