Algebra Foundations for Normal People by Gabriel Beguerie - HTML preview

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Chapter 7

Worked example (no skipped steps)

7.1 Solve 3( ๐‘ฅ + 2) = 18.

Distribute:

3 ๐‘ฅ + 6 = 18

Subtract 6:

3 ๐‘ฅ = 12

Divide by 3:

๐‘ฅ = 4

160

Step-by-step solutions

Worked example (no skipped steps)

7.2 Solve 4(2 ๐‘ฆ โˆ’ 1) = 20.

Distribute:

8 ๐‘ฆ โˆ’ 4 = 20

Add 4:

8 ๐‘ฆ = 24

Divide by 8:

๐‘ฆ = 3

Worked example (no skipped steps)

7.3 Solve 5 โˆ’ 2( ๐‘ฅ + 1) = โˆ’9.

Distribute:

5 โˆ’ 2 ๐‘ฅ โˆ’ 2 = โˆ’9

Combine like terms:

3 โˆ’ 2 ๐‘ฅ = โˆ’9

Subtract 3:

โˆ’2 ๐‘ฅ = โˆ’12

Divide by โˆ’2:

๐‘ฅ = 6

Worked example (no skipped steps)

7.4 Solve 2( ๐‘ฅ โˆ’ 3) + 4 = 10.

Distribute:

2 ๐‘ฅ โˆ’ 6 + 4 = 10

Combine:

2 ๐‘ฅ โˆ’ 2 = 10

Add 2:

2 ๐‘ฅ = 12

Divide by 2:

๐‘ฅ = 6

Chapter 7

161

Worked example (no skipped steps)

7.5 Solve 5 ๐‘ฅ โˆ’ 7 = 2 ๐‘ฅ + 8.

Subtract 2 ๐‘ฅ:

3 ๐‘ฅ โˆ’ 7 = 8

Add 7:

3 ๐‘ฅ = 15

Divide by 3:

๐‘ฅ = 5

Worked example (no skipped steps)

7.6 Solve 6 ๐‘Ž + 3 = 2 ๐‘Ž + 15.

Subtract 2 ๐‘Ž:

4 ๐‘Ž + 3 = 15

Subtract 3:

4 ๐‘Ž = 12

Divide by 4:

๐‘Ž = 3

Worked example (no skipped steps)

7.7 Solve 3 ๐‘ฅ + 4 = 2 ๐‘ฅ + 11.

Subtract 2 ๐‘ฅ:

๐‘ฅ + 4 = 11

Subtract 4:

๐‘ฅ = 7

Worked example (no skipped steps)

7.8 Solve 4( ๐‘ฅ โˆ’ 1) + 3 = 2 ๐‘ฅ + 9.

Distribute:

4 ๐‘ฅ โˆ’ 4 + 3 = 2 ๐‘ฅ + 9

Combine:

4 ๐‘ฅ โˆ’ 1 = 2 ๐‘ฅ + 9

Subtract 2 ๐‘ฅ:

2 ๐‘ฅ โˆ’ 1 = 9

Add 1:

2 ๐‘ฅ = 10

Divide by 2:

๐‘ฅ = 5

162

Step-by-step solutions

Worked example (no skipped steps)

7.9 Solve 2( ๐‘ฅ + 3) โˆ’ 4 = ๐‘ฅ + 9.

Distribute:

2 ๐‘ฅ + 6 โˆ’ 4 = ๐‘ฅ + 9

Combine:

2 ๐‘ฅ + 2 = ๐‘ฅ + 9

Subtract ๐‘ฅ:

๐‘ฅ + 2 = 9

Subtract 2:

๐‘ฅ = 7

Worked example (no skipped steps)

7.10 Solve 3 โˆ’ (2 ๐‘ฅ โˆ’ 5) = 12.

Distribute the negative sign:

3 โˆ’ 2 ๐‘ฅ + 5 = 12

Combine:

8 โˆ’ 2 ๐‘ฅ = 12

Subtract 8:

โˆ’2 ๐‘ฅ = 4

Divide by โˆ’2:

๐‘ฅ = โˆ’2

Worked example (no skipped steps)

7.11 Solve ๐‘ฅ + ๐‘ฅ = 5.

2

3

Multiply every term by 6:

๐‘ฅ

๐‘ฅ

6 ยท

+ 6 ยท

= 6 ยท 5

2

3

Simplify:

3 ๐‘ฅ + 2 ๐‘ฅ = 30

So:

5 ๐‘ฅ = 30

Divide by 5:

