Algebra 1 problem!?!?!?

amelia-richardson_60a322d32eb4camelia-richardson_60a322d32eb4c
in Science & Mathematics

A rental car company has two possible rate plans: $40 a day with unlimited mileage or $15 a day with a charge of $0.50 for each mile driven. How many miles would you need to drive on a three-day business trip to make the two rate plans cost the same?

Update:

Must be solved using algebra! Please use variables and show work.

Comments

  • abhinavabhinav

    40x = 15x + .5y (x is the number of days and y is the number of miles)

    You know x = 3.

    40(3) = 15(3) + .5y

    120 = 45 + .5y

    .5y = 75

    y = 150

    Your answer is 150 miles.

  • mohammed-richardson_60a322d5d045dmohammed-richardson_60a322d5d045d

    We know that 3 days will cost $120 using the first rate plan. So the goal is to take the second rate plan and make it equal to $120.

    - we know that it will be $15 for each of the 3 days and that the total cost must be $120

    15*3 + .5x = 120 where x = the total number of miles

    .5x = 120 - 45

    .5x = 75

    x = 150

  • craig-jones_60a322d6f0516craig-jones_60a322d6f0516

    120-45=65/.5= 130 miles

  • anonymousanonymous

    40 - 15 =25

    25 divided by 0.5= 50 miles

    There is your answer.

Sign In or Register to comment.

© Vanilla Keystone Theme 2024