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Solve a Formula for a Specific Variable

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 $V=\frac{1}{3}\pi {r}^{2}h$ 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

Try it.

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

Solution

Write the formula.
Remove the fraction on the right.
Simplify.
Divide both sides by $\pi {r}^{2}.$

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 $h=\frac{3V}{\pi {r}^{2}}.$

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

Try it.

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

Solution

Write the formula.
Remove the fraction on the right.
Simplify.
Add 32 to both sides.

We can now use the formula $F=\frac{9}{5}C+32$ 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

Try it.

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

Solution

Write the formula.
Isolate the h term by subtracting $2\pi {r}^{2}$ from each side.
Simplify.
Solve for h by dividing both sides by $2\pi r.$
Simplify.

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.

Condensed — the full section is in OpenStax Intermediate Algebra 2e.

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 $A=\frac{1}{2}bh,$ where b = length of the base and h = height.

Example

Try it.

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

Solution

Step 1. Read the problem.
Step 2. Identify what you are looking for.	height of a triangle
Step 3. Name.
Choose a variable to represent it.	Let $h=$ the height.
Draw the figure and label it with the given information.	Area = 126 sq. in.

Step 4. Translate.
Write the appropriate formula.	$\ A=\frac{1}{2}bh$
Substitute in the given information.	$\ 126=\frac{1}{2}\cdot 18\cdot h$
Step 5. Solve the equation.	$\ 126=9h$
Divide both sides by 9.	$\ 14=h$
Step 6. Check. $\begin{array}{lll}A & = & \frac{1}{2}bh \\ 126 & \overset{?}{=} & \frac{1}{2}\cdot 18\cdot 14 \\ 126 & = & 126✓\end{array}$
Step 7. Answer the question.	The height of the triangle is 14 inches.

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.

Condensed — the full section is in OpenStax Intermediate Algebra 2e.

Key Concepts

How To Solve Geometry Applications
1. Read the problem and make sure all the words and ideas are understood.
2. Identify what you are looking for.
3. Name what you are looking for by choosing a variable to represent it. Draw the figure and label it with the given information.
4. Translate into an equation by writing the appropriate formula or model for the situation. Substitute in the given information.
5. Solve the equation using good algebra techniques.
6. Check the answer in the problem and make sure it makes sense.
7. Answer the question with a complete sentence.
The Pythagorean Theorem
- In any right triangle, where a and b are the lengths of the legs, and c is the length of the hypotenuse, the sum of the squares of the lengths of the two legs equals the square of the length of the hypotenuse.

Solve a Formula for a Specific Variable

Solve a Formula for a Specific Variable

In the following exercises, solve the given formula for the specified variable.

Try it.

Solve the formula $C=\pi d$ for d.

Solution

$d=\frac{C}{\pi }$

Try it.

Solve the formula $C=\pi d$ for $\pi .$

Try it.

Solve the formula $V=LWH$ for L.

Solution

$L=\frac{V}{WH}$

Try it.

Solve the formula $V=LWH$ for H.

Try it.

Solve the formula $A=\frac{1}{2}bh$ for b.

Solution

$b=\frac{2A}{h}$

Try it.

Solve the formula $A=\frac{1}{2}bh$ for h.

Try it.

Solve the formula
$A=\frac{1}{2}{d}_{1}{d}_{2}$ for ${d}_{1}.$

Solution

${d}_{1}=\frac{2A}{{d}_{2}}$

Try it.

Solve the formula
$A=\frac{1}{2}{d}_{1}{d}_{2}$ for ${d}_{2}.$

Try it.

Solve the formula
$A=\frac{1}{2}h({b}_{1}+{b}_{2})$ for ${b}_{1}.$

Solution

${b}_{1}=\frac{2A}{h}-{b}_{2}$

Try it.

Solve the formula
$A=\frac{1}{2}h({b}_{1}+{b}_{2})$ for ${b}_{2}.$

Try it.

Solve the formula
$h=54t+\frac{1}{2}a{t}^{2}$ for a.

Solution

$a=\frac{2h-108t}{{t}^{2}}$

Try it.

Solve the formula
$h=48t+\frac{1}{2}a{t}^{2}$ for a.

Try it.

Solve $180=a+b+c$ for a.

Solution

$a=180-b-c$

Try it.

Solve $180=a+b+c$ for c.

Try it.

Solve the formula
$A=\frac{1}{2}pl+B$ for p.

Solution

$p=\frac{2A-2B}{l}$

Try it.

Solve the formula
$A=\frac{1}{2}pl+B$ for l.

Try it.

Solve the formula
$P=2L+2W$ for L.

Solution

$L=\frac{P-2W}{2}$

Try it.

