use the cross products property example 1 write original proportion. 8 15 = x 6 solve the proportion...
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Use the cross products property
EXAMPLE 1
Write original proportion.
8 15 = x 6
Solve the proportion =8 x
615
Cross products property
Simplify.120 = 6x
Divide each side by 6.20 = x
The solution is 20. Check by substituting 20 for x in the original proportion.
ANSWER
=8 x
615
Use the cross products property
EXAMPLE 1
Substitute 20 for x.
CHECK
Cross products property
Simplify. Solution checks.
820
615
=?
8 15 = 20 6?
120 = 120
EXAMPLE 2 Standardized Test Practice
What is the value of x in the proportion = ?4x
83x –
A – 6 B – 3 C 3 D 6
SOLUTION
4x =
8x – 3
Write original proportion.
Cross products property4(x – 3) = x 8
4x – 12 = 8x Simplify.
Subtract 4x from each side. – 12 = 4x
Divide each side by 4.– 3 = x
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
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
x280
= 8 amount of food100 weight of seal
Write a proportion involving two ratios that compare the amount of food with the weight of the seal.
Seals
Write and solve a proportion
EXAMPLE 3
STEP 2Solve the proportion.
8100
x280
= 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 perday.
EXAMPLE 1
Write original proportion.
Solve the proportion. Check your solution.
Cross products property
Simplify.120 = 24a
Divide each side by 24. 5 = a
The solution is 5. Check by substituting 5 for a in the original proportion.
ANSWER
GUIDED PRACTICE for Examples 1,2, and 3
=4 a
2430
=4 a
2430
1.
30 4 = a 24
Use the cross products propertyEXAMPLE 1
Substitute 5 for a.
CHECK
Cross products property
Simplify. Solution checks.
4 5
2430
=?
120 = 120
GUIDED PRACTICE for Examples 1,2, and 3
30 4 = 24 5?
EXAMPLE 2
3x =
2x – 6
Write original proportion.
Cross products property3(x – 6) = 2 x
3x – 18 = 2x Distrubutive property
Subtract 3x from each side. 18 = x
GUIDED PRACTICE for Examples 1,2, and 3
3x =
2x – 6
2.
The value of x is 18. Check by substituting 18 for x in the original proportion.
ANSWER
EXAMPLE 1
Substitute 18 for x.
CHECK
Cross products property
Simplify
318
2 18 – 6
=?
Simplify. Solution checks.
54 – 18 = 36 ?
GUIDED PRACTICE for Examples 1,2, and 3
36 = 36
3(18 – 6) = 18 2?
EXAMPLE 2
m5 =
m – 64
Write original proportion.
Cross products property
4m = 5m – 30 Simplify
Subtract 5m from each side. m = 30
GUIDED PRACTICE for Examples 1,2, and 3
4m5 =
m – 6 3.
m 4 = 5(m – 6)
EXAMPLE 3
4. WHAT IF? In Example 3, suppose the seal weights 260 pounds. How much food should the seal be fed per day?
SOLUTION
STEP 1
8100
x260
= amount of foodweight of seal
Write a proportion involving two ratios that compare the amount of food with the weight of the seal.
GUIDED PRACTICE for Examples 1,2, and 3
Write and solve a proportion
EXAMPLE 3
STEP 2Solve the proportion.
8100
x260
= 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 perday.
GUIDED PRACTICE for Examples 1,2, and 3
SOLUTION
EXAMPLE 4 Use the scale on a map
From the map’s scale, 1 centimeter represents 85 kilometers. On the map, the distance between Cleveland and Cincinnati is about 4.2 centimeters.
Maps
Use a metric ruler and the map of Ohio to estimate the distance between Cleveland and Cincinnati.
EXAMPLE 4 Use the scale on a map
Write and solve a proportion to find the distance d between the cities.
=4.2 d
1 centimeters85 kilometers
Cross products property
d = 357 Simplify.
ANSWER
The actual distance between Cleveland and Cincinnati is about 357 kilometers.
1 d = 85 4.2
SOLUTION
EXAMPLE 4 Use the scale on a mapGUIDED 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.
It is about 212.5 km
EXAMPLE 4 Use the scale on a mapGUIDED PRACTICE for Example 4
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.
Model ships
EXAMPLE 4 Use the scale on a map
Write and solve a proportion to find the length l of the Queen Elizabeth II.
=1.6 l
1600
Cross products property
l = 960 Simplify.
ANSWER
The actual length of the Queen Elizabeth II is about 960 feet.
1 . l = 600 . 1.6
SOLUTION
GUIDED PRACTICE for Example 4