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Use the cross products property

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

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Use the cross products property

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  1. Solve the proportion = 8 6 8 6 15 x x 15 = 8 15 = x 6 ANSWER The solution is20. Check by substituting 20 forxin the original proportion. EXAMPLE 1 Use the cross products property Write original proportion. Cross products property 120 = 6x Simplify. 20=x Divide each side by 6.

  2. 8 6 = 20 15 ? ? 8 15=20 6 120 = 120 EXAMPLE 1 Use the cross products property CHECK Substitute 20 for x. Cross products property Simplify. Solution checks.

  3. 8 What is the value of x in the proportion =? x– 3 3 – 6 – 3 6 B 4 4 8 = A x x x – 3 D C 4(x – 3) =x8 EXAMPLE 2 Standardized Test Practice SOLUTION Write originalproportion. Cross products property 4x – 12 = 8x Simplify. – 12 = 4x Subtract 4xfrom each side. – 3 = x Divide each side by 4.

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

  5. x 280 8 amount of food = weight of seal 100 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.

  6. = 8 280 = 100 x x ANSWER 8 280 100 A280pound seal should be fed22.4pounds of food per day. EXAMPLE 3 Write and solve a proportion STEP 2 Solve the proportion. Write proportion. Cross products property 2240 = 100x Simplify. 22.4 = x Divide each side by 100.

  7. 4 4 a a 24 1. = 30 24 = 30 30 4 = a 24 ANSWER The solution is5. Check by substituting 5 forain the original proportion. EXAMPLE 1 GUIDED PRACTICE for Examples 1,2, and 3 Solve the proportion. Check your solution. Write original proportion. Cross products property 120 = 24a Simplify. 5=a Divide each side by 24.

  8. 24 4 = 30 5 ? 30 4 =24 5 ? 120 = 120 EXAMPLE 1 for Examples 1,2, and 3 Use the cross products property GUIDED PRACTICE CHECK Substitute 5 for a. Cross products property Simplify. Solution checks.

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

  10. 3 18 ? 2 = 18 – 6 3(18 – 6) =18 2 ? ? 36 = 36 54 – 18 = 36 EXAMPLE 1 for Examples 1,2, and 3 GUIDED PRACTICE CHECK Substitute 18 for x. Cross products property Simplify Simplify. Solution checks.

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

  12. x 8 260 amount of food 100 = weight of seal EXAMPLE 3 for Examples 1,2, and 3 GUIDED PRACTICE • 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.

  13. = 8 260 = 100 x x ANSWER 8 260 100 A260pound seal should be fed20.8pounds of food per day. EXAMPLE 3 GUIDED PRACTICE for Examples 1,2, and 3 Write and solve a proportion STEP 2 Solve the proportion. Write proportion. Cross products property 2080 = 100x Simplify. 20.8 = x Divide each side by 100.

  14. Use a metric ruler and the map of Ohio to estimate the distance between Cleveland and Cincinnati. Use the scale on a map EXAMPLE 4 Maps 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.

  15. 1 centimeters = 85 kilometers 1 d = 85 4.2 ANSWER 4.2 The actual distance between Cleveland and Cincinnati is about 357 kilometers. d Use the scale on a map EXAMPLE 4 Write and solve a proportion to find the distance dbetween the cities. Cross products property d = 357 Simplify.

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

  17. 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. Use the scale on a map EXAMPLE 4 for Example 4 GUIDED PRACTICE Model ships

  18. 1 = 600 ANSWER 1.6 The actual length of the Queen Elizabeth II is about 960 feet. l Use the scale on a map EXAMPLE 4 for Example 4 GUIDED PRACTICE SOLUTION Write and solve a proportion to find the length lof the Queen Elizabeth II. 1 . l = 600 . 1.6 Cross products property l = 960 Simplify.

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