Solve the formula
$P=2L+2W$ for W.

In the following exercises, solve for the formula for y.

Try it.

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

Solution

$y=15-8x$

Try it.

Solve the formula
$9x+y=13$ for y.

Try it.

Solve the formula
$-4x+y=-6$ for y.

Solution

$y=-6+4x$

Try it.

Solve the formula
$-5x+y=-1$ for y.

Try it.

Solve the formula
$x-y=-4$ for y.

Solution

$y=4+x$

Try it.

Solve the formula
$x-y=-3$ for y.

Try it.

Solve the formula
$4x+3y=7$ for y.

Solution

$y=\frac{7-4x}{3}$

Try it.

Solve the formula
$3x+2y=11$ for y.

Try it.

Solve the formula
$2x+3y=12$ for y.

Solution

$y=\frac{12-2x}{3}$

Try it.

Solve the formula
$5x+2y=10$ for y.

Try it.

Solve the formula
$3x-2y=18$ for y.

Solution

$y=\frac{18-3x}{-2}$

Try it.

Solve the formula
$4x-3y=12$ for y.

Use Formulas to Solve Geometry Applications

In the following exercises, solve using a geometry formula.

Try it.

A triangular flag has area 0.75 square feet and height 1.5 foot. What is its base?

Solution

1 foot

Try it.

A triangular window has area 24 square feet and height six feet. What is its base?

Try it.

What is the base of a triangle with area 207 square inches and height 18 inches?

Solution

23 inches

Try it.

What is the height of a triangle with area 893 square inches and base 38 inches?

Try it.

The two smaller angles of a right triangle have equal measures. Find the measures of all three angles.

Solution

$45\text{^{\circ}},45\text{^{\circ}},90\text{^{\circ}}$

Try it.

The measure of the smallest angle of a right triangle is $20\text{^{\circ}}$ less than the measure of the next larger angle. Find the measures of all three angles.

Try it.

The angles in a triangle are such that one angle is twice the smallest angle, while the third angle is three times as large as the smallest angle. Find the measures of all three angles.

Solution

$30\text{^{\circ}},60\text{^{\circ}},90\text{^{\circ}}$

Try it.

The angles in a triangle are such that one angle is 20 more than the smallest angle, while the third angle is three times as large as the smallest angle. Find the measures of all three angles.

In the following exercises, use the Pythagorean Theorem to find the length of the hypotenuse.

Try it.

Solution

Try it.

Solution

Try it.

In the following exercises, use the Pythagorean Theorem to find the length of the leg. Round to the nearest tenth if necessary.

Try it.

Solution

Try it.

Solution

Try it.

Solution

$10.2$

Try it.

Solution

$9.8$

Try it.

In the following exercises, solve using a geometry formula.

Try it.

The width of a rectangle is seven meters less than the length. The perimeter is 58 meters. Find the length and width.

Solution

18 meters, 11 meters

Try it.

The length of a rectangle is eight feet more than the width. The perimeter is 60 feet. Find the length and width.

Try it.

The width of the rectangle is 0.7 meters less than the length. The perimeter of a rectangle is 52.6 meters. Find the dimensions of the rectangle.

Solution

$13.5$ m, $12.8$ m

Try it.

The length of the rectangle is 1.1 meters less than the width. The perimeter of a rectangle is 49.4 meters. Find the dimensions of the rectangle.

Try it.

The perimeter of a rectangle of 150 feet. The length of the rectangle is twice the width. Find the length and width of the rectangle.

Solution

25 ft, 50 ft

Try it.

The length of the rectangle is three times the width. The perimeter of a rectangle is 72 feet. Find the length and width of the rectangle.

Try it.

The length of the rectangle is three meters less than twice the width. The perimeter of a rectangle is 36 meters. Find the dimensions of the rectangle.

Solution

l = 11 m, w = 7 m

Try it.

The length of a rectangle is five inches more than twice the width. The perimeter is 34 inches. Find the length and width.

Try it.

The perimeter of a triangle is 39 feet. One side of the triangle is one foot longer than the second side. The third side is two feet longer than the second side. Find the length of each side.

Solution

12 ft, 13 ft, 14 ft

Try it.

The perimeter of a triangle is 35 feet. One side of the triangle is five feet longer than the second side. The third side is three feet longer than the second side. Find the length of each side.

Try it.

One side of a triangle is twice the smallest side. The third side is five feet more than the shortest side. The perimeter is 17 feet. Find the lengths of all three sides.

Solution

3 ft, 6 ft, 8 ft

Try it.

One side of a triangle is three times the smallest side. The third side is three feet more than the shortest side. The perimeter is 13 feet. Find the lengths of all three sides.

