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A telephone company offers two plans with per-minute charges. Plan A involves a monthly rental of $12, and call charges at 7¢ per minute. Plan B involves a monthly rental of $15, and call charges at 5¢ per minute.

Write an inequality in terms of the number of minutes which shows when Plan A is less expensive than Plan B. Solve the inequality, showing the steps in your work.

User IWheelBuy
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2 Answers

3 votes
7m+12<5m+15
7m-5m+12<5m-5m+15
2m+12<15
2m+12-12<15-12
2m<3
2m/2<3/2
m<3/2
User Cocreature
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4 votes

Answer:


12+0.07x< 15+0.05x

Explanation:

Let x be the no. of minutes

Plan A

Monthly rental = $12

Call charges for 1 minute = 7¢

1 cents = 0.01 dollars

So, Call charges for 1 minute = $0.07

Call charges for x minutes = 0.07 x

Total cost of Plan A = 12+0.07x

Plan B

Monthly rental = $15

Call charges for 1 minute = 5¢

1 cents = 0.01 dollars

So, Call charges for 1 minute = $0.05

Call charges for x minutes = 0.05 x

Total cost of Plan B = 15+0.05x

An inequality in terms of the number of minutes which shows when Plan A is less expensive than Plan B=
12+0.07x< 15+0.05x

Solving the inequality :


0.07x-0.05x< 15-12


0.02x< 3


x<(3)/(0.02)


x<150

So, the no. of minutes must be less than 150 for Plan A to be less expensive than Plan B

Hence an inequality in terms of the number of minutes which shows when Plan A is less expensive than Plan B is
12+0.07x< 15+0.05x

User E L Rayle
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7.9k points