1. ## unit problem

Car costs $14,500 and has a rated mileage of 28 miles/gallon. another car costs$21,700 and has a rated mileage of 19 kilometers/liter. the cost of gasoline is $1.25/gal. How many miles do you have to drive for the lower fuel consumption of the second car to compensate for the higher cost of this car? thanks a lot it's driving me crazy nertil 2. Originally Posted by nertil1 Car costs$14,500 and has a rated mileage of 28 miles/gallon. another car costs $21,700 and has a rated mileage of 19 kilometers/liter. the cost of gasoline is$1.25/gal. How many miles do you have to drive for the lower fuel consumption of the second car to compensate for the higher cost of this car?
thanks a lot
it's driving me crazy
nertil
I'll tell you, but first you have to tell me what 19 kilos per liter are in miles per gallon.

3. 44.7 mi/gal

4. Originally Posted by nertil1
44.7 mi/gal
Originally Posted by nertil1
Car costs $14,500 and has a rated mileage of 28 miles/gallon. another car costs$21,700 and has a rated mileage of 19 kilometers/liter. the cost of gasoline is \$1.25/gal. How many miles do you have to drive for the lower fuel consumption of the second car to compensate for the higher cost of this car?
thanks a lot
it's driving me crazy
nertil
Ok then.

What you need to do is write a function to compare the amount of miles gone ( $x$) to the amount of money paid ( $y$) for both cars.

For the first car, you would spend $\frac{1_{\text{gallon}}}{28_{\text{miles}}}\times\ frac{\ 1.25}{1_{\text{gallon}}}=\frac{\ 1.25}{28_{\text{miles}}}=\frac{\ 5}{112_{\text{miles}}}$

or just 5 dollars for every 112 miles

So the equation for the cost of the first car is: $y=\frac{5}{112}x+14500$

Now your job is to find the equation of the second car. And then see when they intersect...

5. Originally Posted by nertil1
where did 5/122 come from?
it's 1.25/28

that's the smallest fraction it could be without having a decimal place (I hate decimal places in fractions)