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EXAMPLE 1 a. –5(–7) = 35 b. –8(2) = –16 c. –12(0) = 0

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Presentation on theme: "EXAMPLE 1 a. –5(–7) = 35 b. –8(2) = –16 c. –12(0) = 0"— Presentation transcript:

1 EXAMPLE 1 a. –5(–7) = 35 b. –8(2) = –16 c. –12(0) = 0
Multiplying Integers a. –5(–7) = 35 The product of two integers with the same sign is positive. b. –8(2) = –16 The product of two integers with different signs is negative. c. –12(0) = 0 The product of an integer and 0 is 0.

2 b. Evaluate xyz when x = 2, y = –4, and z = 6.
EXAMPLE 2 Evaluating Variable Expressions a. Evaluate a2 when a = –3. b. Evaluate xyz when x = 2, y = –4, and z = 6. SOLUTION a. a2 = (–3)2 Substitute –3 for a. = –3(–3) Write –3 as a factor two times. = 9 Multiply –3 and –3. b. xyz = 2(–4)(6) Substitute 2 for x, –4 for y, and 6 for z. = –8(6) Multiply 2 and –4. = –48 Multiply –8 and 6.

3 EXAMPLE 3 Using Integer Multiplication Greenland Most of Greenland is covered with ice that is almost two miles thick in some places. Scientists estimate that 3 feet of this ice melts each year. Find the change in the thickness of the ice after 10 years.

4 Change in ice thickness = –3(10) = –30
EXAMPLE 3 Using Integer Multiplication SOLUTION You can find the total change in the ice thickness by multiplying the yearly change by the number of years. Use – 3 for the yearly change because the thickness of the ice decreases by 3 feet each year. Change in ice thickness = –3(10) = –30 ANSWER The thickness of the ice will decrease 30 feet in 10 years.

5 GUIDED PRACTICE Find the product. 1. –9(2) 3. 5(–5) ANSWER –18 ANSWER
for Examples 1, 2, and 3 Find the product. 1. –9(2) (–5) ANSWER –18 ANSWER –25 2. –3(–4) (–14) ANSWER 12 ANSWER

6 5. Evaluate the expressions x2y and –3xy2 when x = –2 and y = 3.
GUIDED PRACTICE for Examples 1, 2, and 3 5. Evaluate the expressions x2y and –3xy2 when x = –2 and y = 3. ANSWER 12, 54 What If? What is the change in the thickness of the ice in Example 3 after 18 years? 6. ANSWER – 54 ft


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