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Further Techniques in Equation Solving

Module by: Denny Burzynski, Wade Ellis. E-mail the authors

Summary: This module is from Elementary Algebra by Denny Burzynski and Wade Ellis, Jr. In this chapter, the emphasis is on the mechanics of equation solving, which clearly explains how to isolate a variable. The goal is to help the student feel more comfortable with solving applied problems. Ample opportunity is provided for the student to practice translating words to symbols, which is an important part of the "Five-Step Method" of solving applied problems (discussed in modules ((Reference)) and ((Reference))). Objectives of this module: be comfortable with combining techniques in equation solving, be able to recognize identities and contradictions.

Overview

  • Combining Techniques in Equation Solving
  • Recognizing Identities and Contrdictions

Combining Techniques in Equation Solving

In Sections (Reference) and (Reference) we worked with techniques that involved the use of addition, subtraction, multiplication, and division to solve equations. We can combine these techniques to solve more complicated equations. To do so, it is helpful to recall that an equation is solved for a particular variable when all other numbers and/or letters have been disassociated from it and it is alone on one side of the equal sign. We will also note that

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 A

Example 1

Solve 4x−7=9 4x−7=9 for x. x.

4x−7 = 9 First, undo the association between x and 7.  The 7 is associated with x by subtraction. Undo the association by adding 7 to both sides. 4x−7+7 = 9+7 4x = 16 Now, undo the association between x and 4.  The 4 is associated with x by multiplication. Undo the association by dividing both sides by 4. 4 x 4 = 16 4 16−7 = 9 Is this correct? x = 4 4x−7 = 9 First, undo the association between x and 7.  The 7 is associated with x by subtraction. Undo the association by adding 7 to both sides. 4x−7+7 = 9+7 4x = 16 Now, undo the association between x and 4.  The 4 is associated with x by multiplication. Undo the association by dividing both sides by 4. 4 x 4 = 16 4 16−7 = 9 Is this correct? x = 4

Check: 4(4)−7 = 9 Is this correct? 9 = 9 Yes, this is correct. Check: 4(4)−7 = 9 Is this correct? 9 = 9 Yes, this is correct.

Example 2

Solve 3y 4 −5=−11. 3y 4 −5=−11.

3y 4 −5 = −11 −5 is associated with y by subtraction.  Undo the association by adding 5 to both sides. 3y 4 −5+5 = −11+5 3y 4 = −6 4 is associated with y by division.  Undo the association by multiplying both sides by 4. 4⋅ 3y 4 = 4(−6) 4 ⋅ 3y 4 = 4(−6) 3y = −24 3 is associated with y by multiplication.  Undo the association by dividing both sides by 3. 3y 3 = −24 3 3 y 3 = −8 y = −8 3y 4 −5 = −11 −5 is associated with y by subtraction.  Undo the association by adding 5 to both sides. 3y 4 −5+5 = −11+5 3y 4 = −6 4 is associated with y by division.  Undo the association by multiplying both sides by 4. 4⋅ 3y 4 = 4(−6) 4 ⋅ 3y 4 = 4(−6) 3y = −24 3 is associated with y by multiplication.  Undo the association by dividing both sides by 3. 3y 3 = −24 3 3 y 3 = −8 y = −8

Check: 3(−8) 4 −5 = −11 Is this correct? −24 4 −5 = −11 Is this correct? −6−5 = −11 Is this correct? −11 = −11 Yes, this is correct. Check: 3(−8) 4 −5 = −11 Is this correct? −24 4 −5 = −11 Is this correct? −6−5 = −11 Is this correct? −11 = −11 Yes, this is correct.

Example 3

Solve 8a 3b +2m=6m−5 8a 3b +2m=6m−5 for a. a.

8a 3b +2m = 6m-5 2m is  associated with a by addition. Undo the association  by subtracting 2m from both sides. 8a 3b +2m-2m = 6m-5-2m 8a 3b = 4m-5 3b  associated with a by division. Undo the association  by multiplyingboth sides by 3b. (3b)( 8a 3b ) = 3b(4m-5) 8a = 12bm-15b 8 is associated with a by multiplication. Undo the  multiplication by dividingboth sides by 8. 8 a 8 = 12bm-15b 8 a = 12bm-15b 8 8a 3b +2m = 6m-5 2m is  associated with a by addition. Undo the association  by subtracting 2m from both sides. 8a 3b +2m-2m = 6m-5-2m 8a 3b = 4m-5 3b  associated with a by division. Undo the association  by multiplyingboth sides by 3b. (3b)( 8a 3b ) = 3b(4m-5) 8a = 12bm-15b 8 is associated with a by multiplication. Undo the  multiplication by dividingboth sides by 8. 8 a 8 = 12bm-15b 8 a = 12bm-15b 8

Practice Set A

Exercise 1

Solve 3y−1=11 3y−1=11 for y. y.

