Section 4.4
Problem Solving Using
Systems of Equations
Objectives
Assign variables to two unknowns
 Use systems to solve geometry problems
 Use systems to solve number-value
problems
 Use systems to solve interest, uniform
motion, and mixture problems

Objective 1: Assign Variables to Two
Unknowns
 In
previous chapters, many applied
problems were modeled and solved
with an equation in one variable.
 In this section, the application
problems involve two unknowns.
It
is often easier to solve such problems
using a two-variable approach.
Objective 1: Assign Variables to Two
Unknowns

The following steps are helpful when solving problems
involving two unknown quantities.

Problem-Solving Strategy:
1. Analyze the problem by reading it carefully to understand the
given facts. Often a diagram or table will help you visualize the
facts of the problem.
2. Pick different variables to represent two unknown quantities.
Translate the words of the problem to form two equations
involving each of the two variables.
3. Solve the system of equations using graphing, substitution, or
elimination.
4. State the conclusion.
5. Check the results in the words of the problem.
EXAMPLE 1

Motion Pictures
Each year, Academy Award winners are
presented with Oscars. The 13.5-inch
statuette has a base on which a goldplated figure stands. The figure itself is
7.5 inches taller than its base. Find the
height of the figure and the height of the
base.
Analyze the Problem:
 The statuette is a total of 13.5 inches tall.
 The figure is 7.5 inches taller than the base.
 Find the height of the figure and the height of the base.
EXAMPLE 1
Motion Pictures
Form Two Equations Let x = the height
of the figure, in inches, and y = the
height of the base, in inches. We can
translate the words of the problem into
two equations, each involving x and y.
EXAMPLE 1
Motion Pictures
Solution
Since the second equation is solved for x, we will use
substitution to solve the system.
EXAMPLE 1
Motion Pictures
Solution
To find x, substitute 3 for y in the second equation of the system.
State the conclusion: The height of the figure is 10.5
inches and the height of the base is 3 inches.
Check results: The sum of 10.5 inches and 3 inches is 13.5
inches, and the 10.5-inch figure is 7.5 inches taller than the
3-inch base. The results check.
Objective 2: Use Systems to Solve
Geometry Problems
Two angles are said to be complementary
if the sum of their measures is 90°.
 Two angles are said to be supplementary
if the sum of their measures is 180°.

EXAMPLE 3

History
In 1917, James Montgomery Flagg
created the classic I Want You poster to
help recruiting for World War I. The
perimeter of the poster is 114 inches,
and its length is 9 inches less than twice
its width. Find the length and the width
of the poster.
Analyze the Problem:
 The perimeter of the rectangular poster is 114 inches.
 The length is 9 inches less than twice the width.
 Find the length and the width of the poster.
EXAMPLE 3
History
Form Two Equations Let l = the length of the
poster, in inches, and w = the width of the
poster, in inches. The perimeter of a rectangle
is the sum of two lengths and two widths, as
given by the formula P = 2l + 2w, so we have:
If the length of the poster is 9 inches less than twice the width, we have
EXAMPLE 3
History
Solution
Since the second equation is solved for l, we will use
substitution to solve the system.
EXAMPLE 3
History
Solution
To find l, substitute 22 for w in the second equation of the
system.
Conclusion: The length of the poster is 35 inches and the width
is 22 inches.
Check results: The perimeter is 2(35) + 2(22) = 70 + 44 = 114
inches, and 35 inches is 9 inches less than twice 22 inches. The
results check.
Objective 3: Use Systems to Solve
Number-Value Problems

Solve Example 4 on the next few slides.
EXAMPLE 4

Photography
At a school, two picture packages
are available, as shown in the
illustration. Find the cost of a
class picture and the cost of an
individual wallet-size picture.
Analyze the Problem:
 Package 1 contains 1 class picture and 10 wallet-size pictures.
 Package 2 contains 2 class pictures and 15 wallet-size pictures.
 Find the cost of a class picture and the cost of a wallet-size
picture.
EXAMPLE 4
Photography
Form Two Equations Let c = the cost of one class
picture and w = the cost of one wallet-size picture.
We can use the fact that Number × value = total
value to write an equation that models the first
package. We note that (in dollars) the cost of 1
class picture is 1c and the cost of 10 wallet-size
pictures is 10w.
To write an equation that models the second package, we note that (in dollars)
the cost of 2 class pictures is 2c, and the cost of 15 wallet-size pictures is 15w.
The resulting system is:
EXAMPLE 4
Photography
Solution
We can use elimination to solve this system. To eliminate c, we
proceed as follows.
To find c, substitute 1.4 for w in the first equation of the original
system.
EXAMPLE 4
Photography
Solution
Conclusion: A class picture costs $5 and a wallet-size picture
costs $1.40.
Check results: Package 1 has 1 class picture and 10 wallets:
$5 + 10($1.40) = $5 + $14 = $19. Package 2 has 2 class
pictures and 15 wallets: 2($5) + 15($1.40) = $10 + $21 + $31.
The results check.
Objective 4: Use Systems to Solve Interest,
Uniform Motion, and Mixture Problems

Solve Example 6 on the next few slides.
EXAMPLE 6

Boating
A boat traveled 30 miles
downstream in 3 hours and
made the return trip in 5
hours. Find the speed of the
boat in still water and the
speed of the current.
Analyze the Problem Traveling downstream, the speed of
the boat will be faster than it would be in still water.
Traveling upstream, the speed of the boat will be slower
than it would be in still water.
EXAMPLE 6
Boating
Form Two Equations Let s = the speed of the boat in still water
and c = the speed of the current. Then the speed of the boat
going downstream is s + c and the speed of the boat going
upstream is s − c. Using the formula d = rt, we find that 3(s +
c) represents the distance traveled downstream and 5(s − c)
represents the distance traveled upstream. We can organize
the facts of the problem in a table.
EXAMPLE 6

Boating
Since each trip is 30 miles long, the Distance column
of the table helps us to write two equations in two
variables. To write each equation in standard form,
use the distributive property.
EXAMPLE 6
Boating
Solution
To eliminate c, we proceed as follows.
To find c, it appears that the computations will be easiest if we
use 3s + 3c = 30.
EXAMPLE 6
Boating
Solution
Conclusion: The speed of the boat in still water is 8 mph and
the speed of the current is 2 mph.
Check results: With a 2-mph current, the boat’s downstream
speed will be 8 + 2 = 10 mph. In 3 hours, it will travel 10(3) = 30
miles. With a 2-mph current, the boat’s upstream speed will be
8 − 2 = 6 mph. In 5 hours, it will cover 6(5) = 30 miles. The
results check.