Solve a Formula for a Specific Variable

Solve a formula for a specific variable

We have all probably worked with some geometric formulas in our study of mathematics. Formulas are used in so many fields, it is important to recognize formulas and be able to manipulate them easily.

It is often helpful to solve a formula for a specific variable. If you need to put a formula in a spreadsheet, it is not unusual to have to solve it for a specific variable first. We isolate that variable on one side of the equals sign with a coefficient of one and all other variables and constants are on the other side of the equal sign.

Geometric formulas often need to be solved for another variable, too. The formula $V = \tfrac{1}{3}\pi r^2h$ is used to find the volume of a right circular cone when given the radius of the base and height. In the next example, we will solve this formula for the height.

Example. Solve the formula $V = \tfrac{1}{3}\pi r^2h$ for $h$.

We could now use this formula to find the height of a right circular cone when we know the volume and the radius of the base, by using the formula $h = \tfrac{3V}{\pi r^2}$.

In the sciences, we often need to change temperature from Fahrenheit to Celsius or vice versa. If you travel in a foreign country, you may want to change the Celsius temperature to the more familiar Fahrenheit temperature.

Example. Solve the formula $C = \tfrac{5}{9}(F-32)$ for $F$.

We can now use the formula $F = \tfrac{9}{5}C+32$ to find the Fahrenheit temperature when we know the Celsius temperature.

The next example uses the formula for the surface area of a right cylinder.

Example. Solve the formula $S = 2\pi r^2+2\pi rh$ for $h$.

Sometimes we might be given an equation that is solved for $y$ and need to solve it for $x$, or vice versa. In the following example, we’re given an equation with both $x$ and $y$ on the same side and we’ll solve it for $y$.

Example. Solve the formula $8x+7y=15$ for $y$.

Use formulas to solve geometry applications

In this objective we will use some common geometry formulas. We will adapt our problem solving strategy so that we can solve geometry applications. The geometry formula will name the variables and give us the equation to solve.

In addition, since these applications will all involve shapes of some sort, most people find it helpful to draw a figure and label it with the given information. We will include this in the first step of the problem solving strategy for geometry applications.

When we solve geometry applications, we often have to use some of the properties of the figures. We will review those properties as needed.

The next example involves the area of a triangle. The area of a triangle is one-half the base times the height. We can write this as $A=\tfrac{1}{2}bh$, where $b=$ length of the base and $h=$ height.

Example. The area of a triangular painting is 126 square inches. The base is 18 inches. What is the height?

Solution.

1. Read the problem.
2. Identify what you are looking for: height of a triangle.
3. Name. Let $h=$ the height. Draw the figure and label it with the given information: area $=126$ square inches and base $=18$ inches.
4. Translate. Write the appropriate formula and substitute: $126=\tfrac{1}{2}\cdot18\cdot h$.
5. Solve the equation: $126=9h$, so $14=h$.
6. Check: $126\overset{?}{=}\tfrac{1}{2}\cdot18\cdot14$; $126=126\checkmark$.
7. Answer: The height of the triangle is 14 inches.

In the next example, we will work with a right triangle. To solve for the measure of each angle, we need to use two triangle properties. In any triangle, the sum of the measures of the angles is $180^\circ$. We can write this as a formula: $m\angle A+m\angle B+m\angle C=180$. Also, since the triangle is a right triangle, we remember that a right triangle has one $90^\circ$ angle.

Here, we will have to define one angle in terms of another. We will wait to draw the figure until we write expressions for all the angles we are looking for.

Example. The measure of one angle of a right triangle is 40 degrees more than the measure of the smallest angle. Find the measures of all three angles.

Solution.

1. Read the problem.
2. Identify what you are looking for: the measures of all three angles.
3. Name. Let $a=$ first angle, $a+40=$ second angle, and $90=$ third angle (the right angle).
4. Translate: $a+(a+40)+90=180$.
5. Solve: $2a+130=180$, $2a=50$, and $a=25$. The second angle is $25+40=65$; the third angle is $90$.
6. Check: $25+65+90\overset{?}{=}180$; $180=180\checkmark$.
7. Answer: The three angles measure $25^\circ$, $65^\circ$, and $90^\circ$.

The next example uses another important geometry formula. The Pythagorean Theorem tells how the lengths of the three sides of a right triangle relate to each other. Writing the formula in every exercise and saying it aloud as you write it may help you memorize the Pythagorean Theorem.

We will use the Pythagorean Theorem in the next example.

Example. Use the Pythagorean Theorem to find the length of the other leg in the right triangle whose leg is 12 and whose hypotenuse is 13.

Solution.

