Algebra Foundations for Normal People by Gabriel Beguerie - HTML preview

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Chapter 12

Worked example (no skipped steps)

12.1 Use 𝐴 = 𝑙𝑀 with 𝑙 = 8 and 𝑀 = 3.

Substitute the values:

𝐴 = (8)(3)

Multiply:

𝐴 = 24

Answer: 24.

Worked example (no skipped steps)

12.2 Use 𝑑 = π‘Ÿπ‘‘ with π‘Ÿ = 60 and 𝑑 = 2 . 5.

Substitute:

𝑑 = (60)(2 . 5)

Multiply:

𝑑 = 150

Answer: 150.

Worked example (no skipped steps)

12.3 Use 𝑦 = π‘šπ‘₯ + 𝑏 with π‘š = 2, π‘₯ = 5, 𝑏 = βˆ’3.

Substitute carefully:

𝑦 = (2)(5) + (βˆ’3)

Now simplify:

𝑦 = 10 βˆ’ 3 = 7

Answer: 7.

184

Step-by-step solutions

Worked example (no skipped steps)

12.4 Solve for 𝑏 in 𝑦 = π‘šπ‘₯ + 𝑏.

Subtract π‘šπ‘₯ from both sides:

𝑦 βˆ’ π‘šπ‘₯ = 𝑏

So:

𝑏 = 𝑦 βˆ’ π‘šπ‘₯

Answer: 𝑏 = 𝑦 βˆ’ π‘šπ‘₯.

Worked example (no skipped steps)

12.5 Solve for 𝑑 in 𝑑 = π‘Ÿπ‘‘.

Divide both sides by π‘Ÿ:

𝑑

π‘Ÿπ‘‘

=

π‘Ÿ

π‘Ÿ

So:

𝑑

𝑑 = π‘Ÿ

Answer: 𝑑 = 𝑑 (with π‘Ÿ ΜΈ= 0).

π‘Ÿ

Worked example (no skipped steps)

12.6 Solve for 𝑀 in 𝑃 = 2 𝑙 + 2 𝑀.

Subtract 2 𝑙 from both sides:

𝑃 βˆ’ 2 𝑙 = 2 𝑀

Now divide by 2:

𝑃 βˆ’ 2 𝑙

𝑀 =

2

Answer: 𝑀 = 𝑃 βˆ’2 𝑙 .

2

Worked example (no skipped steps)

12.7 Solve for 𝐢 in 𝐹 = 9 𝐢 + 32.

5

Subtract 32:

9

𝐹 βˆ’ 32 =

𝐢

5

Multiply both sides by the reciprocal 5 :

9

5 ( 𝐹 βˆ’ 32) = 𝐢

9

So:

5

𝐢 =

( 𝐹 βˆ’ 32)

9

Answer: 𝐢 = 5 ( 𝐹 βˆ’ 32).

9

Chapter 12

185

Worked example (no skipped steps)

12.8 Solve for π‘Ÿ in 𝐴 = πœ‹π‘Ÿ 2.

Divide by πœ‹:

𝐴 = π‘Ÿ 2

πœ‹

Take the square root of both sides:

βˆšοΈƒ

𝐴

π‘Ÿ =

πœ‹

√︁

Answer: π‘Ÿ =

𝐴 .

πœ‹

Worked example (no skipped steps)

12.9 Solve for π‘₯ in π‘₯βˆ’ π‘Ž = 𝑐.

𝑏

Multiply both sides by 𝑏:

π‘₯ βˆ’ π‘Ž = 𝑏𝑐

Add π‘Ž to both sides:

π‘₯ = 𝑏𝑐 + π‘Ž

Answer: π‘₯ = 𝑏𝑐 + π‘Ž.

Worked example (no skipped steps)

12.10 Solve for 𝑏 in 𝐴 = 1 π‘β„Ž.

2

Multiply both sides by 2:

2 𝐴 = π‘β„Ž

Now divide by β„Ž:

2 𝐴

𝑏 = β„Ž

Answer: 𝑏 = 2 𝐴 (with β„Ž ΜΈ= 0).