Try it.

The perimeter of a rectangular field is 560 yards. The length is 40 yards more than the width. Find the length and width of the field.

Solution

120 yd, 160 yd

Try it.

The perimeter of a rectangular atrium is 160 feet. The length is 16 feet more than the width. Find the length and width of the atrium.

Try it.

A rectangular parking lot has perimeter 250 feet. The length is five feet more than twice the width. Find the length and width of the parking lot.

Solution

40 ft, 85 ft

Try it.

A rectangular rug has perimeter 240 inches. The length is 12 inches more than twice the width. Find the length and width of the rug.

In the following exercises, solve. Approximate answers to the nearest tenth, if necessary.

Try it.

A 13-foot string of lights will be attached to the top of a 12-foot pole for a holiday display as shown. How far from the base of the pole should the end of the string of lights be anchored?

Solution

5 feet

Try it.

am wants to put a banner across her garage door diagonally, as shown, to congratulate her son for his college graduation. The garage door is 12 feet high and 16 feet wide. Approximately how long should the banner be to fit the garage door?

Try it.

Chi is planning to put a diagonal path of paving stones through her flower garden as shown. The flower garden is a square with side 10 feet. What will the length of the path be?

Solution

14.1 feet

Try it.

Brian borrowed a 20-foot extension ladder to use when he paints his house. If he sets the base of the ladder six feet from the house as shown, how far up will the top of the ladder reach?

Condensed — the full section is in OpenStax Intermediate Algebra 2e.

Use the Distance, Rate, and Time Formula

One formula you will use often in algebra and in everyday life is the formula for distance traveled by an object moving at a constant rate. Rate is an equivalent word for “speed.” The basic idea of rate may already familiar to you. Do you know what distance you travel if you drive at a steady rate of 60 miles per hour for 2 hours? (This might happen if you use your car’s cruise control while driving on the highway.) If you said 120 miles, you already know how to use this formula!

We will use the Strategy for Solving Applications that we used earlier in this chapter. When our problem requires a formula, we change Step 4. In place of writing a sentence, we write the appropriate formula. We write the revised steps here for reference.

You may want to create a mini-chart to summarize the information in the problem. See the chart in this first example.

Example

Try it.

Jamal rides his bike at a uniform rate of 12 miles per hour for $3\frac{1}{2}$ hours. What distance has he traveled?

Solution

Step 1. Read the problem.
Step 2. Identify what you are looking for.	distance traveled
Step 3. Name. Choose a variable to represent it.	Let d = distance.
Step 4. Translate: Write the appropriate formula.	$d=rt$

Substitute in the given information.	$d=12\cdot 3\frac{1}{2}$
Step 5. Solve the equation.	$d=42$ miles
Step 6. Check
Does 42 miles make sense?
Jamal rides:

Step 7. Answer the question with a complete sentence.	Jamal rode 42 miles.

Condensed — the full section is in OpenStax Elementary Algebra 2e.

Solve a Formula for a Specific Variable

You are probably familiar with some geometry formulas. A formula is a mathematical description of the relationship between variables. Formulas are also used in the sciences, such as chemistry, physics, and biology. In medicine they are used for calculations for dispensing medicine or determining body mass index. Spreadsheet programs rely on formulas to make calculations. It is important to be familiar with formulas and be able to manipulate them easily.

In and , we used the formula $d=rt$. This formula gives the value of $d$, distance, when you substitute in the values of $r\ \text{and}\ t$, the rate and time. But in , we had to find the value of $t$. We substituted in values of $d\ \text{and}\ r$ and then used algebra to solve for $t$. If you had to do this often, you might wonder why there is not a formula that gives the value of $t$ when you substitute in the values of $d\ \text{and}\ r$. We can make a formula like this by solving the formula $d=rt$ for $t$.

To solve a formula for a specific variable means to isolate that variable on one side of the equals sign with a coefficient of 1. All other variables and constants are on the other side of the equals sign. To see how to solve a formula for a specific variable, we will start with the distance, rate and time formula.

Example

Try it.

Solve the formula $d=rt$ for $t$:

ⓐ when $d=520$ and $r=65$
ⓑ in general

Solution

We will write the solutions side-by-side to demonstrate that solving a formula in general uses the same steps as when we have numbers to substitute.

ⓐ when $d=520$ and $r=65$		ⓑ in general
Write the formula.	$\ d=rt$		Write the formula.	$d=rt$
Substitute.	$520=65t$
Divide, to isolate $t$.	$\frac{520}{65}=\frac{65t}{65}$		Divide, to isolate $t$.	$\frac{d}{r}=\frac{rt}{r}$
Simplify.	$\ 8=t$		Simplify.	$\frac{d}{r}=t$

We say the formula $t=\frac{d}{r}$ is solved for $t$.

