Combine like terms.
Subtract 13.
Divide byThe solution set is. NOW TRY
OBJECTIVE 4 Use the six steps in solving an applied problem.While there
is no one method that allows us to solve all types of applied problems, the following
six steps are helpful.
576
x= 7 - 2.
- 2 x=- 14
- 2 x+ 13 =- 1
SECTION 2.3 Applications of Linear Equations 69
NOW TRY
EXERCISE 2
Decide whether each is an
expressionor an equation.
Simplify any expressions, and
solve any equations.
(a)
(b) 31 x- 52 + 2 x= 1
31 x- 52 + 2 x- 1NOW TRY ANSWERS
- (a)expression;
(b)equation; E^165 F
5 x- 16Solving an Applied Problem
Step 1 Readthe problem, several times if necessary. What information is
given? What is to be found?
Step 2 Assign a variableto represent the unknown value. Use a sketch, dia-
gram, or table, as needed. Write down what the variable represents. If
necessary, express any other unknown values in terms of the variable.
Step 3 Write an equationusing the variable expression(s).
Step 4 Solvethe equation.
Step 5 State the answer.Label it appropriately. Does it seem reasonable?
Step 6 Checkthe answer in the words of the originalproblem.Solving a Perimeter ProblemThe length of a rectangle is 1 cm more than twice the width. The perimeter of the rec-
tangle is 110 cm. Find the length and the width of the rectangle.
Step 1 Readthe problem. What must be found? The
length and width of the rectangle. What is
given? The length is 1 cm more than twice
the width and the perimeter is 110 cm.
Step 2 Assign a variable.Let Then
Make a sketch, as in
FIGURE 4.
Step 3 Write an equation.Use the formula for the perimeter of a rectangle.
Perimeter of a rectangle
Let andStep 4 Solvethe equation obtained in Step 3.
Distributive property
Combine like terms.
Subtract 2.
Combine like terms.Divide by 6.18 = W
108
6
=
6 W
6
108 = 6 W
110 - 2 = 6 W+ 2 - 2
110 = 6 W+ 2
110 = 4 W+ 2 + 2 W
110 = 212 W+ 12 + 2 W L= 2 W+ 1 P=110.
P= 2 L+ 2 W
2 W + 1 = the length.
W= the width.
EXAMPLE 3
2 W + 1WFIGURE 4We also
need to find
the length.