Algebra Foundations for Normal People by Gabriel Beguerie - HTML preview

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Chapter 5

Worked example (no skipped steps)

5.1 Simplify 5 ๐‘ฅ + 2 ๐‘ฅ.

5 ๐‘ฅ + 2 ๐‘ฅ = 7 ๐‘ฅ

Worked example (no skipped steps)

5.2 Simplify 7 ๐‘Ž โˆ’ 3 + 2 ๐‘Ž + 8.

Combine like terms:

7 ๐‘Ž + 2 ๐‘Ž = 9 ๐‘Ž,

โˆ’3 + 8 = 5

So:

9 ๐‘Ž + 5

152

Step-by-step solutions

Worked example (no skipped steps)

5.3 Expand 4( ๐‘ฅ + 3).

4 ๐‘ฅ + 12

Worked example (no skipped steps)

5.4 Expand โˆ’3( ๐‘ฆ โˆ’ 2).

โˆ’3 ๐‘ฆ + 6

Worked example (no skipped steps)

5.5 Expand and simplify 2(3 ๐‘ฅ + 1) โˆ’ 4.

2(3 ๐‘ฅ + 1) โˆ’ 4 = 6 ๐‘ฅ + 2 โˆ’ 4 = 6 ๐‘ฅ โˆ’ 2

Worked example (no skipped steps)

5.6 Expand and simplify โˆ’2(4 ๐‘ฅ โˆ’ 5) + 3.

โˆ’8 ๐‘ฅ + 10 + 3 = โˆ’8 ๐‘ฅ + 13

Worked example (no skipped steps)

5.7 Decide whether 3 ๐‘ฅ and 3 are like terms.

No. One has a variable and one does not.

Worked example (no skipped steps)

5.8 Simplify 6 ๐‘š โˆ’ 2 ๐‘š + 9 โˆ’ 4.

6 ๐‘š โˆ’ 2 ๐‘š = 4 ๐‘š,

9 โˆ’ 4 = 5

So the result is

4 ๐‘š + 5

Worked example (no skipped steps)

5.9 Simplify 4 ๐‘ฅ + 7 โˆ’ 2 ๐‘ฅ + 3.

Combine the like terms with ๐‘ฅ:

4 ๐‘ฅ โˆ’ 2 ๐‘ฅ = 2 ๐‘ฅ

Combine the constants:

7 + 3 = 10

So the expression becomes:

2 ๐‘ฅ + 10

Chapter 5

153

Answer: 2 ๐‘ฅ + 10.

Worked example (no skipped steps)

5.10 Expand 3( ๐‘ฅ โˆ’ 5).

Distribute the 3 to each term inside the parentheses:

3( ๐‘ฅ โˆ’ 5) = 3 ยท ๐‘ฅ + 3 ยท (โˆ’5)

So:

3( ๐‘ฅ โˆ’ 5) = 3 ๐‘ฅ โˆ’ 15

Answer: 3 ๐‘ฅ โˆ’ 15.

Worked example (no skipped steps)

5.11 Simplify โˆ’2(4 ๐‘ฆ โˆ’ 1) + 3 ๐‘ฆ.

Distribute โˆ’2 first:

โˆ’2(4 ๐‘ฆ โˆ’ 1) = โˆ’8 ๐‘ฆ + 2

Now add the remaining 3 ๐‘ฆ:

โˆ’8 ๐‘ฆ + 2 + 3 ๐‘ฆ

Combine the like terms:

โˆ’8 ๐‘ฆ + 3 ๐‘ฆ = โˆ’5 ๐‘ฆ

So the result is:

โˆ’5 ๐‘ฆ + 2

Answer: โˆ’5 ๐‘ฆ + 2.

Worked example (no skipped steps)

5.12 Evaluate 2 ๐‘Ž + 5 ๐‘ when ๐‘Ž = 3 and ๐‘ = โˆ’1.

Substitute the values:

2 ๐‘Ž + 5 ๐‘ = 2(3) + 5(โˆ’1)

Multiply:

2(3) = 6 ,

5(โˆ’1) = โˆ’5

Now add:

6 + (โˆ’5) = 1

Answer: 1.

154

Step-by-step solutions

Worked example (no skipped steps)

5.13 Simplify 3 ๐‘ฅ + 5 โˆ’ 2 ๐‘ฅ + 7.

Combine like terms:

3 ๐‘ฅ โˆ’ 2 ๐‘ฅ = ๐‘ฅ

5 + 7 = 12

So:

๐‘ฅ + 12

Answer: ๐‘ฅ + 12.

Worked example (no skipped steps)

5.14 Distribute and simplify 4( ๐‘ฅ โˆ’ 3).

4( ๐‘ฅ โˆ’ 3) = 4 ๐‘ฅ โˆ’ 12

Answer: 4 ๐‘ฅ โˆ’ 12.

Worked example (no skipped steps)

5.15 Distribute and simplify โˆ’2(5 ๐‘ฆ + 1).

โˆ’2(5 ๐‘ฆ + 1) = โˆ’10 ๐‘ฆ โˆ’ 2

Answer: โˆ’10 ๐‘ฆ โˆ’ 2.

Worked example (no skipped steps)

5.16 Simplify 2(3 ๐‘Ž โˆ’ 4) + 5 ๐‘Ž.

Distribute first:

2(3 ๐‘Ž โˆ’ 4) = 6 ๐‘Ž โˆ’ 8

Then combine like terms:

6 ๐‘Ž โˆ’ 8 + 5 ๐‘Ž = 11 ๐‘Ž โˆ’ 8

Answer: 11 ๐‘Ž โˆ’ 8.

Worked example (no skipped steps)

5.17 Simplify 7 โˆ’ (2 ๐‘ฅ โˆ’ 5).

Distribute the negative sign:

7 โˆ’ 2 ๐‘ฅ + 5

Combine constants:

12 โˆ’ 2 ๐‘ฅ

Answer: 12 โˆ’ 2 ๐‘ฅ.

Chapter 6

155

Worked example (no skipped steps)

5.18 Combine like terms: 6 ๐‘š โˆ’ 2 + 4 ๐‘š + 9.

6 ๐‘š + 4 ๐‘š = 10 ๐‘š

โˆ’2 + 9 = 7

Answer: 10 ๐‘š + 7.

Worked example (no skipped steps)

5.19 Simplify 3(2 ๐‘ + 5) โˆ’ 4( ๐‘ โˆ’ 1).

Distribute:

6 ๐‘ + 15 โˆ’ 4 ๐‘ + 4

Combine like terms:

2 ๐‘ + 19

Answer: 2 ๐‘ + 19.

Worked example (no skipped steps)

5.20 Simplify 5 โˆ’ 3( ๐‘ฅ โˆ’ 2).

Distribute the โˆ’3:

5 โˆ’ 3 ๐‘ฅ + 6 = 11 โˆ’ 3 ๐‘ฅ

Answer: 11 โˆ’ 3 ๐‘ฅ.

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