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Solving Linear Equations and Inequalities: Solving Equations

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

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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 able to identify various types of equations, understand the meaning of solutions and equivalent equations, be able to solve equations of the form x + a = b and x - a = b, be familiar with and able to solve literal equations.

Note: Your browser may not currently support MathML. See our browser support page for additional details. You can always view the correct math in the PDF version.

Overview

  • Types of Equations
  • Solutions and Equivalent Equations
  • Literal Equations
  • Solving Equations of the Form x+a=b x+a=b and x−a=b x−a=b

Types of Equations

Identity

Some equations are always true. These equations are called identities. Identities are equations that are true for all acceptable values of the variable, that is, for all values in the domain of the equation.

 5x=5x 5x=5x is true for all acceptable values of x x .
y+1=y+1 y+1=y+1 is true for all acceptable values of y y .
2+5=72+5=7 is true, and no substitutions are necessary.

Contradiction

Some equations are never true. These equations are called contradictions. Contradictions are equations that are never true regardless of the value substituted for the variable.

x=x+1 x=x+1 is never true for any acceptable value of x x .
 0 · k=14 0 · k=14 is never true for any acceptable value of k k .
 2=12=1 is never true.

Conditional Equation

The truth of some equations is conditional upon the value chosen for the variable. Such equations are called conditional equations. Conditional equations are equations that are true for at least one replacement of the variable and false for at least one replacement of the variable.

 x+6=11 x+6=11 is true only on the condition that x=5x=5 .
 y−7=−1y−7=−1 is true only on the condition that y=6 y=6.

Solutions and Equivalent Equations

Solutions and Solving an Equation

The collection of values that make an equation true are called solutions of the equation. An equation is solved when all its solutions have been found.

Equivalent Equations

Some equations have precisely the same collection of solutions. Such equations are called equivalent equations. The equations
 2x+1=7, 2x=6 and x=3 2x+1=7, 2x=6 and x=3
are equivalent equations because the only value that makes each one true is 3.

Sample Set A

Tell why each equation is an identity, a contradiction, or conditional.

Example 1

The equation x−4=6 x−4=6 is a conditional equation since it will be true only on the condition that x=10x=10.

Example 2

The equation x−2=x−2x−2=x−2 is an identity since it is true for all values of x x . For example,

if x = 5, 5−2 = 5−2 is true x = −7, −7−2 = −7−2 is true if x = 5, 5−2 = 5−2 is true x = −7, −7−2 = −7−2 is true

Example 3

The equation a+5=a+1 a+5=a+1 is a contradiction since every value of a a produces a false statement. For example,

if a = 8, 8+5 = 8+1 is false if a = −2, −2+5 = −2+1 is false if a = 8, 8+5 = 8+1 is false if a = −2, −2+5 = −2+1 is false

Practice Set A

For each of the following equations, write "identity," "contradiction," or "conditional." If you can, find the solution by making an educated guess based on your knowledge of arithmetic.

Exercise 1

x+1=10 x+1=10

Solution

conditional, x=9 x=9

Exercise 2

y−4=7 y−4=7

Solution

conditional, y=11 y=11

Exercise 3

5a=25 5a=25

Solution

conditional, a=5 a=5

Exercise 4

x 4 =9 x 4 =9

Solution

conditional, x=36 x=36

Exercise 5

18 b =6 18 b =6

Solution

conditional, b=3 b=3

Exercise 6

y−2=y−2 y−2=y−2

Solution

identity

Exercise 7

x+4=x−3 x+4=x−3

Solution

contradiction

Exercise 8

x+x+x=3x x+x+x=3x

Solution

identity

Exercise 9

8x=0 8x=0

Solution

conditional, x=0 x=0

Exercise 10

m−7=−5 m−7=−5

Solution

conditional, m=2 m=2

Literal Equations

Literal Equations

Some equations involve more than one variable. Such equations are called literal equations.

An equation is solved for a particular variable if that variable alone equals an expression that does not contain that particular variable.

The following equations are examples of literal equations.

  1. y=2x+7 y=2x+7 . It is solved for y y .
  2. d=rt d=rt . It is solved for d d .
  3. I=prt I=prt . It is solved for I I .
  4. z= x−u s z= x−u s . It is solved for z z .
  5. y+1=x+4 y+1=x+4 . This equation is not solved for any particular variable since no variable is isolated.

Solving Equation of the form x+a=b and x−a=bx+a=b and x−a=b

Recall that the equal sign of an equation indicates that the number represented by the expression on the left side is the same as the number represented by the expression on the right side.

This is ​the this number same as number ↓ ↓ ↓ x = 6 x+2 = 8 x−1 = 5 This is ​the this number same as number ↓ ↓ ↓ x = 6 x+2 = 8 x−1 = 5

This suggests the following procedures:

  1. We can obtain an equivalent equation (an equation having the same solutions as the original equation) by adding the same number to both sides of the equation.
  2. We can obtain an equivalent equation by subtracting the same number from both sides of the equation.

