A phone company offers two monthly plans. Plan A costs plus an additional for each minute of calls. Plan B costs plus an additional for each minute of calls.

Question

A phone company offers two monthly plans. Plan A costs $25

 plus an additional $0.12

 for each minute of calls. Plan B costs $19

 plus an additional $0.14

 for each minute of calls.

Answer 1

I would set this up as an inequality for Plan A to be less than or equal to Plan B..

A ≤B

Substitute the respective expressions, in minutes, m. for A and for B.

A=$25+$0.12m

B=$19+$0.14m

Solve for m to establish what range of minutes cause A ≤B to be true, A less costly or the same as Plan B

The range of minutes that make A ≤B false would be the range of minutes causing plan A to be more costly than Plan B.

A>B.

The value of m that makes A=B would be "a wash".

A=B

Your solutions will be in the form of

either m≤$a or m≥$a for A≤B

and

either m<$a or m>$a for A>B

and

m=$a for A=B

where a is the specific number in your solution for m in each case of A compared with B.

Finally put your answer in the words of the problem rather than algebraically:

For Example, but not the actual solution to the problem, which I leave to the student to do without my doing the problem for the student to copy:

"Plan A is less costly than Plan B when the number of minutes is greater than or equal to 25."

Answer 2

A= 25+.12m

B= 19+.14m

25+.12m= 19+.14m

6 =.02m

m=6/.02=300 minutes

m>300, A is cheaper

m<300, B is cheaper

m=300 they cost the same