Example

Try it.

Solve the formula $A=\frac{1}{2}bh$ for $h$:

ⓐ when $A=90$ and $b=15$ ⓑ in general

Solution

ⓐ when $A=90$ and $b=15$	ⓑ in general
Write the formula.		Write the formula.
Substitute.
Clear the fractions.		Clear the fractions.
Simplify.		Simplify.
Solve for $h$.		Solve for $h$.

We can now find the height of a triangle, if we know the area and the base, by using the formula $h=\frac{2A}{b}$.

Condensed — the full section is in OpenStax Elementary Algebra 2e.

Key Concepts

To Solve an Application (with a formula)
1. Read the problem. Make sure all the words and ideas are understood.
2. Identify what we are looking for.
3. Name what we are looking for. Choose a variable to represent that quantity.
4. Translate into an equation. Write the appropriate formula for the situation. Substitute in the given information.
5. Solve the equation using good algebra techniques.
6. Check the answer in the problem and make sure it makes sense.
7. Answer the question with a complete sentence.
Distance, Rate and Time
For an object moving at a uniform (constant) rate, the distance traveled, the elapsed time, and the rate are related by the formula: $d=rt$ where d = distance, r = rate, t = time.
To solve a formula for a specific variable means to get that variable by itself with a coefficient of 1 on one side of the equation and all other variables and constants on the other side.

Solve a Formula for a Specific Variable

Use the Distance, Rate, and Time Formula

In the following exercises, solve.

Try it.

Steve drove for $8\frac{1}{2}$ hours at 72 miles per hour. How much distance did he travel?

Try it.

Socorro drove for $4\frac{5}{6}$ hours at 60 miles per hour. How much distance did she travel?

Solution

290 miles

Try it.

Yuki walked for $1\frac{3}{4}$ hours at 4 miles per hour. How far did she walk?

Try it.

Francie rode her bike for $2\frac{1}{2}$ hours at 12 miles per hour. How far did she ride?

Solution

30 miles

Try it.

Connor wants to drive from Tucson to the Grand Canyon, a distance of 338 miles. If he drives at a steady rate of 52 miles per hour, how many hours will the trip take?

Try it.

Megan is taking the bus from New York City to Montreal. The distance is 380 miles and the bus travels at a steady rate of 76 miles per hour. How long will the bus ride be?

Solution

5 hours

Try it.

Aurelia is driving from Miami to Orlando at a rate of 65 miles per hour. The distance is 235 miles. To the nearest tenth of an hour, how long will the trip take?

Try it.

Kareem wants to ride his bike from St. Louis to Champaign, Illinois. The distance is 180 miles. If he rides at a steady rate of 16 miles per hour, how many hours will the trip take?

Solution

11.25 hours

Try it.

Javier is driving to Bangor, 240 miles away. If he needs to be in Bangor in 4 hours, at what rate does he need to drive?

Try it.

Alejandra is driving to Cincinnati, 450 miles away. If she wants to be there in 6 hours, at what rate does she need to drive?

Solution

75 mph

Try it.

Aisha took the train from Spokane to Seattle. The distance is 280 miles and the trip took 3.5 hours. What was the speed of the train?

Try it.

Philip got a ride with a friend from Denver to Las Vegas, a distance of 750 miles. If the trip took 10 hours, how fast was the friend driving?

Solution

75 mph

Solve a Formula for a Specific Variable

In the following exercises, use the formula $d=rt$.

Try it.

Solve for $t$ⓐ when $d=350$ and $r=70$ ⓑ in general

Try it.

Solve for $t$ ⓐ when $d=240\ \text{and}\ r=60$ ⓑ in general

Solution

ⓐ $t=4$ ⓑ $t=\frac{d}{r}$

Try it.

Solve for $t$ ⓐ when $d=510\ \text{and}\ r=60$ ⓑ in general

Try it.

Solve for $t$
ⓐ when $d=175\ \text{and}\ r=50$
ⓑ in general

Solution

ⓐ $t=3.5$ ⓑ $t=\frac{d}{r}$

Try it.

Solve for $r$
ⓐ when $d=204\ \text{and}\ t=3$ ⓑ in general

Try it.

Solve for $r$ⓐ when $d=420\ \text{and}\ t=6$ⓑ in general

Solution

ⓐ $r=70$ ⓑ $r=\frac{d}{t}$

Try it.

Solve for $r$ⓐ when $d=160\ \text{and}\ t=2.5$ⓑ in general

Try it.

