Lesson Objectives

- Demonstrate an understanding of how to solve a linear equation in one variable
- Learn how to solve an equation for a specified variable

## How to Solve Literal Equations (Solve for a Specified Variable)

As we continue our study of Algebra, we will often encounter formulas. These formulas are equations where variables are used to describe a mathematical relationship. In some cases, we will be asked to manipulate the formula by solving the equation for a specified variable. When we solve a formula for a specified variable, we generally refer to this as solving literal equations. As an example, let’s consider the distance formula:

d = r • t

We previously worked with this formula in our Algebra 1 course. The distance formula is used for solving motion word problems. This formula tells us that the amount of distance traveled is equal to the rate of speed multiplied by the amount of time traveled. Suppose a road trip involves us driving at an average speed of 85 mph for 10 hours. How could we calculate the distance traveled?

d = r • t

We can plug in for rate and time since they are known:

d = 85 • 10

d = 850

The distance traveled was 850 miles.

Now let's think about another scenario. Let's suppose we know the distance and the rate, but don't know the time. In this case, we can solve our formula for time: $$d=rt$$ Divide each side by r since this is multiplying t: $$\frac{d}{r}=\frac{rt}{r}$$ $$\require{cancel}\frac{d}{r}=\frac{\cancel{r}t}{\cancel{r}}$$ $$t=\frac{d}{r}$$ Using our same scenario, let's plug in 850 (850 miles) for the distance and 85 (85 mph) for the rate. This should give us a time of 10 (10 hours): $$t=\frac{850}{85}$$ $$t=10$$ We can see that time is 10 (10 hours), just as we would expect.

Example 1: Solve for the specified variable and answer the given question.

The trip to New Zantar was a total distance of 1215 miles and took 27 hours. Solve the distance formula for rate and find the average rate of speed for the trip.

$$d=r \cdot t$$ To solve for rate, we can divide both sides of the equation by t: $$r=\frac{d}{t}$$ Now that we have solved the equation for r, we can plug in for distance (d) and time (t):

d = 1215, r = 27 $$r=\frac{1215}{27}$$ $$r=45$$ Since r is 45, this means the average rate of speed for the trip was 45 mph.

We commonly see problems in Geometry that use formulas for area and perimeter. One such formula is used to find the perimeter of a rectangle.

P = 2L + 2W

Where P is the perimeter of the rectangle, L is the length, and W is the width.

Suppose we saw a rectangle with a length of 8 feet and a width of 6 feet. What is the perimeter of this rectangle? To answer this question, we can plug into our perimeter formula:

L = 8, W = 6

P = 2(8) + 2(6)

P = 16 + 12

P = 28

The perimeter of the rectangle is 28 feet.

Now suppose we knew only the width and the perimeter. How could we find the length? We would start by solving our formula for L: $$P=2L + 2W$$ Since we want L by itself, we first need to isolate the "2L": $$P - 2W=2L + 2W - 2W$$ $$P - 2W=2L$$ Now we can divide each side by 2, the coefficient of L: $$\frac{2L}{2}=\frac{P - 2W}{2}$$ $$\frac{\cancel{2}L}{\cancel{2}}=\frac{P - 2W}{2}$$ $$L=\frac{P - 2W}{2}$$ Since we know the values for perimeter and width, we can plug this into our formula that is solved for L:

P = 28, W = 6 $$L=\frac{28 - 2(6)}{2}$$ $$L=\frac{28 - 12}{2}$$ $$L=\frac{16}{2}$$ $$L=8$$ We see that L is equal to 8 (8 feet), which is what we would expect.

Example 2: Solve for the specified variable and answer the given question.

A rectangle has a length of 12 inches and a perimeter of 44 inches. Solve the perimeter formula for width and find the width of this rectangle. $$P=2L + 2W$$ To solve for W, start by isolating the "2W" term: $$P - 2L=2L - 2L + 2W$$ $$P - 2L=2W$$ $$2W=P - 2L$$ Divide each side by 2, the coefficient of W: $$\frac{2W}{2}=\frac{P - 2L}{2}$$ $$\frac{\cancel{2}W}{\cancel{2}}=\frac{P - 2L}{2}$$ $$W=\frac{P - 2L}{2}$$ Now that we have solved our formula for W, we can plug in for length and perimeter.

