McGraw Hill Glencoe Algebra 1, 2012
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McGraw Hill Glencoe Algebra 1, 2012 View details
5. Inequalities Involving Absolute Value
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Exercise 7 Page 314

The absolute value of the difference between the stock trading price and $70.85 cannot be greater than $0.75.

{m | 70.10≤ m≤ 71.60}

Practice makes perfect
To calculate the difference between the purchase price of Jerome's stock, $70.85, and one of the price fluctuations, we would subtract the original price from the current price. Current price-Original price=Price fluctuation If we call the current price m and substitute the given original price, we can specify our expression. m-70.85=price fluctuation

We are told that when the stock drops below Jerome's original purchasing price, it drops no more than $0.75. Furthermore, when the stock rises it rises no more than $0.75. Therefore, m-70.85 is greater than or equal to - 0.75 and less than or equal to 0.75. We can express this inequality in two different ways. Absolute Value:& |m-70.85|≤ 0.75 Compound:& - 0.75 ≤ m-70.85 ≤ 0.75 Note that there is a third way to write this and type of compound inequality. - 0.75≤ m-70.85 and m-70.85 ≤ 0.75 Let's isolate m in both of these cases before recombining them to form the final solution set.

Case 1

Let's look at the case for the price dropping, - 0.75≤ m-70.85.
- 0.75≤ m-70.85
70.10≤ m
This inequality is true when m is greater than or equal to 70.10.

Case 2

Let's look at the case for the price rising, m-70.85≤ 0.75.
m-70.85≤ 0.75
m ≤ 71.60
This inequality is true when m is less than or equal to 71.60.

Range of Stock Prices

The solution to this type of compound inequality is the intersection of the solution sets. First Solution Set:& 70.10 ≤ m Second Solution Set:& m ≤ 71.60 Intersecting Solution Set:& 70.10 ≤ m ≤ 71.60 The value of Jerome's stock lies in the interval $70.10≤ m≤ $71.60. We can then write the solution set in set-builder notation, as {m | 70.10≤ m≤ 71.60}.