Solve for $r$ⓐ when $d=180\ \text{and}\ t=4.5$ⓑ in general

Solution

ⓐ $r=40$ ⓑ $r=\frac{d}{t}$

In the following exercises, use the formula $A=\frac{1}{2}bh$.

Try it.

Solve for $b$ⓐ when $A=126\ \text{and}\ h=18$ⓑ in general

Try it.

Solve for $h$
ⓐ when $A=176\ \text{and}\ b=22$ⓑ in general

Solution

ⓐ $h=16$ ⓑ $h=\frac{2A}{b}$

Try it.

Solve for $h$ⓐ when $A=375\ \text{and}\ b=25$ⓑ in general

Try it.

Solve for $b$ⓐ when $A=65\ \text{and}\ h=13$ⓑ in general

Solution

ⓐ $b=10$ ⓑ $b=\frac{2A}{h}$

In the following exercises, use the formula I = Prt.

Try it.

Solve for the principal, P forⓐ $I=\text{\$}5,480,r=4\%,$$t=7\ \text{years}\$ⓑ in general

Try it.

Solve for the principal, P for
ⓐ $I=\text{\$}3,950,r=6\%,$$t=5\ \text{years}\$ⓑ in general

Solution

ⓐ $P=\text{\$}13,166.67$ ⓑ $P=\frac{I}{rt}$

Try it.

Solve for the time, t for ⓐ $I=\text{\$}2,376,P=\text{\$}9,000,$$r=4.4\%$ⓑ in general

Try it.

Solve for the time, t for
ⓐ $I=\text{\$}624,P=\text{\$}6,000,$$r=5.2\%$ⓑ in general

Solution

ⓐ $t=2$ years ⓑ $t=\frac{I}{\text{Pr}}$

In the following exercises, solve.

Try it.

Solve the formula $2x+3y=12$ for yⓐ when $x=3$ⓑ in general

Try it.

Solve the formula $5x+2y=10$ for yⓐ when $x=4$ⓑ in general

Solution

ⓐ $y=-5$ ⓑ $y=\frac{10-5x}{2}$

Try it.

Solve the formula $3x-y=7$ for yⓐ when $x=-2$ⓑ in general

Try it.

Solve the formula $4x+y=5$ for yⓐ when $x=-3$ⓑ in general

Solution

ⓐ $y=17$ ⓑ $y=5-4x$

Try it.

Solve $a+b=90$ for $b$.

Try it.

Solve $a+b=90$ for $a$.

Solution

$a=90-b$

Try it.

Solve $180=a+b+c$ for $a$.

Try it.

Solve $180=a+b+c$ for $c$.

Solution

$c=180-a-b$

Try it.

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

Try it.

Solve the formula $9x+y=13$ for y.

Solution

$y=13-9x$

Try it.

Solve the formula $-4x+y=-6$ for y.

Try it.

Solve the formula $-5x+y=-1$ for y.

Solution

$y=-1+5x$

Try it.

Solve the formula $4x+3y=7$ for y.

Try it.

Solve the formula $3x+2y=11$ for y.

Solution

$y=\frac{11-3x}{2}$

Try it.

Solve the formula $x-y=-4$ for y.

Try it.

Solve the formula $x-y=-3$ for y.

Solution

$y=3+x$

Try it.

Solve the formula $P=2L+2W$ for $L$.

Try it.

Solve the formula $P=2L+2W$ for $W$.

Solution

$W=\frac{P-2L}{2}$

Try it.

Solve the formula $C=\pi d$ for $d$.

Try it.

Solve the formula $C=\pi d$ for $\pi$.

Solution

$\pi =\frac{C}{d}$

Try it.

Solve the formula $V=LWH$ for $L$.

Try it.

Solve the formula $V=LWH$ for $H$.

Solution

$H=\frac{V}{LW}$

Condensed — the full section is in OpenStax Elementary Algebra 2e.

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

Evaluate $2(x+3)$ when $x=5.$
If you missed this problem, review .

Jiżvelaw it-tweġiba

$16$
The length of a rectangle is three less than the width. Let w represent the width. Write an expression for the length of the rectangle.
If you missed this problem, review .

Jiżvelaw it-tweġiba

$w-3$
Evaluate $\frac{1}{2}bh$ when $b=14$ and $h=9.$
If you missed this problem, review .

Jiżvelaw it-tweġiba

$63$
Solve the formula $V=\frac{1}{3}\pi {r}^{2}h$ for h.

Jiżvelaw it-tweġiba

Write the formula.
Remove the fraction on the right.
Simplify.
Divide both sides by $\pi {r}^{2}.$
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 $h=\frac{3V}{\pi {r}^{2}}.$
Use the formula $A=\frac{1}{2}bh$ to solve for b.