L = 12, P = 44

$$W=\frac{P - 2L}{2}$$ $$W=\frac{44 - 2(12)}{2}$$ $$W=\frac{44 - 24}{2}$$ $$W=\frac{20}{2}$$ $$W=10$$ Since W is 10, this tells us the width of the rectangle is 10 inches.

We will see a variety of situations in which we must solve for a specified variable. In general, we can follow a three-step procedure for solving a literal equation:

Example 3: Solve the following equation for z. $$2m + az=z - 3u$$ Step 1) Move all variable terms that contain "z" to the left side and all other terms to the right side: $$2m - 2m + az=z - 3u - 2m$$ $$az=z - 3u - 2m$$ $$az - z=z - z - 3u - 2m$$ $$az - z=- 3u - 2m$$ Step 2) Let's use the distributive property to combine like terms:

Since z is common to each term on the left side of the equation, "z" can be factored out: $$z(a - 1)=-3u - 2m$$ Step 3) Divide both sides by the coefficient of z. In this case, the quantity (a - 1) is multiplying z, we can isolate z by dividing both sides by this quantity: $$\frac{z(a - 1)}{(a - 1)}=\frac{-3u - 2m}{(a - 1)}$$ $$\frac{z\cancel{(a - 1)}}{\cancel{(a - 1)}}=\frac{-3u - 2m}{(a - 1)}$$ $$z=\frac{-3u - 2m}{a - 1}$$

d = r • t

We previously worked with this formula in our Algebra 1 course. The distance formula is used for solving motion word problems. This formula tells us that the amount of distance traveled is equal to the rate of speed multiplied by the amount of time traveled. Suppose a road trip involves us driving at an average speed of 85 mph for 10 hours. How could we calculate the distance traveled?

d = r • t

We can plug in for rate and time since they are known:

d = 85 • 10

d = 850

The distance traveled was 850 miles.

Now let's think about another scenario. Let's suppose we know the distance and the rate, but don't know the time. In this case, we can solve our formula for time: $$d=rt$$ Divide each side by r since this is multiplying t: $$\frac{d}{r}=\frac{rt}{r}$$ $$\require{cancel}\frac{d}{r}=\frac{\cancel{r}t}{\cancel{r}}$$ $$t=\frac{d}{r}$$ Using our same scenario, let's plug in 850 (850 miles) for the distance and 85 (85 mph) for the rate. This should give us a time of 10 (10 hours): $$t=\frac{850}{85}$$ $$t=10$$ We can see that time is 10 (10 hours), just as we would expect.

Example 1: Solve for the specified variable and answer the given question.

The trip to New Zantar was a total distance of 1215 miles and took 27 hours. Solve the distance formula for rate and find the average rate of speed for the trip.

$$d=r \cdot t$$ To solve for rate, we can divide both sides of the equation by t: $$r=\frac{d}{t}$$ Now that we have solved the equation for r, we can plug in for distance (d) and time (t):

d = 1215, r = 27 $$r=\frac{1215}{27}$$ $$r=45$$ Since r is 45, this means the average rate of speed for the trip was 45 mph.

We commonly see problems in Geometry that use formulas for area and perimeter. One such formula is used to find the perimeter of a rectangle.

P = 2L + 2W

Where P is the perimeter of the rectangle, L is the length, and W is the width.

Suppose we saw a rectangle with a length of 8 feet and a width of 6 feet. What is the perimeter of this rectangle? To answer this question, we can plug into our perimeter formula:

L = 8, W = 6

P = 2(8) + 2(6)

P = 16 + 12

P = 28

The perimeter of the rectangle is 28 feet.