1. Read the problem.
2. Identify what you are looking for: the length of the leg of the triangle.
3. Name. Let $a=$ the leg of the triangle. Label side $a$.
4. Translate. Write the appropriate formula. Substitute.

5. Solve the equation. Isolate the variable term. Use the definition of square root. Simplify.
6. Check: $5^2+12^2\overset{?}{=}13^2$; $25+144\overset{?}{=}169$; $169=169\checkmark$.
7. Answer: The length of the leg is 5.

The next example is about the perimeter of a rectangle. Since the perimeter is just the distance around the rectangle, we find the sum of the lengths of its four sides—the sum of two lengths and two widths. We can write it as $P=2L+2W$ where $L$ is the length and $W$ is the width. To solve the example, we will need to define the length in terms of the width.

Example. The length of a rectangle is six centimeters more than twice the width. The perimeter is 96 centimeters. Find the length and width.

Solution.

1. Read the problem.
2. Identify what we are looking for: the length and the width.
3. Name. Choose a variable to represent the width. The length is six more than twice the width. Let $W=$ width and $2W+6=$ length. Draw the figure and label it with the given information: $P=96$ cm.
4. Translate. Write the appropriate formula. Substitute in the given information.

5. Solve the equation. The length is $2W+6=2(14)+6=34$.
6. Check: $96\overset{?}{=}2\cdot34+2\cdot14$; $96=96\checkmark$.
7. Answer: The length is 34 cm and the width is 14 cm.

The next example is about the perimeter of a triangle. Since the perimeter is just the distance around the triangle, we find the sum of the lengths of its three sides. We can write this as $P=a+b+c$, where $a$, $b$, and $c$ are the lengths of the sides.

Example. One side of a triangle is three inches more than the first side. The third side is two inches more than twice the first. The perimeter is 29 inches. Find the length of the three sides of the triangle.

Solution.

1. Read the problem.
2. Identify what we are looking for: the lengths of the three sides of a triangle.
3. Name. Choose a variable to represent the length of the first side. Let $x=$ length of first side, $x+3=$ length of second side, and $2x+2=$ length of third side. Draw the figure and label it with the given information.
4. Translate. Write the appropriate formula. Substitute in the given information: $P=a+b+c$ and $29=x+(x+3)+(2x+2)$.
5. Solve the equation: $29=4x+5$, $24=4x$, and $6=x$, the length of the first side. The second side is $x+3=6+3=9$. The third side is $2x+2=2\cdot6+2=14$.
6. Check: $29\overset{?}{=}6+9+14$; $29=29\checkmark$.
7. Answer: The lengths of the sides of the triangle are 6, 9, and 14 inches.

Example. The perimeter of a rectangular soccer field is 360 feet. The length is 40 feet more than the width. Find the length and width.

Solution.

1. Read the problem.
2. Identify what we are looking for: the length and width of the soccer field.
3. Name. Choose a variable to represent it. The length is 40 feet more than the width. Draw the figure and label it with the given information. Let $w=$ width and $w+40=$ length; perimeter $=360$ feet.
4. Translate. Write the appropriate formula and substitute: $P=2L+2W$ and $360=2(w+40)+2w$.
5. Solve the equation: $360=2w+80+2w$, $360=4w+80$, $280=4w$, and $70=w$, the width of the field. The length of the field is $w+40=70+40=110$.
6. Check: $360\overset{?}{=}2(110)+2(70)$; $360=360\checkmark$.
7. Answer: The length of the soccer field is 110 feet and the width is 70 feet.

Applications of these geometric properties can be found in many everyday situations as shown in the next example.

Example. Kelvin is building a gazebo and wants to brace each corner by placing a 10-inch piece of wood diagonally as shown. How far from the corner should he fasten the wood if he wants the distances from the corner to be equal? Approximate to the nearest tenth of an inch.

Solution.

1. Read the problem.
2. Identify what we are looking for: the distance from the corner that the bracket should be attached.
3. Name. Choose a variable to represent it. Draw the figure and label it with the given information. Let $x=$ the distance from the corner.
4. Translate. Write the appropriate formula and substitute.

5. Solve the equation. Isolate the variable. Use the definition of square root. Simplify. Approximate to the nearest tenth.
6. Check: $a^2+b^2=c^2$ and $(7.1)^2+(7.1)^2\approx10^2$, yes.
7. Answer: Kelvin should fasten each piece of wood approximately 7.1 inches from the corner.

Key terms: Pythagorean Theorem.

Adapted from OpenStax Intermediate Algebra 2e, Section 2.3, by Lynn Marecek and Andrea Honeycutt Mathis, © OpenStax, licensed under CC BY-NC-SA 4.0. Access the original for free at openstax.org. Changes: adapted the section to this site’s format and converted Try It exercises into interactive checks.