Jiżvelaw it-tweġiba

$b=\frac{2A}{h}$
Use the formula $A=\frac{1}{2}bh$ to solve for h.

Jiżvelaw it-tweġiba

$h=\frac{2A}{b}$
Solve the formula $C=\frac{5}{9}(F-32)$ for F.

Jiżvelaw it-tweġiba

Write the formula.
Remove the fraction on the right.
Simplify.
Add 32 to both sides.
We can now use the formula $F=\frac{9}{5}C+32$ to find the Fahrenheit temperature when we know the Celsius temperature.
Solve the formula $F=\frac{9}{5}C+32$ for C.

Jiżvelaw it-tweġiba

$C=\frac{5}{9}(F-32)$
Solve the formula $A=\frac{1}{2}h(b+B)$ for b.

Jiżvelaw it-tweġiba

$b=\frac{2A-Bh}{h}$
Solve the formula $S=2\pi {r}^{2}+2\pi rh$ for h.

Jiżvelaw it-tweġiba

Write the formula.
Isolate the h term by subtracting $2\pi {r}^{2}$ from each side.
Simplify.
Solve for h by dividing both sides by $2\pi r.$
Simplify.
Solve the formula $A=P+Prt$ for t.

Jiżvelaw it-tweġiba

$t=\frac{A-P}{\text{P}r}$
Solve the formula $A=P+Prt$ for r.

Jiżvelaw it-tweġiba

$r=\frac{A-P}{\text{P}t}$
Solve the formula $8x+7y=15$ for y.

Jiżvelaw it-tweġiba

We will isolate y on one side of the equation.
Subtract $8x$ from both sides to isolate the term with y.
Simplify.
Divide both sides by 7 to make the coefficient of y one.
Simplify.
Solve the formula $4x+7y=9$ for y.

Jiżvelaw it-tweġiba

$y=\frac{9-4x}{7}$
Solve the formula $5x+8y=1$ for y.

Jiżvelaw it-tweġiba

$y=\frac{1-5x}{8}$

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

Jiżvelaw it-tweġiba

Step 1. Read the problem.
Step 2. Identify what you are looking for.	height of a triangle
Step 3. Name.
Choose a variable to represent it.	Let $h=$ the height.
Draw the figure and label it with the given information.	Area = 126 sq. in.

Step 4. Translate.
Write the appropriate formula.	$\ A=\frac{1}{2}bh$
Substitute in the given information.	$\ 126=\frac{1}{2}\cdot 18\cdot h$
Step 5. Solve the equation.	$\ 126=9h$
Divide both sides by 9.	$\ 14=h$
Step 6. Check. $\begin{array}{lll}A & = & \frac{1}{2}bh \\ 126 & \overset{?}{=} & \frac{1}{2}\cdot 18\cdot 14 \\ 126 & = & 126✓\end{array}$
Step 7. Answer the question.	The height of the triangle is 14 inches.

The area of a triangular church window is 90 square meters. The base of the window is 15 meters. What is the window’s height?

Jiżvelaw it-tweġiba

The window’s height is 12 meters.
A triangular tent door has an area of 15 square feet. The height is five feet. What is the length of the base?

Jiżvelaw it-tweġiba

The length of the base is 6 feet.

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.

Jiżvelaw it-tweġiba

Step 1. Read the problem.
Step 2. Identify what you are looking for.	the measures of all three angles
Step 3. Name. Choose a variable to represent it.	$\begin{array}{lll}\text{Let}\ a & = & {1}^{\text{st}}\ \text{angle.} \\ a+40 & = & {2}^{\text{nd}}\ \text{angle} \\ 90 & = & {3}^{\text{rd}}\ \text{angle (the right angle)}\end{array}$
Draw the figure and label it with the given information.
Step 4. Translate.
Write the appropriate formula.
Substitute into the formula.
Step 5. Solve the equation.
Step 6. Check. $\begin{array}{lll}25+65+90 & \overset{?}{=} & 180 \\ 180 & = & 180✓\end{array}$
Step 7. Answer the question.	The three angles measure $25\text{^{\circ}},65\text{^{\circ}},$ and $90\text{^{\circ}}.$

The measure of one angle of a right triangle is 50 more than the measure of the smallest angle. Find the measures of all three angles.

Jiżvelaw it-tweġiba

The measures of the angles are 20°, 70°, and 90°.
The measure of one angle of a right triangle is 30 more than the measure of the smallest angle. Find the measures of all three angles.

Jiżvelaw it-tweġiba

The measures of the angles are 30°, 60°, and 90°.