Solution

y=4 y=4

Exercise 2

Solve 5m 2 +6=1 5m 2 +6=1 for m. m.

Solution

m=−2 m=−2

Exercise 3

Solve 2n+3m=4 2n+3m=4 for n. n.

Solution

n= 4−3m 2 n= 4−3m 2

Exercise 4

Solve 9k 2h +5=p−2 9k 2h +5=p−2 for k. k.

Solution

k= 2hp−14h 9 k= 2hp−14h 9

Sometimes when solving an equation it is necessary to simplify the expressions composing it.

Sample Set B

Example 4

Solve 4x+1−3x=(−2)(4) 4x+1−3x=(−2)(4) for x. x.

4x+1−3x = (−2)(4) x+1 = −8 x = −9 4x+1−3x = (−2)(4) x+1 = −8 x = −9

Check: 4(−9)+1−3(−9) = −8 Is this correct? −36+1+27 = −8 Is this correct? −8 = −8 Yes, this is correct. Check: 4(−9)+1−3(−9) = −8 Is this correct? −36+1+27 = −8 Is this correct? −8 = −8 Yes, this is correct.

Example 5

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

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

Check: 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. Check: 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.

Exercise 5

16x−3−15x=8 16x−3−15x=8 for x. x.

Solution

x=11 x=11

Exercise 6

4(y−5)−3y=−1 4(y−5)−3y=−1 for y. y.

Solution

y=19 y=19

Exercise 7

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

Solution

a=−2 a=−2

Exercise 8

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

Solution

a= 10+5m 6 a= 10+5m 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 (Reference) and (Reference).

Sample Set C

Example 6

Solve 6x−4=2x+8 6x−4=2x+8 for x. x.

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

Check: 6(3)−4 = 2(3)+8 Is this correct? 18−4 = 6+8 Is this correct? 14 = 14 Yes, this is correct. Check: 6(3)−4 = 2(3)+8 Is this correct? 18−4 = 6+8 Is this correct? 14 = 14 Yes, this is correct.

Example 7

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

6−18x+1 = 2x−[3x−21−20] −18x+7 = 2x−[3x−41] −18x+7 = 2x−3x+41 −18x+7 = −x+41 To isolate x on the right side, add 18x to both sides. −18x+7+18x = −x+41+18x 7 = 17x+41 Subtract 41 from both sides. 7−41 = 17x+41−41 −34 = 17x 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 6−18x+1 = 2x−[3x−21−20] −18x+7 = 2x−[3x−41] −18x+7 = 2x−3x+41 −18x+7 = −x+41 To isolate x on the right side, add 18x to both sides. −18x+7+18x = −x+41+18x 7 = 17x+41 Subtract 41 from both sides. 7−41 = 17x+41−41 −34 = 17x 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

Check: 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. Check: 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

Exercise 9

Solve 8a+5=3a−5 8a+5=3a−5 for a. a.

Solution

a=−2 a=−2

Exercise 10

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

Solution

y=−1 y=−1

Exercise 11

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

Solution

k=5 k=5

Recognizing Identities and Contradictions

As we noted in Section (Reference), 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

Example 8

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

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

9x+12−9x = 12 12 = 12 9x+12−9x = 12 12 = 12

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

Example 9

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

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

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

Exercise 12

6x+3(1−2x)=3 6x+3(1−2x)=3

Solution

identity, 3=3 3=3

Exercise 13

−8m+4(2m−7)=28 −8m+4(2m−7)=28

Solution

contradiction, −28=28 −28=28

Exercise 14

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

Solution

identity, 0=0 0=0

Exercise 15

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

Solution

contradiction, −22=6 −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.

