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 [latex]30[/latex] min, or [latex]\dfrac{1}{2}[/latex] h at rate [latex]r[/latex]. His drive home takes [latex]40[/latex] min, or [latex]\dfrac{2}{3}[/latex] h, and his speed averages [latex]10 mi/h[/latex] less than the morning drive. Both trips cover distance [latex]d[/latex]. A table, such as the one below, is often helpful for keeping track of information in these types of problems.

Write two equations, one for each trip.

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

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

The distance between home and work is [latex]20 mi[/latex].Note that we could have cleared the fractions in the equation by multiplying both sides of the equation by the LCD to solve for [latex]r[/latex].

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.

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

The dimensions are [latex]L=15[/latex] ft and [latex]W=12[/latex] ft.

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 [latex]6[/latex] 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.

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

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. 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.

First, isolate the term with w by subtracting 2l from both sides of the equation.

[latex]\displaystyle \begin{array}{l}\,\,\,\,\,\,\,\,\,\,P\,=\,\,\,\,2l+2w\\\underline{\,\,\,\,\,-2l\,\,\,\,\,-2l\,\,\,\,\,\,\,\,\,\,\,}\\P-2l=\,\,\,\,\,\,\,\,\,\,\,\,\,2w\end{array}[/latex]

Next, clear the coefficient of w by dividing both sides of the equation by [latex]2[/latex].

[latex]\displaystyle \begin{array}{l}\underline{P-2l}=\underline{2w}\\\,\,\,\,\,\,2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2\\ \,\,\,\dfrac{P-2l}{2}\,\,=\,\,w\\\,\,\,\,\,\,\,\,\,\,\,w=\dfrac{P-2l}{2}\end{array}[/latex]

You can rewrite the equation so the isolated variable is on the left side.

[latex]w=\dfrac{P-2l}{2}[/latex]

[latex]\begin{array}{l}\,\,\,\,\,\,\,\,A=\dfrac{1}{2}\normalsize bh\\\\\left(2\right)A=\left(2\right)\dfrac{1}{2}\normalsize bh\\\\\,\,\,\,\,\,2A=bh\\\\\,\,\,\,\,\,\,\dfrac{2A}{h}=\frac{bh}{h}\\\\\,\,\,\,\,\,\,\,\dfrac{2A}{h}=\frac{b\cancel{h}}{\cancel{h}}\end{array}[/latex]

Write the equation with the desired variable on the left-hand side as a matter of convention:

[latex]b=\dfrac{2A}{h}[/latex]

[latex]\begin{array}{l}\,\,\,\,\,\,\,\,A=\dfrac{1}{2}\normalsize bh\\\\\left(2\right)A=\left(2\right)\dfrac{1}{2}\normalsize bh\\\\\,\,\,\,\,\,2A=bh\\\\\,\,\,\,\,\,\,\dfrac{2A}{b}=\frac{bh}{b}\\\\\,\,\,\,\,\,\,\,\dfrac{2A}{b}=\frac{h\cancel{b}}{\cancel{b}}\end{array}[/latex]

Write the equation with the desired variable on the left-hand side as a matter of convention:

[latex]h=\dfrac{2A}{b}[/latex]

Substitute the given temperature in[latex]{}^{\circ}{C}[/latex] into the conversion formula:

[latex]12=\left(F-32\right)\cdot\dfrac{5}{9}[/latex]

Isolate the variable F to obtain the equivalent temperature.

[latex]\begin{array}{r}12=\left(F-32\right)\cdot\dfrac{5}{9}\\\\\left(\dfrac{9}{5}\normalsize\right)12=F-32\,\,\,\,\,\,\,\,\,\,\,\,\,\\\\\left(\dfrac{108}{5}\normalsize\right)12=F-32\,\,\,\,\,\,\,\,\,\,\,\,\,\\\\21.6=F-32\,\,\,\,\,\,\,\,\,\,\,\,\,\\\underline{+32\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+32}\,\,\,\,\,\,\,\,\,\,\,\,\\\\53.6={}^{\circ}{F}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\end{array}[/latex]

To isolate the variable F, it would be best to clear the fraction involving F first. Multiply both sides of the equation by [latex]\displaystyle \frac{9}{5}[/latex].

[latex]\begin{array}{l}\\\,\,\,\,\left(\dfrac{9}{5}\normalsize\right)C=\left(F-32\right)\left(\dfrac{5}{9}\normalsize\right)\left(\dfrac{9}{5}\normalsize\right)\\\\\,\,\,\,\,\,\,\,\,\,\,\,\dfrac{9}{5}\normalsize C=F-32\end{array}[/latex]

Add 32 to both sides.

[latex]\begin{array}{l}\dfrac{9}{5}\normalsize\,C+32=F-32+32\\\\\dfrac{9}{5}\normalsize\,C+32=F\\F=\dfrac{9}{5}\normalsize C+32\end{array}[/latex]