Question:
The cost of a car rental is $42 per day plus 20 cents per mile. you are on a daily budget of $98. write and solve an inequality to find the greatest distance you can drive each day while staying within your budget. use pencil and paper. find 2 other two-step inequalities with the same solutions.
Answer:
(a) The greatest distance is 280 miles
(b) [tex]30 + 0.10 m \le 58[/tex] and [tex]5 m-400 \le 1000[/tex]
Step-by-step explanation:
Given
[tex]d = \$42[/tex] ---->daily rate
[tex]m = 20\ cents[/tex] -- amount per mile
[tex]m = \$0.20[/tex] --- in dollars
[tex]Budget = \$98[/tex]
Solving (a): Inequality for the scenario
Because you are on a budget, it means that the spending can not exceed $98 (i.e. less than or equal to $98).
So, the inequality is:
[tex]42 * days + 0.20 * miles \le Budget[/tex]
Represent days with d, miles with m and substitute 98 for Budget
[tex]42d + 0.20m \le 98[/tex]
Because it is daily, then d = 1
[tex]42 * 1+ 0.20 * m \le 98[/tex]
[tex]42+ 0.20m \le 98[/tex]
Collect like terms
[tex]0.20m \le 98 - 42[/tex]
[tex]0.20m \le 56[/tex]
[tex]m \le \frac{56}{0.20}[/tex]
[tex]m \le 280[/tex]
The greatest distance is 280 miles
Solving (b): Two other inequalities with the same solution as: [tex]m \le 280[/tex]
To do this, all you need to do is to perform the same operation on both sides of the inequality.
[tex]m \le 280[/tex]
Multiply both sides by 0.10
[tex]0.10 * m \le 280 * 0.10[/tex]
[tex]0.10 m \le 28[/tex]
Add 30 to both sides
[tex]30 + 0.10 m \le 28+30[/tex]
[tex]30 + 0.10 m \le 58[/tex] --- equation (1)
Another possible equation is as follows:
[tex]m \le 280[/tex]
Multiply both sides by 5
[tex]5 * m \le 280 * 5[/tex]
[tex]5 m \le 1400[/tex]
Subtract 400 from both sides
[tex]5 m-400 \le 1400-400[/tex]
[tex]5 m-400 \le 1000[/tex] --- equation (2)
