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Pearson Algebra 1 Common Core, 2011
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Pearson Algebra 1 Common Core, 2011 View details
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Exercise 24 Page 73

Organize the different cases you can have for the statement under different conditions. Then test which of them make the statement true.

| a -b | = | a | - | b | is true when | a | ≥ | b | and both numbers have the same sign.

Practice makes perfect

We are asked to find when the absolute value of a difference of two numbers is equal to the difference of their absolute values. Let's write this algebraically so that we can better visualize the meaning of the statement. | a -b | = | a | - | b | If we try to analyze the statement, we will see that it can get confusing due to the many possibilities. One way to solve this is to organize all these different cases using example values for each case.

Condition | a | > | b |

First, we will consider that the absolute value of the first number is greater than the absolute value of the second. Under this condition, we can still have different cases: both numbers are positive, both are negative, or one is positive and one is negative. Let's work out the case when both are negative using the example values a=-8 and b=-3.
| a -b |
SubstituteII

a= -8, b= -3

| -8 -( -3) |
SubNeg

a-(- b)=a+b

| -8 +3 |
AddTerms

Add terms

| -5 |
AbsNeg

|-5|=5

5
Now let's work out the right-hand side of the statement.
| a | - | b |
SubstituteII

a= -8, b= -3

| -8 | - | -3 |
AbsNeg

|-a|=a

8 - 3
SubTerm

Subtract term

5
We will continue to do this for all the possible different cases.
Case | a -b | | a | - | b | | a -b | ? = | a | - | b |
a, b <0 | -8 - (-3) | = 5 | -8 | - | (-3) | = 5 âś“
a>0, b <0 | 8 - (-3) | = 11 | 8 | - | (-3) | = 5 *
a, b >0 | 8 - 3 | = 5 | 8 | - | 3 | = 5 âś“
a<0, b >0 | -8 - 3 | = 11 | -8 | - | 3 | = 5 *

Condition | b | > | a |

Now, let's consider the possibilities when | b | > | a |.

Case | a -b | | a | - | b | | a -b | ? = | a | - | b |
a, b <0 | -3 - (-8)| = 5 | -3|- | (-8)| = -5 *
a>0, b <0 |3 - (-8) | = 11 | 3 | - | (-8) | = -5 *
a, b >0 |3 - 8 | = 5 | 3 | - |8 | = -5 *
a<0, b >0 | -3 - 8 | = 11 | -3 | - | 8 | = -5 *

We can see that there is no possible case that makes the statement true when | b | > | a |.

Condition | a | = | b |

Finally let's consider | a | = | b |.

Case | a -b | | a | - | b | | a -b | ? = | a | - | b |
a, b <0 | -3 - (-3) | = 0 | -3 | - | (-3) | = 0 âś“
a>0, b <0 | 3 - (-3) | = 6 | 3 | - | (-3) | = 0 *
a, b >0 | 3 - 3 | = 0 | 3 | - | 3 | = 0 âś“
a<0, b >0 | -3 - 3 | = 6 | -3 | - | 3 | = 0 *

Conclusions

Looking at the results from the previous sections we can see that the statement is true only when

  • | a | > | b | and both numbers have the same sign.
  • | a | = | b | and both numbers have the same sign.

We can summarize these results and say that the statement is true when | a | ≥ | b | and both numbers have the same sign.

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arrow_right Exercises p.73
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Exercises 2 (Page 73)
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Exercises 4 (Page 73)
Exercises 5 (Page 73)
Exercises 6 (Page 73)
Exercises 7 (Page 73)
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Exercises 10 (Page 73)
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Exercises 12 (Page 73)
Exercises 13 (Page 73)
Exercises 14 (Page 73)
Exercises 15 (Page 73)
Exercises 16 (Page 73)
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Exercises 18 (Page 73)
Exercises 19 (Page 73)
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Exercises 22 (Page 73)
Exercises 23 (Page 73)
Exercises 24 (Page 73)