Now suppose we knew only the width and the perimeter. How could we find the length? We would start by solving our formula for L: $$P=2L + 2W$$ Since we want L by itself, we first need to isolate the "2L": $$P - 2W=2L + 2W - 2W$$ $$P - 2W=2L$$ Now we can divide each side by 2, the coefficient of L: $$\frac{2L}{2}=\frac{P - 2W}{2}$$ $$\frac{\cancel{2}L}{\cancel{2}}=\frac{P - 2W}{2}$$ $$L=\frac{P - 2W}{2}$$ Since we know the values for perimeter and width, we can plug this into our formula that is solved for L:

P = 28, W = 6 $$L=\frac{28 - 2(6)}{2}$$ $$L=\frac{28 - 12}{2}$$ $$L=\frac{16}{2}$$ $$L=8$$ We see that L is equal to 8 (8 feet), which is what we would expect.

Example 2: Solve for the specified variable and answer the given question.

A rectangle has a length of 12 inches and a perimeter of 44 inches. Solve the perimeter formula for width and find the width of this rectangle. $$P=2L + 2W$$ To solve for W, start by isolating the "2W" term: $$P - 2L=2L - 2L + 2W$$ $$P - 2L=2W$$ $$2W=P - 2L$$ Divide each side by 2, the coefficient of W: $$\frac{2W}{2}=\frac{P - 2L}{2}$$ $$\frac{\cancel{2}W}{\cancel{2}}=\frac{P - 2L}{2}$$ $$W=\frac{P - 2L}{2}$$ Now that we have solved our formula for W, we can plug in for length and perimeter.

L = 12, P = 44

$$W=\frac{P - 2L}{2}$$ $$W=\frac{44 - 2(12)}{2}$$ $$W=\frac{44 - 24}{2}$$ $$W=\frac{20}{2}$$ $$W=10$$ Since W is 10, this tells us the width of the rectangle is 10 inches.

We will see a variety of situations in which we must solve for a specified variable. In general, we can follow a three-step procedure for solving a literal equation:

- Move all variable terms containing the specified variable to one side and all other terms to the other
- When necessary, use the distributive property to combine terms with the specified variable
- Divide both sides by the coefficient of the specified variable

Example 3: Solve the following equation for z. $$2m + az=z - 3u$$ Step 1) Move all variable terms that contain "z" to the left side and all other terms to the right side: $$2m - 2m + az=z - 3u - 2m$$ $$az=z - 3u - 2m$$ $$az - z=z - z - 3u - 2m$$ $$az - z=- 3u - 2m$$ Step 2) Let's use the distributive property to combine like terms:

Since z is common to each term on the left side of the equation, "z" can be factored out: $$z(a - 1)=-3u - 2m$$ Step 3) Divide both sides by the coefficient of z. In this case, the quantity (a - 1) is multiplying z, we can isolate z by dividing both sides by this quantity: $$\frac{z(a - 1)}{(a - 1)}=\frac{-3u - 2m}{(a - 1)}$$ $$\frac{z\cancel{(a - 1)}}{\cancel{(a - 1)}}=\frac{-3u - 2m}{(a - 1)}$$ $$z=\frac{-3u - 2m}{a - 1}$$

#### Skills Check:

Example #1

Solve the equation for n. $$8t + dn=n + 17w$$

Please choose the best answer.

A

$$n=\frac{d - 1}{17w - 8t}$$

B

$$n=136w - dt$$

C

$$n=\frac{25w}{t + d}$$

D

$$n=\frac{17w - 8t}{d - 1}$$

E

$$n=\frac{w}{t + 25d}$$

Example #2

Solve the equation for p. $$18p + 10p=jp - q$$

Please choose the best answer.

A

$$p=\frac{q + j}{28 - j}$$

B

$$p=-\frac{q}{28 - j}$$

C

$$p=-\frac{180 - j}{q + j}$$

D

$$p=\frac{28j}{q}$$

E

$$p=\frac{14j + 1}{q}$$

Example #3

Solve the equation for w. $$j=\frac{11z - r}{17w}$$

Please choose the best answer.

A

$$w=\frac{11z - r}{17j}$$

B

$$w=\frac{17z + r}{11z - r + 11}$$

C

$$w=\frac{187 + r}{z - j}$$

D

$$w=\frac{28rz}{j + z}$$

E

$$w=\frac{11 + rz}{j - z}$$

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