A certain car model cost Birr 20,000 with a gasoline engine and Br 25,000 with a diesel engine. The number of miles per gallon of fuel for cars with these two engines is 25 and 30, respectively. Assume that the price of both types of fuel is Birr 1.50 per gallon.

Required:
a) Drive the equation for the cost of driving a gasoline powered car.
b) Drive the equation for the cost of driving a diesel powered car.
c) Find the break-even point, that is, find the mileage at which the diesel-powered car becomes more economical than the gasoline powered car.

number of miles driven = x

cost to drive the gas car = 20000 + x/20 * 1.5 = 3x/40 + 20000
cost to drive the diesel - 25000 + x/30*1.5 = x/20 + 25000

3x/40 + 20000 = x/20 + 25000
3x/40 - x/20 = 5000
x/40 = 5000
x = 20,000

oops, copy error !!

I used 20 instead of 25 in line #2

Make the necessary changes, or else solve oobleck's equation

good

A family has two cars. The first car has a fuel efficiency of 20 miles per gallon of gas and the second has a fuel efficiency of 30 miles per gallon of gas. During one particular week, the two cars went a combined total of 1300 miles, for a total gas consumption of 50 gallons. How many gallons were consumed by each of the two cars that week?

a) The cost of driving a gasoline powered car can be calculated using the equation:

Cost(gasoline) = (Miles driven / Miles per gallon) * Price of fuel per gallon

b) The cost of driving a diesel powered car can be calculated using the equation:
Cost(diesel) = (Miles driven / Miles per gallon) * Price of fuel per gallon

c) To find the break-even point, we need to set the cost of driving a diesel powered car equal to the cost of driving a gasoline powered car and solve for the mileage:
(Miles driven / 25) * 1.50 = (Miles driven / 30) * 1.50

To answer these questions, we need to consider the cost of purchasing the car, the cost of fuel, and the fuel efficiency of each engine type.

a) To drive the equation for the cost of driving a gasoline-powered car, we need to take into account the cost of the car and the cost of fuel. The cost of the car is given as Birr 20,000. The cost of fuel is Birr 1.50 per gallon, and the fuel efficiency is 25 miles per gallon.

Let x be the number of miles driven. The cost of fuel for a gasoline-powered car can be calculated by dividing the number of miles driven (x) by the fuel efficiency (25) and then multiplying it by the cost of fuel (Birr 1.50 per gallon).
Therefore, the equation for the cost of driving a gasoline-powered car is:
Cost of driving a gasoline-powered car = Birr 20,000 + (x / 25) * 1.50

b) Similarly, the equation for the cost of driving a diesel-powered car can be derived. The cost of the car is Birr 25,000. The cost of fuel is still Birr 1.50 per gallon, but the fuel efficiency is 30 miles per gallon.

Let x be the number of miles driven. The cost of fuel for a diesel-powered car can be calculated by dividing the number of miles driven (x) by the fuel efficiency (30) and then multiplying it by the cost of fuel (Birr 1.50 per gallon).
Therefore, the equation for the cost of driving a diesel-powered car is:
Cost of driving a diesel-powered car = Birr 25,000 + (x / 30) * 1.50

c) To find the break-even point, we need to determine the mileage at which the diesel-powered car becomes more economical than the gasoline-powered car.

We can set the two equations for the cost of driving each type of car equal to each other and solve for x (the number of miles driven):
Birr 20,000 + (x / 25) * 1.50 = Birr 25,000 + (x / 30) * 1.50

Simplifying the equation:
(x / 25) * 1.50 - (x / 30) * 1.50 = 25,000 - 20,000

Combining like terms:
(1.50/25 - 1.50/30) * x = 5,000

Simplifying the coefficients:
0.06x = 5,000

Solving for x:
x = 5,000 / 0.06

Therefore, the break-even point is reached after driving approximately 83,333 miles. At this mileage, the diesel-powered car becomes more economical than the gasoline-powered car.

"derive" not "drive"

cost in $/mi is just mi/(mi/gal) * $/gal
so, the cost of driving x miles is
(a) 20000 + x/25 * 1.50 = 0.06x + 20000
do the same for (b) and then solve for (c)