๐‘ฅ = 6

Chapter 7

163

Worked example (no skipped steps)

7.12 Solve ๐‘ฅ + 5 = 9.

3

Subtract 5:

๐‘ฅ = 4

3

Multiply by 3:

๐‘ฅ = 12

Worked example (no skipped steps)

7.13 Solve 2 ๐‘ฅโˆ’1 = 5.

3

Multiply by 3:

2 ๐‘ฅ โˆ’ 1 = 15

Add 1:

2 ๐‘ฅ = 16

Divide by 2:

๐‘ฅ = 8

Worked example (no skipped steps)

7.14 Solve ๐‘ฅ+4 = 7.

2

Multiply by 2:

๐‘ฅ + 4 = 14

Subtract 4:

๐‘ฅ = 10

Worked example (no skipped steps)

7.15 Solve 2 ๐‘ฅ + 5 = 2 ๐‘ฅ โˆ’ 1.

Subtract 2 ๐‘ฅ from both sides:

5 = โˆ’1

This is false, so the equation has no solution.

Worked example (no skipped steps)

7.16 Solve 4( ๐‘ฅ + 1) = 4 ๐‘ฅ + 4.

Distribute:

4 ๐‘ฅ + 4 = 4 ๐‘ฅ + 4

Subtract 4 ๐‘ฅ:

4 = 4

This is always true, so the equation has infinitely many solutions.

164

Step-by-step solutions

Worked example (no skipped steps)

7.17 Solve 3( ๐‘ฅ โˆ’ 2) = 3 ๐‘ฅ + 5.

Distribute:

3 ๐‘ฅ โˆ’ 6 = 3 ๐‘ฅ + 5

Subtract 3 ๐‘ฅ:

โˆ’6 = 5

This is false, so there is no solution.

Worked example (no skipped steps)

7.18 Solve 7 โˆ’ 2( ๐‘ฆ + 3) = 1.

Distribute:

7 โˆ’ 2 ๐‘ฆ โˆ’ 6 = 1

Combine:

1 โˆ’ 2 ๐‘ฆ = 1

Subtract 1:

โˆ’2 ๐‘ฆ = 0

Divide by โˆ’2:

๐‘ฆ = 0

Worked example (no skipped steps)

7.19 Solve 5( ๐‘ฅ โˆ’ 2) = 3 ๐‘ฅ + 6.

Distribute:

5 ๐‘ฅ โˆ’ 10 = 3 ๐‘ฅ + 6

Subtract 3 ๐‘ฅ:

2 ๐‘ฅ โˆ’ 10 = 6

Add 10:

2 ๐‘ฅ = 16

Divide by 2:

๐‘ฅ = 8

Worked example (no skipped steps)

7.20 Solve 3 ๐‘ฅ+6 = 8.

3

Multiply by 3:

3 ๐‘ฅ + 6 = 24

Subtract 6:

3 ๐‘ฅ = 18

Divide by 3:

๐‘ฅ = 6

Chapter 7

165

Worked example (no skipped steps)

7.21 Solve 2(3 ๐‘š โˆ’ 1) + 5 = 15.

Distribute:

6 ๐‘š โˆ’ 2 + 5 = 15

Combine:

6 ๐‘š + 3 = 15

Subtract 3:

6 ๐‘š = 12

Divide by 6:

๐‘š = 2

Worked example (no skipped steps)

7.22 Solve 4 โˆ’ 3(2 ๐‘ฅ โˆ’ 1) = โˆ’11.

Distribute:

4 โˆ’ 6 ๐‘ฅ + 3 = โˆ’11

Combine:

7 โˆ’ 6 ๐‘ฅ = โˆ’11

Subtract 7:

โˆ’6 ๐‘ฅ = โˆ’18

Divide by โˆ’6:

๐‘ฅ = 3

Worked example (no skipped steps)

7.23 Decide whether 4 ๐‘Ž + 7 = 4 ๐‘Ž โˆ’ 2 has one solution, no solution, or infinitely many solutions.

Subtract 4 ๐‘Ž from both sides:

7 = โˆ’2

This is false, so there is no solution.

166

Step-by-step solutions

Worked example (no skipped steps)

7.24 Decide whether 3( ๐‘ โˆ’ 2) = 3 ๐‘ โˆ’ 6 has one solution, no solution, or infinitely many solutions.

Distribute:

3 ๐‘ โˆ’ 6 = 3 ๐‘ โˆ’ 6

Subtract 3 ๐‘:

โˆ’6 = โˆ’6

This is always true, so there are infinitely many solutions.

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