Respuesta :
You may solve this problem using a system of equations. Let the daily fee be x and the amount charged per mile be y.
This makes the equation for the first road trip:
226 = 4x + 350y
and the equation for the second road trip:
52 = x + 75y
rendering the following system of equations:
226 = 4x + 350y
52 = x + 75y
To solve this system using the substitution method, first isolate x in the second equation.
52 = x + 75y
52 - 75y = x
x = 52 - 75y
Substitute 52 - 75y for x into the first equation and solve algebraically for y.
226 = 4x + 350y
226 = 4(52 - 75y) + 350y
226 = 208 - 300y + 350y
226 = 208 + 50y
18 = 50y
18/50 = y
9/25 = y
y = 9/25
Substitute 9/25 for y into either of the original equations and solve algebraically for x.
52 = x + 75y
52 = x + 75(9/25)
52 = x + 675/25
52 = x + 27
25 = x
x = 25
Answer:
The company charged a daily fee of $25 and a they charged $0.36 per mile to rent the vehicle.
This makes the equation for the first road trip:
226 = 4x + 350y
and the equation for the second road trip:
52 = x + 75y
rendering the following system of equations:
226 = 4x + 350y
52 = x + 75y
To solve this system using the substitution method, first isolate x in the second equation.
52 = x + 75y
52 - 75y = x
x = 52 - 75y
Substitute 52 - 75y for x into the first equation and solve algebraically for y.
226 = 4x + 350y
226 = 4(52 - 75y) + 350y
226 = 208 - 300y + 350y
226 = 208 + 50y
18 = 50y
18/50 = y
9/25 = y
y = 9/25
Substitute 9/25 for y into either of the original equations and solve algebraically for x.
52 = x + 75y
52 = x + 75(9/25)
52 = x + 675/25
52 = x + 27
25 = x
x = 25
Answer:
The company charged a daily fee of $25 and a they charged $0.36 per mile to rent the vehicle.