We've updated our
Privacy Policy effective December 15. Please read our updated Privacy Policy and tap

Study Guides > College Algebra CoRequisite Course

Using and Manipulating Formulas

Learning Outcomes

  • Recognize formulas for perimeter, area, and volume
  • Recognize the formula that relates distance, rate, and time
  • Rearrange formulas to isolate specific variables
Many real-world applications (problems that can be modeled by a mathematical equation and solved for an unknown quantity) involve formulas that describe relationships between quantities. For example, wall-to-wall carpeting is sold by the square foot. To find out how much carpeting would cover the floor of a 12' x 15' foot room, we can multiply the length and width of the room to discover how many square feet of carpeting is required.  12×15=18012 \times 15 = 180. So, we would need to purchase 180 square feet of carpeting. Examples of formulas that commonly occur in applications include:
  • the perimeter of a rectangle of length L and width W
    • P=2L+2WP=2L+2W
  • the area of a rectangular region of length L and width W
    • A=LWA=LW
  • the volume of a rectangular solid with length L, width W, and height H
    • V=LWHV=LWH
  • the distance dd covered when traveling at a constant rate rr for some time tt
    • d=rtd=rt.
Formulas such as these may be used to solve problems by substituting known values and solving for an unknown value. You should know these formulas and be able to recognize when to apply them to a problem.

Isolate Variables in Formulas

Sometimes, we need to isolate one of the variables in a formula in order to solve for the unknown. In the carpeting example above, we found that it would take 180 square feet of carpet to cover a 12' x 15' room. We knew the dimensions of the room and solved for how many square feet of carpet we needed, But what if, instead, we had tiles that we could use to cover the floor and we knew the length of the room, but needed to know what width room could be fully tiled? Let's say you have a banquet hall with one wall that can slide to create a room of variable width. The length of the room (the length of the movable wall) is 85 feet. If you have 4,590 square feet of tile, what must the width be to ensure the entire room is fully tiled? Use the formula A=LWA=LW, but solve it for W, width. AL=W\dfrac{A}{L}=W. Then, substitute what you know: 459085=W=54\dfrac{4590}{85} = W = 54. It looks like we need to slide the wall out to create a width of 54 feet in order to use all the tile. . This technique, isolating a variable of choice in any formula, is especially helpful if you have to perform the same calculation repeatedly, or you are having a computer perform the calculation repeatedly, for different values of the unknown variable.

Example: isolate a variable in a formula

Isolate the variable for width ww from the formula for the perimeter of a rectangle:  

P=2(l)+2(w){P}=2\left({l}\right)+2\left({w}\right).

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

          P =    2l+2w     −2l     −2l           ‾P−2l=             2w \displaystyle \begin{array}{l}\,\,\,\,\,\,\,\,\,\,P\,=\,\,\,\,2l+2w\\\underline{\,\,\,\,\,-2l\,\,\,\,\,-2l\,\,\,\,\,\,\,\,\,\,\,}\\P-2l=\,\,\,\,\,\,\,\,\,\,\,\,\,2w\end{array}

Next, clear the coefficient of w by dividing both sides of the equation by 22.

P−2l‾=2w‾      2                2   P−2l2  =  w           w=P−2l2\displaystyle \begin{array}{l}\underline{P-2l}=\underline{2w}\\\,\,\,\,\,\,2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2\\ \,\,\,\Large\frac{P-2l}{2}\,\,=\,\,w\\\,\,\,\,\,\,\,\,\,\,\,w=\Large\frac{P-2l}{2}\end{array}

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

w=P−2l2w=\Large\frac{P-2l}{2}

Example: isolate a variable in a formula

  Use the multiplication and division properties of equality to isolate the variable b given A=12bhA=\Large\frac{1}{2}\normalsize bh

Answer:

        A=12bh(2)A=(2)12bh      2A=bh       2Ah=bhh        2Ah=bhh\begin{array}{l}\,\,\,\,\,\,\,\,A=\Large\frac{1}{2}\normalsize bh\\\\\left(2\right)A=\left(2\right)\Large\frac{1}{2}\normalsize bh\\\\\,\,\,\,\,\,2A=bh\\\\\,\,\,\,\,\,\,\Large\frac{2A}{h}=\frac{bh}{h}\\\\\,\,\,\,\,\,\,\,\Large\frac{2A}{h}=\frac{b\cancel{h}}{\cancel{h}}\end{array}

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

b=2Ahb=\Large\frac{2A}{h}

Use the multiplication and division properties of equality to isolate the variable h given A=12bhA=\Large\frac{1}{2}\normalsize bh

Answer:

        A=12bh(2)A=(2)12bh      2A=bh       2Ab=bhb        2Ab=hbb\begin{array}{l}\,\,\,\,\,\,\,\,A=\Large\frac{1}{2}\normalsize bh\\\\\left(2\right)A=\left(2\right)\Large\frac{1}{2}\normalsize bh\\\\\,\,\,\,\,\,2A=bh\\\\\,\,\,\,\,\,\,\Large\frac{2A}{b}=\frac{bh}{b}\\\\\,\,\,\,\,\,\,\,\Large\frac{2A}{b}=\frac{h\cancel{b}}{\cancel{b}}\end{array}

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

h=2Abh=\Large\frac{2A}{b}

Think About It

Isolate the variable for height, hh, from the formula for the surface area of a cylinder, s=2πrh+2πr2s=2\pi rh+2\pi r^{2}. In this example, the variable h is buried pretty deeply in the formula for surface area of a cylinder. Using the order of operations, it can be isolated. Before you look at the solution, use the box below to write down what you think is the best first step to take to isolate h. [practice-area rows="1"][/practice-area]

Answer: Isolate the term containing the variable h by subtracting 2πr22\pi r^{2}from both sides.