Use the Pythagorean Theorem to find the length of the other leg in

Jiżvelaw it-tweġiba

Step 1. Read the problem.
Step 2. Identify what you are looking for.	the length of the leg of the triangle
Step 3. Name.
Choose a variable to represent it.	Let a = the leg of the triangle.
Label side a.
Step 4. Translate.
Write the appropriate formula. Substitute.	$\begin{array}{lll}{a}^{2}+{b}^{2} & = & {c}^{2} \\ {a}^{2}+{12}^{2} & = & {13}^{2}\end{array}$
Step 5. Solve the equation. Isolate the variable term. Use the definition of square root. Simplify.	$\begin{array}{lll}{a}^{2}+{144}^{} & = & 169 \\ {a}^{2} & = & 25 \\ a & = & \sqrt{25} \\ a & = & 5\end{array}$
Step 6. Check.
Step 7. Answer the question.	The length of the leg is 5.

Use the Pythagorean Theorem to find the length of the leg in the figure.

Jiżvelaw it-tweġiba

The length of the leg is 8.
Use the Pythagorean Theorem to find the length of the leg in the figure.

Jiżvelaw it-tweġiba

The length of the leg is 12.

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

Jiżvelaw it-tweġiba

Step 1. Read the problem.
Step 2. Identify what we are looking for.	the length and the width
Step 3. Name. Choose a variable to represent the width. The length is six more than twice the width.	Let $\ w=$ width. $2w+6=$ length $P=96$ cm
Step 4. Translate.
Write the appropriate formula.
Substitute in the given information.
Step 5. Solve the equation.
Step 6. Check. $\begin{array}{lll}P & = & 2L+2W \\ 96 & \overset{?}{=} & 2\cdot 34+2\cdot 14 \\ 96 & = & 96✓\end{array}$
Step 7. Answer the question.	The length is 34 cm and the width is 14 cm.

The length of a rectangle is seven more than twice the width. The perimeter is 110 inches. Find the length and width.

Jiżvelaw it-tweġiba

The length is 39 inches and the width is 16 inches.
The width of a rectangle is eight yards less than twice the length. The perimeter is 86 yards. Find the length and width.

Jiżvelaw it-tweġiba

The length is 17 yards and the width is 26 yards.

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.

Jiżvelaw it-tweġiba

Step 1. Read the problem.
Step 2. Identify what we are looking for.	the lengths of the three sides of a triangle
Step 3. Name. Choose a variable to represent the length of the first side.	$\begin{array}{lll}\text{Let}\ x & = & \text{length of}\ {1}^{\text{st}}\ \text{side.} \\ x+3 & = & \text{length of}\ {2}^{\text{nd}}\ \text{side} \\ 2x+2 & = & \text{length of}\ {3}^{\text{rd}}\ \text{side}\end{array}$
Step 4. Translate. Write the appropriate formula. Substitute in the given information.
Step 5. Solve the equation.
Step 6. Check. $\ 29\overset{?}{=}6+9+14$ $\ 29=29✓$
Step 7. Answer the question.	The lengths of the sides of the triangle are 6, 9, and 14 inches.

One side of a triangle is seven inches more than the first side. The third side is four inches less than three times the first. The perimeter is 28 inches. Find the length of the three sides of the triangle.

Jiżvelaw it-tweġiba

The lengths of the sides of the triangle are 5, 11 and 12 inches.
One side of a triangle is three feet less than the first side. The third side is five feet less than twice the first. The perimeter is 20 feet. Find the length of the three sides of the triangle.

Jiżvelaw it-tweġiba

The lengths of the sides of the triangle are 4, 7 and 9 feet.

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

Jiżvelaw it-tweġiba

Step 1. Read the problem.
Step 2. Identify what we are looking for.	the length and width of the soccer field
Step 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. $w+40=$ length
Step 4. Translate. Write the appropriate formula and substitute.
Step 5. Solve the equation.
Step 6. Check. $\begin{array}{lll}P & = & 2L+2W \\ 360 & \overset{?}{=} & 2(110)+2(70) \\ 360 & = & 360✓\end{array}$
Step 7. Answer the question.	The length of the soccer field is 110 feet and the width is 70 feet.

The perimeter of a rectangular swimming pool is 200 feet. The length is 40 feet more than the width. Find the length and width.

Jiżvelaw it-tweġiba

The length of the swimming pool is 70 feet and the width is 30 feet.
The length of a rectangular garden is 30 yards more than the width. The perimeter is 300 yards. Find the length and width.

Jiżvelaw it-tweġiba

The length of the garden is 90 yards and the width is 60 yards.

Kelvin is building a gazebo and wants to brace each corner by placing a 10” piece of wood diagonally as shown.

