Applications of Linear Equations

Example 3

Solving an Application Using a Formula

It takes Andrew $30$ min to drive to work in the morning. He drives home using the same route, but it takes $10$ min longer, and he averages $10 \, \mathrm{mi} / \mathrm{h}$ less than in the morning. How far does Andrew drive to work?

Analysis

Note that we could have cleared the fractions in the equation by multiplying both sides of the equation by the LCD to solve for $r$.

$\begin{aligned} r\left(\frac{1}{2}\right) &=(r-10)\left(\frac{2}{3}\right) \\ 6 \times r\left(\frac{1}{2}\right) &=6 \times(r-10)\left(\frac{2}{3}\right) \\ 3 r &=4(r-10) \\ 3 r &=4 r-40 \\-r &=-40 \\ r &=40 \end{aligned}$

Try It #3

On Saturday morning, it took Jennifer $3.6 \, h$ to drive to her mother's house for the weekend. On Sunday evening, due to heavy traffic, it took Jennifer $4 \, h$ to return home. Her speed was $5$ mi/h slower on Sunday than on Saturday. What was her speed on Sunday?

Example 4

Solving a Perimeter Problem

The perimeter of a rectangular outdoor patio is $54 \, \mathrm{ft}$. The length is $3 \, \mathrm{ft}$ greater than the width. What are the dimensions of the patio?

Solution

The perimeter formula is standard: $P=2 L+2 W$. We have two unknown quantities, length, and width. However, we can write the length in terms of the width as $L=W+3$. 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 3.

Figure 3

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

The dimensions are $L=15 \, \mathrm{ft}$ and $W=12 \, \mathrm{ft}$.

Try It #4

Find the dimensions of a rectangle given that the perimeter is $110 \, \mathrm{~cm}$ and the length is $1 \mathrm{~cm}$ more than twice the width.

Example 5

Solving an Area Problem

The perimeter of a tablet of graph paper is $48$ in. The length is $6$ in. more than the width. Find the area of the graph paper.

Solution

The standard formula for area is $A=L W$; 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 $L=W+6$. Substitute the value of the perimeter and the expression for length into the perimeter formula and find the length.

$\begin{aligned} P &=2 L+2 W \\ 48 &=2(W+6)+2 W \\ 48 &=2 W+12+2 W \\ 48 &=4 W+12 \\ 36 &=4 W \\ 9 &=W \\(9+6) &=L \\ 15 &=L \end{aligned}$

Now, we find the area given the dimensions of $L=15$ in. and $W=9$ in.

$\begin{aligned} A &=L W \\ A &=15(9) \\ &=135 \, \text { in. }^{2} \end{aligned}$

The area is $135$ in. $^{2}$.

Try It #5

A game room has a perimeter of $70 \, \mathrm{ft}$. The length is five more than twice the width. How many $\mathrm{ft}^{2}$ of new carpeting should be ordered?

Example 6

Solving a Volume Problem

Find the dimensions of a shipping box given that the length is twice the width, the height is $8$ inches, and the volume is $1,600 \, in.^{3}$.

Solution

The formula for the volume of a box is given as $V=L W H$, the product of length, width, and height. We are given that $L=2 W$, and $H=8$. The volume is 1,600 cubic inches.

$\begin{aligned} V &=L W H \\ 1,600 &=(2 W) W(8) \\ 1,600 &=16 W^{2} \\ 100 &=W^{2} \\ 10 &=W \end{aligned}$

The dimensions are $L=20$ in., $W=10$ in., and $H=8$ in.

Analysis

Note that the square root of $W^{2}$ would result in a positive and a negative value. However, because we are describing width, we can use only the positive result.