S  =2πrh+2πr2−2πr2             −2πr2‾S−2πr2    =    2πrh              \begin{array}{r}S\,\,=2\pi rh+2\pi r^{2} \\ \underline{-2\pi r^{2}\,\,\,\,\,\,\,\,\,\,\,\,\,-2\pi r^{2}}\\S-2\pi r^{2}\,\,\,\,=\,\,\,\,2\pi rh\,\,\,\,\,\,\,\,\,\,\,\,\,\,\end{array}

Next, isolate the variable h by dividing both sides of the equation by 2πr2\pi r.

S−2πr22πr=2πrh2πrS−2πr22πr=h          \begin{array}{r}\Large\frac{S-2\pi r^{2}}{2\pi r}=\frac{2\pi rh}{2\pi r} \\\\\Large\frac{S-2\pi r^{2}}{2\pi r}=h\,\,\,\,\,\,\,\,\,\,\end{array}

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

h=S−2πr22πrh=\Large\frac{S-2\pi r^{2}}{2\pi r}

Hopefully, you can see the value in being able to isolate a variable of interest! Here is another formula for which the skill is particularly useful. The formula for converting from the Fahrenheit temperature scale to the Celsius scale is given by C=(F−−32)⋅59C=\left(F--32\right)\cdot\Large\frac{5}{9}. This formula takes a temperature in Fahrenheit, FF, and coverts it to an equivalent temperature in Celsius, CC.

Example: use a formula to convert fahrenheit to celsius

Given a temperature of 12∘C12^{\circ}{C}, find the equivalent in ∘F{}^{\circ}{F}.

Answer: Substitute the given temperature in∘C{}^{\circ}{C} into the conversion formula:

12=(F−32)⋅5912=\left(F-32\right)\cdot\Large\frac{5}{9}

Isolate the variable F to obtain the equivalent temperature.

12=(F−32)⋅59(95)12=F−32             (1085)12=F−32             21.6=F−32             +32                +32‾            53.6=∘F                        \begin{array}{r}12=\left(F-32\right)\cdot\Large\frac{5}{9}\\\\\left(\Large\frac{9}{5}\normalsize\right)12=F-32\,\,\,\,\,\,\,\,\,\,\,\,\,\\\\\left(\Large\frac{108}{5}\normalsize\right)12=F-32\,\,\,\,\,\,\,\,\,\,\,\,\,\\\\21.6=F-32\,\,\,\,\,\,\,\,\,\,\,\,\,\\\underline{+32\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+32}\,\,\,\,\,\,\,\,\,\,\,\,\\\\53.6={}^{\circ}{F}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\end{array}

But what if we have the temperature in Celsius already, and wish to convert it to Fahrenheit? We can solve the formula for FF. This will give us a formula that takes a temperature on the Celsius scale and converts it to an equivalent Fahrenheit temperature.

Example: isolate a variable to convert celsius to fahrenheit

Solve the formula shown below for converting from the Fahrenheit scale to the Celsius scale for F. Then use the new formula to convert a temperature of 21∘C21^{\circ}{C} to Fahrenheit. C=(F−−32)⋅59C=\left(F--32\right)\cdot\Large\frac{5}{9}

Answer: To isolate the variable F, it would be best to clear the fraction involving F first. Multiply both sides of the equation by 95 \displaystyle \frac{9}{5}.

    (95)C=(F−32)(59)(95)            95C=F−32\begin{array}{l}\\\,\,\,\,\left(\Large\frac{9}{5}\normalsize\right)C=\left(F-32\right)\left(\Large\frac{5}{9}\normalsize\right)\left(\Large\frac{9}{5}\normalsize\right)\\\\\,\,\,\,\,\,\,\,\,\,\,\,\Large\frac{9}{5}\normalsize C=F-32\end{array}

Add 32 to both sides.

95 C+32=F−32+3295 C+32=FF=95C+32\begin{array}{l}\Large\frac{9}{5}\normalsize\,C+32=F-32+32\\\\\Large\frac{9}{5}\normalsize\,C+32=F\\F=\Large\frac{9}{5}\normalsize C+32\end{array}

And we can convert 21∘C21^{\circ}{C}  by substituting it into the new formula: F=95(21)+32F=\dfrac{9}{5}(21)+32 21∘C=69.8∘F21^{\circ}{C}=69.8^{\circ}{F}

Licenses & Attributions