Answer:
![200 + 0.07m \leq 230](https://img.qammunity.org/2021/formulas/mathematics/high-school/a6794jkc4qfj1z0yodg9galervvyexh5iw.png)
A maximum of
428 miles is the distance what Divya can travel.
Explanation:
Given that:
Rent to be paid for the car in the weekend = $200
Charges to be paid per mile = $0.07
Total money available with Divya = $230
To find:
The inequality as per her limitations and solution to the problem.
Solution:
Let the number of miles for which Divya can drive =
miles
Charges for one mile = $0.07
Charges for
miles = $0.07
![m](https://img.qammunity.org/2021/formulas/mathematics/middle-school/id5t45gvtu7zzm4ok70kfwzrd93p5069g2.png)
Total charges for renting and
miles = Rental charges + Operational charges
Total charges for renting and
miles = $200 + $0.07
![m](https://img.qammunity.org/2021/formulas/mathematics/middle-school/id5t45gvtu7zzm4ok70kfwzrd93p5069g2.png)
These are charges must be lesser than equal to the amount of money available with Divya.
Therefore, we can write:
![200 + 0.07m \leq 230](https://img.qammunity.org/2021/formulas/mathematics/high-school/a6794jkc4qfj1z0yodg9galervvyexh5iw.png)
Subtracting 200 from both the sides:
![0.07m \leq 30](https://img.qammunity.org/2021/formulas/mathematics/high-school/8dwlmloojlbrlnf63s3we2cpjc2qa1627h.png)
Dividing both sides with 0.07:
![m \leq (30)/(0.07)\\m\leq 428.6](https://img.qammunity.org/2021/formulas/mathematics/high-school/s8onefk8v4ti6wd879tbtofofkf6muemp1.png)
Therefore, a maximum of
428 miles is the distance what Divya can travel.