Module Two:

Literal Equations and Word Problems

Equations containing more than one variable are called literal equations. A literal equation can be solved for one of the variables using the same procedures discussed in module one(Solving Equations and Inequalities).

Solving Literal Equations

Example 1.

Solve for y: by – ax = c

Step 1. by = ax + c

Step 2. y =

or y =

Example 2.

Solve for t: c – dt = mt

Step 1. c = mt + dt

Step 2. c = t(m + d)

Step 3.

Formulas are real world applications of literal equations.

Example 3.

The formula to convert Celsius temperature to Fahrenheit temperature is F = . Solve this formula for C.

Step 1. 5(F) = 5

Step 2. 5F = 9C + 160

Step 3. 5F – 160 = 9C

Step 4. or or

In general, formulas may be written in various forms.

Example 4. The distance an object falls in a vacuum is determined by the formula . Solve the formula for t.

Step 1.

Step 2. 2s = gt²

Step 3.

Step 4.

Since t in this formula represents time, we should choose only the positive square root and therefore .

In the following examples, we will review using algebra to model problems expressed in words.

Word Problems

Example 5.

Joe spends d dollars on four equally priced CDs. If Joe buys an additional fifteen CDs at the same price, how much more will he spend?

Step 1. If 4 CD’s cost d dollars, then one CD costs dollars

Step 2. Therefore, fifteen CDs cost dollars

or

Example 6.

There are 35 students in the class. On a particular day x students are absent. Write an algebraic expression describing the number of students who are present.

Step 1. Since the total number of students is 35, the number of students present is 35 minus the number of students absent. Therefore 35 – x represents the number of students present.

Example 7.

Five-eighths of the members in a club go on a bus trip. If 40 members go on the trip, how many members are in the club?

Step 1. Let x represent the number of club members

Step 2. An equation that models the information is:

Step 3. 5x = 320

Step 4. x = 64

Step 5. There are 64 club members.

Example 8.

The area of a rectangle is 70 sq. inches. The width is three inches shorter than the length. If w is the width of the rectangle, determine the equation that can be used to solve for the length.

Step 1. Let w represent the width

Step 2. w + 3 represents the length

Step 3. Since area = (length)(width), w(w + 3) represents the area

Step 4. An equation which models the given information is

w(w + 3) = 70

or w² + 3w = 70

Practice Problems.

1. Solve for x: y = mx + b

Solution for problem 1.

y = mx + b

y – b = mx

2. Solve for h:

Solution for problem 2.

3

3. If you have x dimes and (x + 4) nickels, write an algebraic expression representing the amount of money you have in cents.

Solution for problem 3.

10x represents the amount of money in dimes

5(x+4) represents the amount of money in nickels

10x + 5(x + 4) represents the money you have in cents

4. On Monday a teacher noticed of her class wore blue jeans. If she counted 20 students wearing blue jeans, how many students were in class that day?

Solution for problem 4.

Let x represent the number of students in the class.

represents the number of students wearing blue jeans.

The equation modeling the information is :

x = 25 students were in class that day

5. A pool is enclosed by fencing that forms a rectangle whose area is 540 square feet. The width of the fenced area is 12 feet less than the length. Determine the equation of the fenced area.

Solution for problem 5.

Let x = length of the rectangle

then x – 12 = the width of the rectangle

since

area = (length)(width)

the fenced-in area is represented by the equation 540 = x(x – 12)