# One airplane costs four times as much as a certain car. Two such planes cost $6000 more than six of the cars. Find the cost of each.

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One airplane costs four times as much as a certain car. Two such planes cost $6000 more than six of the cars:

Let x represent the cost of a car. Let y represent the cost of an airplane.

Then y=4x (One airplane costs as much as 4 cars.)

2y=6x+6000 (Two airplanes cost 6000 more than 6 of the cars.) ==> y=3x+3000

We have a system of linear equations in two variables:

y=4x

y=3x+3000

We can use substitution: substitute the expression 3x+3000 for the y in the first equation:

3x+3000=4x

3000=x ==> y=12000

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The cost of the car is $3000 and the cost of the airplane is $12000

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Check: one car costs $3000 and an airplane costs 4 times as much or $12000.

Two airplanes (2(12000)=$24000) costs 6000 more than 6 times the price of a car; 24000=6000+6(3000)=6000+18000.

One airplane costs four times as much as a certain car. Two such planes cost $6000 more than six of the cars. Find the cost of each.

so P=4C

2P=6000+6C

Multiply the (p=4c) by 2

2P=8C

- 2P=6C+6000

2C-6000=0

2C=6000

C=3000

P=4C

P=12000