How far from the corner should he fasten the wood if wants the distances from the corner to be equal? Approximate to the nearest tenth of an inch.

Jiżvelaw it-tweġiba

Step 1. Read the problem.
Step 2. Identify what we are looking for.	the distance from the corner that the bracket should be attached
Step 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.
Step 4. Translate. Write the appropriate formula and substitute.	${a}^{2}+{b}^{2}={c}^{2}$ ${x}^{2}+{x}^{2}={10}^{2}$
Step 5. Solve the equation. Isolate the variable. Use the definition of square root. Simplify. Approximate to the nearest tenth.	$\begin{array}{lll} \\ \\ 2{x}^{2} & = & 100 \\ {x}^{2} & = & 50 \\ x & = & \sqrt{50} \\ x & \approx & 7.1\end{array}$
Step 6. Check. $\begin{array}{lll}{a}^{2}+{b}^{2} & = & {c}^{2} \\ {(7.1)}^{2}+{(7.1)}^{2} & \approx & {10}^{2}\ \text{Yes.}\end{array}$
Step 7. Answer the question.	Kelvin should fasten each piece of wood approximately 7.1” from the corner.

John puts the base of a 13-foot ladder five feet from the wall of his house as shown in the figure. How far up the wall does the ladder reach?

Jiżvelaw it-tweġiba

The ladder reaches 12 feet.
Randy wants to attach a 17-foot string of lights to the top of the 15 foot mast of his sailboat, as shown in the figure. How far from the base of the mast should he attach the end of the light string?

Jiżvelaw it-tweġiba

He should attach the lights 8 feet from the base of the mast.
Solve the formula $C=\pi d$ for d.

Jiżvelaw it-tweġiba

$d=\frac{C}{\pi }$
Solve the formula $C=\pi d$ for $\pi .$
Solve the formula $V=LWH$ for L.

Jiżvelaw it-tweġiba

$L=\frac{V}{WH}$
Solve the formula $V=LWH$ for H.

Symbols used here

\pi

pi
Ratio of a circle's circumference to its diameter, 3.14159…

f'(x),\ \frac{dy}{dx}

derivative
Instantaneous rate of change; slope of the graph.

^\circ

degrees
1/360 of a full turn. 180° = π radians.

\pm

plus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.

\neq

not equal
The two sides are different.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

|x|

absolute value / modulus
Distance from zero: |−3| = 3. For a complex number, distance from the origin.

imaginary unit
i² = −1.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\mathbb{N},\ \mathbb{Z},\ \mathbb{Q},\ \mathbb{R},\ \mathbb{C}

number sets
Naturals, integers, rationals, reals, complex numbers.

How to: Solve a Formula for a Specific Variable

Solve a formula for a specific variable
Use formulas to solve geometry applications
In any right triangle, where

Questions people ask

What does it mean to solve an equation?

To find every value of the unknown that makes both sides equal. Each step is an operation applied to both sides that keeps the solution set the same, until the unknown stands alone.

Why do I sometimes get two answers?

A quadratic can cross the axis twice, so it can have two solutions. A degree-n polynomial has up to n. The graph shows where each one comes from.

How do I know whether to factor or use the quadratic formula?

Try factoring for a few seconds: look for two numbers that multiply to a·c and add to b. If nothing obvious appears, the discriminant b² − 4ac tells you how many real roots there are, and the formula finds them without guessing.

Ipprova tiegħek stess

Parts of this page are adapted from OpenStax Elementary Algebra 2e (CC BY-NC-SA 4.0), OpenStax Intermediate Algebra 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

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ⓐ when \(d=520\) and \(r=65\)		ⓑ in general
Write the formula.	\(\ d=rt\)		Write the formula.	\(d=rt\)
Substitute.	\(520=65t\)
Divide, to isolate \(t\).	\(\frac{520}{65}=\frac{65t}{65}\)		Divide, to isolate \(t\).	\(\frac{d}{r}=\frac{rt}{r}\)
Simplify.	\(\ 8=t\)		Simplify.	\(\frac{d}{r}=t\)

Write the formula.
Isolate the h term by subtracting \(2\pi {r}^{2}\) from each side.
Simplify.
Solve for h by dividing both sides by \(2\pi r.\)
Simplify.

ⓐ when \(A=90\) and \(b=15\)	ⓑ in general
Write the formula.		Write the formula.
Substitute.
Clear the fractions.		Clear the fractions.
Simplify.		Simplify.
Solve for \(h\).		Solve for \(h\).

We will isolate y on one side of the equation.
Subtract \(8x\) from both sides to isolate the term with y.
Simplify.
Divide both sides by 7 to make the coefficient of y one.
Simplify.