Use the cross products property

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Presentation transcript:

Use the cross products property EXAMPLE 1 Use the cross products property Solve the proportion = 8 x 6 15 = 8 x 6 15 Write original proportion. 8 15 = x 6 Cross products property 120 = 6x Simplify. 20 = x Divide each side by 6. The solution is 20. Check by substituting 20 for x in the original proportion. ANSWER

Use the cross products property EXAMPLE 1 Use the cross products property CHECK 8 20 6 15 = ? Substitute 20 for x. 8 15 = 20 6 ? Cross products property 120 = 120 Simplify. Solution checks.

Standardized Test Practice EXAMPLE 2 Standardized Test Practice What is the value of x in the proportion = ? 4 x 8 3 x – C 3 A – 6 B – 3 D 6 SOLUTION 4 x = 8 – 3 Write original proportion. 4(x – 3) = x 8 Cross products property 4x – 12 = 8x Simplify. – 12 = 4x Subtract 4x from each side. – 3 = x Divide each side by 4.

EXAMPLE 2 Standardized Test Practice The value of x is – 3. The correct answer is B. ANSWER A B C D

Write and solve a proportion EXAMPLE 3 Write and solve a proportion Seals Each day, the seals at an aquarium are each fed 8 pounds of food for every 100 pounds of their body weight. A seal at the aquarium weighs 280 pounds. How much food should the seal be fed per day ? SOLUTION STEP 1 Write a proportion involving two ratios that compare the amount of food with the weight of the seal. x 280 = 8 amount of food 100 weight of seal

Write and solve a proportion EXAMPLE 3 Write and solve a proportion STEP 2 Solve the proportion. 8 100 x 280 = Write proportion. 8 280 = 100 x Cross products property 2240 = 100x Simplify. 22.4 = x Divide each side by 100. ANSWER A 280 pound seal should be fed 22.4 pounds of food per day.

Solve the proportion. Check your solution. EXAMPLE 1 GUIDED PRACTICE for Examples 1,2, and 3 Solve the proportion. Check your solution. = 4 a 24 30 1. = 4 a 24 30 Write original proportion. 30 4 = a 24 Cross products property 120 = 24a Simplify. 5 = a Divide each side by 24. The solution is 5. Check by substituting 5 for a in the original proportion. ANSWER

Use the cross products property for Examples 1,2, and 3 GUIDED PRACTICE Use the cross products property for Examples 1,2, and 3 CHECK 4 5 24 30 = ? Substitute 5 for a. 30 4 = 24 5 ? Cross products property 120 = 120 Simplify. Solution checks.

The value of x is 18. Check by substituting 18 for x in the EXAMPLE 2 GUIDED PRACTICE for Examples 1,2, and 3 3 x = 2 – 6 2. 3 x = 2 – 6 Write original proportion. 3(x – 6) = 2 x Cross products property 3x – 18 = 2x Distrubutive property 18 = x Subtract 3x from each side. The value of x is 18. Check by substituting 18 for x in the original proportion. ANSWER

EXAMPLE 1 GUIDED PRACTICE for Examples 1,2, and 3 CHECK 3 18 2 18 – 6 = ? Substitute 18 for x. 3(18 – 6) = 18 2 ? Cross products property 54 – 18 = 36 ? Simplify 36 = 36 Simplify. Solution checks.

EXAMPLE 2 GUIDED PRACTICE for Examples 1,2, and 3 4 m 5 = m – 6 3. m 5 Write original proportion. m 4 = 5(m – 6) Cross products property 4m = 5m – 30 Simplify m = 30 Subtract 5m from each side.

EXAMPLE 3 GUIDED PRACTICE for Examples 1,2, and 3 WHAT IF? In Example 3, suppose the seal weights 260 pounds. How much food should the seal be fed per day? SOLUTION STEP 1 Write a proportion involving two ratios that compare the amount of food with the weight of the seal. 8 100 x 260 = amount of food weight of seal

Write and solve a proportion for Examples 1,2, and 3 GUIDED PRACTICE Write and solve a proportion for Examples 1,2, and 3 STEP 2 Solve the proportion. 8 100 x 260 = Write proportion. 8 260 = 100 x Cross products property 2080 = 100x Simplify. 20.8 = x Divide each side by 100. ANSWER A 260 pound seal should be fed 20.8 pounds of food per day.

EXAMPLE 4 Use the scale on a map Maps Use a metric ruler and the map of Ohio to estimate the distance between Cleveland and Cincinnati. SOLUTION From the map’s scale, 1 centimeter represents 85 kilometers. On the map, the distance between Cleveland and Cincinnati is about 4.2 centimeters.

Use the scale on a map EXAMPLE 4 Write and solve a proportion to find the distance d between the cities. = 4.2 d 1 centimeters 85 kilometers 1 d = 85 4.2 Cross products property d = 357 Simplify. ANSWER The actual distance between Cleveland and Cincinnati is about 357 kilometers.

EXAMPLE 4 Use the scale on a map GUIDED PRACTICE for Example 4 5. Use a metric ruler and the map in Example 4 to estimate the distance (in kilometers) between Columbus and Cleveland. SOLUTION It is about 212.5 km

EXAMPLE 4 Use the scale on a map GUIDED PRACTICE for Example 4 Model ships 6. The ship model kits sold at a hobby store have a scale of 1 ft : 600 ft. A completed model of the Queen Elizabeth II is 1.6 feet long. Estimate the actual length of the Queen Elizabeth II.

The actual length of the Queen Elizabeth II is about 960 feet. EXAMPLE 4 Use the scale on a map GUIDED PRACTICE for Example 4 SOLUTION Write and solve a proportion to find the length l of the Queen Elizabeth II. 1 600 = 1.6 l 1 . l = 600 . 1.6 Cross products property l = 960 Simplify. ANSWER The actual length of the Queen Elizabeth II is about 960 feet.