Exercise 16

3x+1=16 3x+1=16

Solution

x=5 x=5

Exercise 17

6y−4=20 6y−4=20

Exercise 18

4a−1=27 4a−1=27

Solution

a=7 a=7

Exercise 19

3x+4=40 3x+4=40

Exercise 20

2y+7=−3 2y+7=−3

Solution

y=−5 y=−5

Exercise 21

8k−7=−23 8k−7=−23

Exercise 22

5x+6=−9 5x+6=−9

Solution

x=−3 x=−3

Exercise 23

7a+2=−26 7a+2=−26

Exercise 24

10y−3=−23 10y−3=−23

Solution

y=−2 y=−2

Exercise 25

14x+1=−55 14x+1=−55

Exercise 26

x 9 +2=6 x 9 +2=6

Solution

x=36 x=36

Exercise 27

m 7 −8=−11 m 7 −8=−11

Exercise 28

y 4 +6=12 y 4 +6=12

Solution

y=24 y=24

Exercise 29

x 8 −2=5 x 8 −2=5

Exercise 30

m 11 −15=−19 m 11 −15=−19

Solution

m=−44 m=−44

Exercise 31

k 15 +20=10 k 15 +20=10

Exercise 32

6+ k 5 =5 6+ k 5 =5

Solution

k=−5 k=−5

Exercise 33

1− n 2 =6 1− n 2 =6

Exercise 34

7x 4 +6=−8 7x 4 +6=−8

Solution

x=−8 x=−8

Exercise 35

−6m 5 +11=−13 −6m 5 +11=−13

Exercise 36

3k 14 +25=22 3k 14 +25=22

Solution

k=−14 k=−14

Exercise 37

3(x−6)+5=−25 3(x−6)+5=−25

Exercise 38

16(y−1)+11=−85 16(y−1)+11=−85

Solution

y=−5 y=−5

Exercise 39

6x+14=5x−12 6x+14=5x−12

Exercise 40

23y−19=22y+1 23y−19=22y+1

Solution

y=20 y=20

Exercise 41

−3m+1=3m−5 −3m+1=3m−5

Exercise 42

8k+7=2k+1 8k+7=2k+1

Solution

k=−1 k=−1

Exercise 43

12n+5=5n−16 12n+5=5n−16

Exercise 44

2(x−7)=2x+5 2(x−7)=2x+5

Solution

contradiction

Exercise 45

−4(5y+3)+5(1+4y)=0 −4(5y+3)+5(1+4y)=0

Exercise 46

3x+7=−3−(x+2) 3x+7=−3−(x+2)

Solution

x=−3 x=−3

Exercise 47

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

Exercise 48

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

Solution

x= 19 14 x= 19 14

Exercise 49

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

Exercise 50

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

Solution

k=3 k=3

Exercise 51

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

Exercise 52

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

Solution

m=2 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.

Exercise 53

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

Exercise 54

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

Solution

R=112 R=112

Exercise 55

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

Exercise 56

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

Solution

S x 2 =F· S y 2 ;  S x 2 =7.1604 S x 2 =F· S y 2 ;  S x 2 =7.1604

Exercise 57

Solve p= nRT V p= nRT V for R. R.

Exercise 58

Solve x=4y+7 x=4y+7 for y. y.

Solution

y= x−7 4 y= x−7 4

Exercise 59

Solve y=10x+16 y=10x+16 for x. x.

Exercise 60

Solve 2x+5y=12 2x+5y=12 for y. y.

Solution

y= −2x+12 5 y= −2x+12 5

Exercise 61

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

Exercise 62

Solve m= 2n−h 5 m= 2n−h 5 for n. n.

Solution

n= 5m+h 2 n= 5m+h 2

Exercise 63

Solve t= Q+6P 8 t= Q+6P 8 for P. P.

Exercise 64

Solve A star = □ +9j Δ = □ +9j Δ for j j .

Solution

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

Exercise 65

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

Exercise 66

((Reference)) Simplify (x+3) 2 (x−2) 3 (x−2) 4 (x+3). (x+3) 2 (x−2) 3 (x−2) 4 (x+3).

Solution

( x+3 ) 3 ( x−2 ) 7 ( x+3 ) 3 ( x−2 ) 7

Exercise 67

((Reference)) Find the product. (x−7)(x+7). (x−7)(x+7).

Exercise 68

((Reference)) Find the product. (2x−1) 2 . (2x−1) 2 .

Solution

4 x 2 −4x+1 4 x 2 −4x+1

Exercise 69

((Reference)) Solve the equation y−2=−2. y−2=−2.

Exercise 70

((Reference)) Solve the equation 4x 5 =−3. 4x 5 =−3.

Solution

x= −15 4 x= −15 4

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