Example 2 from Algebra and Trigonometry, 2.3 Models and Applications

Setting Up a Linear Equation to Solve a Real-World Application

There are two cell phone companies that offer different packages. Company A charges a monthly service fee of $34 plus $.05/min talk-time. Company B charges a monthly service fee of $40 plus $.04/min talk-time.

ⓐWrite a linear equation that models the packages offered by both companies.
ⓑ If the average number of minutes used each month is 1,160, which company offers the better plan?
ⓒIf the average number of minutes used each month is 420, which company offers the better plan?
ⓓHow many minutes of talk-time would yield equal monthly statements from both companies?

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

1
Solution
1. ⓐThe model for Company A can be written as $A = 0.05 x + 34.$ This includes the variable cost of $0.05 x$ plus the monthly service charge of $34. Company B’s package charges a higher monthly fee of $40, but a lower variable cost of $0.04 x .$ Company B’s model can be written as $B = 0.04 x + $ 40.$
2. ⓑ
  If the average number of minutes used each month is 1,160, we have the following:
  
  $\begin{matrix} Company A & = & 0.05 (1, 160) + 34 \\ = & 58 + 34 \\ = & 92 \\ Company B & = & 0.04 (1, 160) + 40 \\ = & 46.4 + 40 \\ = & 86.4 \end{matrix}$
  So, Company B offers the lower monthly cost of $86.40 as compared with the $92 monthly cost offered by Company A when the average number of minutes used each month is 1,160.
3. ⓒ
  If the average number of minutes used each month is 420, we have the following:
  
  $\begin{matrix} Company A & = & 0.05 (420) + 34 \\ = & 21 + 34 \\ = & 55 \\ Company B & = & 0.04 (420) + 40 \\ = & 16.8 + 40 \\ = & 56.8 \end{matrix}$
  If the average number of minutes used each month is 420, then Company A offers a lower monthly cost of $55 compared to Company B’s monthly cost of $56.80.
4. ⓓ
  To answer the question of how many talk-time minutes would yield the same bill from both companies, we should think about the problem in terms of $(x, y)$ coordinates: At what point are both the x-value and the y-value equal? We can find this point by setting the equations equal to each other and solving for x.
  
  $\begin{matrix} 0.05 x + 34 & = & 0.04 x + 40 \\ 0.01 x & = & 6 \\ x & = & 600 \end{matrix}$
  Check the x-value in each equation.
  
  $\begin{matrix} 0.05 (600) + 34 & = & 64 \\ 0.04 (600) + 40 & = & 64 \end{matrix}$
  
  Therefore, a monthly average of 600 talk-time minutes renders the plans equal. See Figure 2
Figure 2

Your turn

The book's Try It right after this example: same idea, new numbers. Only the answer is given.

Find a linear equation to model this real-world application: It costs ABC electronics company $2.50 per unit to produce a part used in a popular brand of desktop computers. The company has monthly operating expenses of $350 for utilities and $3,300 for salaries. What are the company’s monthly expenses?

How did it go?

Your turn

The book's Try It right after this example: same idea, new numbers. Only the answer is given.

Solution

Your turn

Solution

Your turn