β„Ž

Worked example (no skipped steps)

12.11 Solve for π‘₯ in 𝑧 = π‘₯βˆ’ 𝑦 .

3

Multiply both sides by 3:

3 𝑧 = π‘₯ βˆ’ 𝑦

Add 𝑦 to both sides:

π‘₯ = 3 𝑧 + 𝑦

Answer: π‘₯ = 3 𝑧 + 𝑦.

186

Step-by-step solutions

Worked example (no skipped steps)

12.12 Use 𝑃 = 2 𝑙 + 2 𝑀 with 𝑙 = 7 and 𝑀 = 4.

Substitute:

𝑃 = 2(7) + 2(4)

Simplify:

𝑃 = 14 + 8 = 22

Answer: 22.

Worked example (no skipped steps)

12.13 Use 𝐼 = 𝑃 π‘Ÿπ‘‘ with 𝑃 = 500, π‘Ÿ = 0 . 08, 𝑑 = 2.

Substitute:

𝐼 = (500)(0 . 08)(2)

Multiply step by step:

500 Β· 0 . 08 = 40

40 Β· 2 = 80

Answer: 80.

Worked example (no skipped steps)

12.14 Solve for 𝑃 in 𝐼 = 𝑃 π‘Ÿπ‘‘.

Divide both sides by π‘Ÿπ‘‘:

𝐼

𝑃 = π‘Ÿπ‘‘

Answer: 𝑃 = 𝐼 (with π‘Ÿ ΜΈ= 0 and 𝑑 ΜΈ= 0).

π‘Ÿπ‘‘

Worked example (no skipped steps)

12.15 Solve for β„Ž in 𝐴 = 1 π‘β„Ž.

2

Multiply both sides by 2:

2 𝐴 = π‘β„Ž

Divide both sides by 𝑏:

2 𝐴

β„Ž = 𝑏

Answer: β„Ž = 2 𝐴 (with 𝑏 ΜΈ= 0).

𝑏

Chapter 12

187

Worked example (no skipped steps)

12.16 Solve for π‘₯ in 𝑦 = π‘Žπ‘₯ + 𝑏.

Subtract 𝑏:

𝑦 βˆ’ 𝑏 = π‘Žπ‘₯

Divide by π‘Ž:

𝑦 βˆ’ 𝑏

π‘₯ =

π‘Ž

Answer: π‘₯ = π‘¦βˆ’ 𝑏 (with π‘Ž ΜΈ= 0).

π‘Ž

Worked example (no skipped steps)

12.17 Solve for π‘Ÿ in 𝐢 = 2 πœ‹π‘Ÿ.

Divide both sides by 2 πœ‹:

𝐢

π‘Ÿ = 2 πœ‹

Answer: π‘Ÿ = 𝐢 .

2 πœ‹

Worked example (no skipped steps)

12.18 Use 𝐢 = 2 πœ‹π‘Ÿ with π‘Ÿ = 3.

Substitute:

𝐢 = 2 πœ‹(3)

Simplify:

𝐢 = 6 πœ‹

Answer: 6 πœ‹.

Worked example (no skipped steps)

12.19 Use 𝑠 = 𝑒𝑑 + 1 π‘Žπ‘‘ 2 with 𝑒 = 4, π‘Ž = 2, 𝑑 = 3.

2

Substitute:

1

𝑠 = (4)(3) +

(2)(32)

2

Simplify each part:

(4)(3) = 12

32 = 9

1 (2)(9) = 1 Β· 9 = 9

2

Now add:

𝑠 = 12 + 9 = 21

Answer: 21.

188

Step-by-step solutions

Worked example (no skipped steps)

12.20 Solve for π‘Ž in 𝑠 = 𝑒𝑑 + 1 π‘Žπ‘‘ 2.

2

Subtract 𝑒𝑑 from both sides:

1

𝑠 βˆ’ 𝑒𝑑 =

π‘Žπ‘‘ 2

2

Multiply both sides by 2:

2( 𝑠 βˆ’ 𝑒𝑑) = π‘Žπ‘‘ 2

Now divide by 𝑑 2:

2( 𝑠 βˆ’ 𝑒𝑑)

π‘Ž =

𝑑 2

Answer: π‘Ž = 2( π‘ βˆ’ 𝑒𝑑) (with 𝑑 ΜΈ= 0).

𝑑 2

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