We can use these results to isolate x x , thus solving for x x .

Example 4: Solving x+a=bx+a=b for xx

x+a = b The a is associated with x by addition. Undo the association x+a−a = b−a by subtracting a from both sides. x+0 = b−a a−a=0 and 0 is the additive identity. x+0=x. x = b−a This equation is equivalent to the first equation, and it is solved for x. x+a = b The a is associated with x by addition. Undo the association x+a−a = b−a by subtracting a from both sides. x+0 = b−a a−a=0 and 0 is the additive identity. x+0=x. x = b−a This equation is equivalent to the first equation, and it is solved for x.

Example 5: Solving x−a=bx−a=b for xx

x−a = b The a is associated with x by subtraction. Undo the association x−a+a = b+a by adding a to both sides. x+0 = b+a −a+a=0 and 0 is the additive identity. x+0=x. x = b+a This equation is equivalent to the first equation, and it is solved for x. x−a = b The a is associated with x by subtraction. Undo the association x−a+a = b+a by adding a to both sides. x+0 = b+a −a+a=0 and 0 is the additive identity. x+0=x. x = b+a This equation is equivalent to the first equation, and it is solved for x.

Example 6: Method for Solving x+a=bx+a=b and x−a=bx−a=b for xx

To solve the equation x+a=b x+a=b for x x , subtract a a from both sides of the equation.
 To solve the equation x−a=b x−a=b for x x , add a a to both sides of the equation.

Sample Set B

Example 7

Solve x+7=10 x+7=10 for x x .

x+7 = 10 7 is associated with x by addition. Undo the association  x+7−7 = 10−7 by subtracting 7 from both sides. x+0 = 3 7−7=0 and 0 is the additive identity. x+0=x. x = 3 x is isolated, and the equation x=3 is equivalent to the original equation x+7=10.  Therefore, these two equation have the same solution. The solution to x=3 is clearly 3. Thus, the solution to x+7=10 is also 3. x+7 = 10 7 is associated with x by addition. Undo the association  x+7−7 = 10−7 by subtracting 7 from both sides. x+0 = 3 7−7=0 and 0 is the additive identity. x+0=x. x = 3 x is isolated, and the equation x=3 is equivalent to the original equation x+7=10.  Therefore, these two equation have the same solution. The solution to x=3 is clearly 3. Thus, the solution to x+7=10 is also 3.

Check: Substitute 3 for x x in the original equation. x+7 = 10 3+7 = 10 Is this correct? 10 = 10 Yes, this is correct. x+7 = 10 3+7 = 10 Is this correct? 10 = 10 Yes, this is correct.

Example 8

Solve m−2=−9 m−2=−9 for m m .

m−2 = −9 2 is associated with m by subtraction. Undo the association  m−2+2 = −9+2 by adding 2 from both sides. m+0 = −7 −2+2=0 and 0 is the additive identity. m+0=m. m = −7 m−2 = −9 2 is associated with m by subtraction. Undo the association  m−2+2 = −9+2 by adding 2 from both sides. m+0 = −7 −2+2=0 and 0 is the additive identity. m+0=m. m = −7

Check: Substitute −7 −7 for m m in the original equation. m−2 = −9 −7−2 = −9 Is this correct? −9 = −9 Yes, this is correct. m−2 = −9 −7−2 = −9 Is this correct? −9 = −9 Yes, this is correct.

Example 9

Use a calculator to solve this equation. Solve y−2.181=−16.915 y−2.181=−16.915 for y y .

y−2.181 = −16.915 y−2.181+2.181 = −16.915+2.181 y = −14.734 y−2.181 = −16.915 y−2.181+2.181 = −16.915+2.181 y = −14.734

On the Calculator
 Type 16.915 Press +/− Press + Type 2.181 Press = Display reads: −14.734 Type 16.915 Press +/− Press + Type 2.181 Press = Display reads: −14.734

Example 10

Solve y+m=s y+m=s for y y .

y+m = s m is associated with y by addition. Undo the association  y+m−m = s−m by subtracting m from both sides. y+0 = s−m m−m=0 and 0 is the additive identity. y+0=y. y = s−m y+m = s m is associated with y by addition. Undo the association  y+m−m = s−m by subtracting m from both sides. y+0 = s−m m−m=0 and 0 is the additive identity. y+0=y. y = s−m

Check: Substitute s−m s−m for y y in the original equation. y+m = s s−m+m = s Is this correct? s = s True Yes, this is correct. y+m = s s−m+m = s Is this correct? s = s True Yes, this is correct.

