Many applications are solved using known formulas. The problem is stated, a formula is identified, the known quantities are substituted into the formula, the equation is solved for the unknown, and the problem’s question is answered. Sometimes, these problems involve two equations representing two unknowns, which can be written using one equation in one variable by expressing one unknown in terms of the other.
Examples of formulas include the perimeter of a rectangle, [latex]P=2L+2W[/latex]; the area of a rectangular region, [latex]A=LW[/latex]; and the volume of a rectangular solid, [latex]V=LWH[/latex].The distance [latex]d[/latex] covered when traveling at a constant rate [latex]r[/latex] for some time [latex]t[/latex] is given by the formula [latex]d=rt[/latex].It takes Andrew [latex]30[/latex] minutes to drive to work in the morning. He drives home using the same route, but it takes [latex]10[/latex] minutes longer, and he averages [latex]10[/latex] mi/h less than in the morning. How far does Andrew drive to work?
Step 1: Define the Variables Let:
[latex]d[/latex] be the distance to work (in miles).
[latex]r[/latex] be the morning driving speed (in miles per hour).
Morning trip time: [latex]30[/latex] minutes.
Afternoon trip time, which is [latex]10[/latex] minutes longer: [latex]40[/latex] minutes.
Step 2: Convert Times to Hours
Since the rate is in miles per hour, convert the time from minutes to hours:
Morning trip time [latex]= \frac{30}{60} = \frac{1}{2}[/latex] hours
Afternoon trip time [latex]= \frac{40}{60} = \frac{2}{3}[/latex] hours
Thus, Andrew drives [latex]20[/latex] miles to work.
How were the fractions handled in the example above?
Recall that when solving multi-step equations, it is helpful to multiply by the LCD to clear the denominators from the equation. But it is also permissible to use operations on fractions to combine like terms. The example above demonstrates both.
In the text of the solution when solving for [latex]r[/latex] the first time, the parentheses were eliminated using the distributive property, then operations on fractions were used to combine like terms.
Later, in the analysis of the solution, the LCD between the denominators [latex]2 \text{ and } 3[/latex] was multiplied on both sides of the equation to cancel out the denominators so that operations on fractions were not necessary. Do you see how the LCD wasn’t actually multiplied through on both sides, but that the denominators cancelled out, resulting in a linear equation in one variable without denominators?
Both methods are equally correct unless your instructor requires you to specifically demonstrate knowledge of one or the other.
Which do you prefer?
How To: Solve Multi-Step Equations
(Optional) Multiply to clear any fractions or decimals.
Simplify each side by clearing parentheses and combining like terms.
Add or subtract to isolate the variable term—you may have to move a term with the variable.
Multiply or divide to isolate the variable.
Check the solution.
The perimeter of a rectangular outdoor patio is [latex]54[/latex] ft. The length is [latex]3[/latex] ft. greater than the width. What are the dimensions of the patio?
Given: [latex]P = 54[/latex]ft. and [latex]L = W+3[/latex] ft.The relationship can be expressed and solved as follows:
[latex]\begin{align*} \text{Given the perimeter formula:} && P &= 2L + 2W \\ \text{Substitute } P=54 \text{ and } L = W + 3: && 54 &= 2(W + 3) + 2W \\ \text{Expand and simplify:} && 54 &= 2W + 6 + 2W \\ \text{Combine like terms:} && 54 &= 4W + 6 \\ \text{Solve for } W: && 48 &= 4W \\ && W &= 12 \\ \text{Solve for } L: && L &= W + 3 = 12 + 3 = 15 \end{align*}[/latex]
Thus, the dimensions are [latex]L=15[/latex] ft and [latex]W=12[/latex] ft.
The perimeter of a tablet of graph paper is [latex]48[/latex] inches[latex]^2[/latex]. The length is [latex]6[/latex] inches more than the width. Find the area of the graph paper.
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.
[latex]\\[/latex]
We know that the length is [latex]6[/latex] inches more than the width, so we can write length as [latex]L=W+6[/latex].
[latex]\\[/latex]
Substitute the value of the perimeter and the expression for length into the perimeter formula and find the length.
The area is [latex]135[/latex] inches[latex]^2[/latex].
Find the dimensions of a shipping box given that the length is twice the width, the height is [latex]8[/latex] inches, and the volume is [latex]1,600[/latex] in.3.
The formula for the volume of a box is given as [latex]V=LWH[/latex], the product of length, width, and height. We are given that [latex]L=2W[/latex], and [latex]H=8[/latex]. The volume is [latex]1,600[/latex] cubic inches.
The dimensions are [latex]L=20[/latex] in., [latex]W=10[/latex] in., and [latex]H=8[/latex] in.
Analysis of the Solution
Note that the square root of [latex]{W}^{2}[/latex] would result in a positive and a negative value. However, because we are describing width, we can use only the positive result.
The relationship can be expressed and solved as follows: