Let [latex]x[/latex] equal the first number. Then, as the second number exceeds the first by [latex]17[/latex], we can write the second number as [latex]x+17.[/latex] The sum of the two numbers is [latex]31[/latex]. We usually interpret the word is as an equal sign.

[latex]\begin{array}{ccc}\hfill x+\left(x+17\right)& =& 31\hfill \\ \hfill 2x+17& =& 31\phantom{\rule{2em}{0ex}}\text{Simplify and solve}\text{.}\hfill \\ \hfill 2x& =& 14\hfill \\ \hfill x& =& 7\hfill \\ \hfill\phantom{\rule{1em}{0ex}} x+17& =& 7+17\hfill \\ & =& 24\hfill \end{array}[/latex]

The two numbers are [latex]7[/latex] and [latex]24.[/latex]

The model for Company A can be written as [latex]A=0.05x+34.[/latex] This includes the variable cost of [latex]0.05x[/latex] plus the monthly service charge of [latex]\$34[/latex]. Company B’s package charges a higher monthly fee of [latex]\$40[/latex], but a lower variable cost of [latex]0.04x.[/latex] Company B’s model can be written as [latex]B=0.04x+\text{\$}40.[/latex]

If the average number of minutes used each month is [latex]1{,}160[/latex], we have the following:

[latex]\begin{array}{ccc}\hfill \text{Company }A& =& 0.05\left(1{,}160\right)+34\hfill \\ & =& 58+34\hfill \\ & =& 92\hfill \\ \phantom{\rule{1em}{0ex}}\hfill \text{Company }B& =& 0.04\left(1{,}160\right)+40\hfill \\ & =& 46.4+40\hfill \\ \hfill & =& 86.4\hfill \end{array}[/latex]

So, Company B offers the lower monthly cost of [latex]\$86.40[/latex] as compared with the [latex]\$92[/latex] monthly cost offered by Company A when the average number of minutes used each month is [latex]1{,}160[/latex].

If the average number of minutes used each month is [latex]420[/latex], we have the following:

[latex]\begin{array}{ccc}\hfill \text{Company }A& =& 0.05\left(420\right)+34\hfill \\ & =& 21+34\hfill \\ & =& 55\hfill \\ \phantom{\rule{1em}{0ex}}\hfill \text{Company }B& =& 0.04\left(420\right)+40\hfill \\ & =& 16.8+40\hfill \\ & =& 56.8\hfill \end{array}[/latex]

If the average number of minutes used each month is [latex]420[/latex], then Company A offers a lower monthly cost of [latex]\$55[/latex] compared to Company B’s monthly cost of [latex]\$56.80[/latex].

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 [latex]\left(x,y\right)[/latex] coordinates: At what point are both the [latex]x[/latex]-value and the [latex]y[/latex]-value equal? We can find this point by setting the equations equal to each other and solving for [latex]x[/latex].

[latex]\begin{array}{ccc}\hfill 0.05x+34& =& 0.04x+40\hfill \\ \hfill 0.01x& =& 6\hfill \\ \hfill x& =& 600\hfill \end{array}[/latex]

Check the [latex]x[/latex]-value in each equation.

[latex]\begin{array}{ccc}\hfill 0.05\left(600\right)+34& =& 64\hfill \\ \hfill 0.04\left(600\right)+40& =& 64\hfill \end{array}[/latex]

Therefore, a monthly average of [latex]600[/latex] talk-time minutes renders the plans equal. See Figure 4.

This is a distance problem, so we can use the formula [latex]d=rt[/latex] where distance equals rate multiplied by time. Note that when rate is given in mi/h, time must be expressed in hours. Consistent units of measurement are key to obtaining a correct solution.

First, we identify the known and unknown quantities. Andrew’s morning drive to work takes 30 min, or [latex]\frac{1}{2}[/latex] h at rate [latex]r.[/latex] His drive home takes 40 min, or [latex]\frac{2}{3}[/latex] h, and his speed averages 10 mi/h less than the morning drive. Both trips cover distance [latex]d.[/latex] A table, such as Table 2, is often helpful for keeping track of information in these types of problems.

Table 2

Write two equations, one for each trip.

[latex]\begin{array}{cccc}\hfill d& =& r\left(\frac{1}{2}\right)\hfill & \phantom{\rule{2em}{0ex}}\text{To work}\hfill \\ \hfill d& =& \left(r-10\right)\left(\frac{2}{3}\right)\hfill & \phantom{\rule{2em}{0ex}}\text{To home}\hfill \end{array}[/latex]

As both equations equal the same distance, we set them equal to each other and solve for [latex]r[/latex].

We have solved for the rate of speed to work, 40 mph. Substituting 40 into the rate on the return trip yields 30 mi/h. Now we can answer the question. Substitute the rate back into either equation and solve for [latex]d[/latex].

[latex]\begin{array}{cc}\hfill d& = 40\left(\frac{1}{2}\right)\hfill \\ & = 20\hfill \end{array}[/latex]

The distance between home and work is 20 mi.

The perimeter formula is standard: [latex]P=2L+2W[/latex]. We have two unknown quantities, length and width. However, we can write the length in terms of the width as [latex]L=W+3[/latex]. Substitute the perimeter value and the expression for length into the formula. It is often helpful to make a sketch and label the sides, as in Figure 5.

Now we can solve for the width and then calculate the length.

[latex]\begin{array}{cc}\hfill P& = 2L+2W\hfill \\ \hfill 54& = 2\left(W+3\right)+2W\hfill \\ \hfill 54& = 2W+6+2W\hfill \\ \hfill 54& = 4W+6\hfill \\ \hfill 48& = 4W\hfill \\ \hfill 12& = W\hfill \\ \hfill \left(12+3\right)& = L\hfill \\ \hfill 15& = L\hfill \end{array}[/latex]

The dimensions are [latex]L=15\text{ ft}[/latex] and [latex]W=12\text{ ft}[/latex].

The standard formula for area is [latex]A=LW[/latex]; however, we will solve the problem using the perimeter formula. The reason we use the perimeter formula is because we know enough information about the perimeter that the formula will allow us to solve for one of the unknowns. As both perimeter and area use length and width as dimensions, they are often used together to solve a problem such as this one.

We know that the length is 6 in. more than the width, so we can write length as [latex]L=W+6.[/latex] Substitute the value of the perimeter and the expression for length into the perimeter formula and find the length.

[latex]\begin{array}{cc}\hfill P& = 2L+2W\hfill \\ \hfill 48& = 2\left(W+6\right)+2W\hfill \\ \hfill 48& = 2W+12+2W\hfill \\ \hfill 48& = 4W+12\hfill \\ \hfill 36& = 4W\hfill \\ \hfill 9& = W\hfill \\ \hfill \left(9+6\right)& = L\hfill \\ \hfill 15& = L\hfill \end{array}[/latex]

Now, we find the area given the dimensions of [latex]L=15\text{ in.}[/latex] and [latex]W=9\text{ in.}[/latex]

[latex]\begin{array}{cc}\hfill A& = LW\hfill \\ \hfill A& = 15\left(9\right)\hfill \ & = \hfill 135\text{ in}{\text{.}}^{2}\end{array}[/latex]

The area is [latex]135\text{ in}^2.[/latex]