Example 11

Solve k−3h=−8h+5 k−3h=−8h+5 for k k .

k−3h = −8h+5 3h is associated with k by subtraction. Undo the association  k−3h+3h = −8h+5+3h by adding 3h to both sides. k+0 = −5h+5 −3h+3h=0 and 0 is the additive identity. k+0=k. k = −5h+5 k−3h = −8h+5 3h is associated with k by subtraction. Undo the association  k−3h+3h = −8h+5+3h by adding 3h to both sides. k+0 = −5h+5 −3h+3h=0 and 0 is the additive identity. k+0=k. k = −5h+5

Practice Set B

Exercise 11

Solve y−3=8 for y. y−3=8 for y.

Solution

y=11 y=11

Exercise 12

Solve x+9=−4 for x. x+9=−4 for x.

Solution

x=−13 x=−13

Exercise 13

Solve m+6=0 for m. m+6=0 for m.

Solution

m=−6 m=−6

Exercise 14

Solve g−7.2=1.3 for g. g−7.2=1.3 for g.

Solution

g=8.5 g=8.5

Exercise 15

solve f+2d=5d for f. f+2d=5d for f.

Solution

f=3d f=3d

Exercise 16

Solve x+8y=2y−1 for x. x+8y=2y−1 for x.

Solution

x=−6y−1 x=−6y−1

Exercise 17

Solve y+4x−1=5x+8 for y. y+4x−1=5x+8 for y.

Solution

y=x+9 y=x+9

Exercises

For the following problems, classify each of the equations as an identity, contradiction, or conditional equation.

Exercise 18

m+6=15 m+6=15

Solution

conditional

Exercise 19

y−8=−12 y−8=−12

Exercise 20

x+1=x+1 x+1=x+1

Solution

identity

Exercise 21

k−2=k−3 k−2=k−3

Exercise 22

g+g+g+g=4g g+g+g+g=4g

Solution

identity

Exercise 23

x+1=0 x+1=0

For the following problems, determine which of the literal equations have been solved for a variable. Write "solved" or "not solved."

Exercise 24

y=3x+7 y=3x+7

Solution

solved

Exercise 25

m=2k+n−1 m=2k+n−1

Exercise 26

4a=y−6 4a=y−6

Solution

not solved

Exercise 27

hk=2k+h hk=2k+h

Exercise 28

2a=a+1 2a=a+1

Solution

not solved

Exercise 29

5m=2m−7 5m=2m−7

Exercise 30

m=m m=m

Solution

not solved

For the following problems, solve each of the conditional equations.

Exercise 31

h−8=14 h−8=14

Exercise 32

k+10=1 k+10=1

Solution

k=−9 k=−9

Exercise 33

m−2=5 m−2=5

Exercise 34

y+6=−11 y+6=−11

Solution

y=−17 y=−17

Exercise 35

y−8=−1 y−8=−1

Exercise 36

x+14=0 x+14=0

Solution

x=−14 x=−14

Exercise 37

m−12=0 m−12=0

Exercise 38

g+164=−123 g+164=−123

Solution

g=−287 g=−287

Exercise 39

h−265=−547 h−265=−547

Exercise 40

x+17=−426 x+17=−426

Solution

x=−443 x=−443

Exercise 41

h−4.82=−3.56 h−4.82=−3.56

Exercise 42

y+17.003=−1.056 y+17.003=−1.056

Solution

y=−18.059 y=−18.059

Exercise 43

k+1.0135=-6.0032 k+1.0135=-6.0032

Exercise 44

Solve n+m=4 for n. n+m=4 for n.

Solution

n=4−m n=4−m

Exercise 45

Solve P+3Q−8=0 for P. P+3Q−8=0 for P.

Exercise 46

Solve a+b−3c=d−2f for b. a+b−3c=d−2f for b.

Solution

b=−a+3c+d−2f b=−a+3c+d−2f

Exercise 47

Solve x−3y+5z+1=2y−7z+8 for x. x−3y+5z+1=2y−7z+8 for x.

Exercise 48

Solve 4a−2b+c+11=6a−5b for c. 4a−2b+c+11=6a−5b for c.

Solution

c=2a−3b−11 c=2a−3b−11

Exercises for Review

Exercise 49

((Reference)) Simplify (4 x 5 y 2 ) 3 (4 x 5 y 2 ) 3 .

Exercise 50

((Reference)) Write 20 x 3 y 7 5 x 5 y 3 20 x 3 y 7 5 x 5 y 3 so that only positive exponents appear.

Solution

4 y 4 x 2 4 y 4 x 2

Exercise 51

((Reference)) Write the number of terms that appear in the expression 5 x 2 +2x−6+(a+b) 5 x 2 +2x−6+(a+b) , and then list them.

Exercise 52

((Reference)) Find the product. (3x−1) 2 (3x−1) 2 .

Solution

9 x 2 −6x+1 9 x 2 −6x+1

Exercise 53

((Reference)) Specify the domain of the equation y= 5 x−2 y= 5 x−2 .

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