* Set up a linear equation to solve a real-world application. * Use a formula to solve a real-world application.

1. 2.3 Linear Models and
Applications
Chapter 2 Equations and Inequalities

2. Concepts & Objectives
 Objectives for this section:
 Set up a linear equation to solve a real-world
application.
 Use a formula to solve a real-world application.

3. Solving Applied Problems
Step 1 Read the problem carefully until you understand
what is given and what is to be found.
Step 2 Assign a variable to represent the unknown value,
using diagrams or tables as needed. Express any
other unknown values in terms of the variable.
Step 3 Write an equation using the variable expression(s).
Step 4 Solve the equation
Step 5 State the answer to the problem. Does it seem
reasonable?
Step 6 Check your answer to the original problem.

4. Verbal to Math Operations
 When solving real-world problems, there are certain
expressions that we can translate directly into math.
Verbal (Written) Math Operations
One number exceeds another by a x + a
Twice a number 2x
One number is a more than another
number
x + a
One number is a less than twice
another number
2x ‒ a
The product of a number and a,
decreased by b
ax ‒ b

5. Geometry Problems
Example: If the length of each side of a square is increased
by 3 cm, the perimeter of the new square is 40 cm more
than twice the length of each side of the original square.
Find the dimensions of the original square.
 Let x represent the original length of the side.
 The new length would then be x + 3, and the new
perimeter would be 40 + 2x.
 The equation would then be comparing the perimeters:
 
  
4 3 40 2
x x

6. Geometry Problems (cont.)
 Solve the equation:
 The original length of a side is 14 cm.
 To check: the new length is 14+3 = 17 cm. The new
perimeter would be 417 = 68 cm. 40 + 214 = 68 cm,
which matches.
 
4 3 40 2
x x
  
4 12 40 2
x x
  
2 28
x 
14
x 

7. Motion Problems
 In a motion problem, the components distance, rate, and
time are usually denoted by d, r, and t, respectively. (The
rate is also called speed or velocity. In these problems,
rate is understood to be constant.) These variables are
related by the equation
.
d rt


8. Motion Problems (cont.)
Example: Maria and Eduardo are traveling to a business
conference. The trip takes 2 hours for Maria and 2.5 hours
for Eduardo, since he lives 40 miles farther away. Eduardo
travels 5 mph faster than Maria. Find their average rates.

9. Motion Problems (cont.)
Example: Maria and Eduardo are traveling to a business
conference. The trip takes 2 hours for Maria and 2.5 hours
for Eduardo, since he lives 40 miles farther away. Eduardo
travels 5 mph faster than Maria. Find their average rates.
We also know that Eduardo’s distance is 40 more than
Maria’s.
r (mph) t (hours) d = rt
Maria x 2 2x
Eduardo x + 5 2.5 2.5x + 5

10. Motion Problems (cont.)
Eduardo’s distance = 40 more than Maria’s
This means that Maria’s rate is 55 mph and Eduardo’s is
55 + 5 = 60 mph.
 
2.5 5 2 40
x x
  
2.5 12.5 2 40
x x
  
.5 27.5
x 
55
x 

11. Investment Problems
 In mixed investment problems, multiply each principal
by the interest rate to find the amount of interest
earned.
 Remember that if a certain total amount is invested, say
$100,000, then if one quantity is x, the other quantity is
100000 – x.

12. Investment Problems (cont.)
Example: Last year, Owen earned a total of $1456 in
interest from two investments. He invested a total of
$28,000; part of it he invested at 4.8% and the rest at 5.5%.
How much did he invest at each rate?

13. Investment Problems (cont.)
Example: Last year, Owen earned a total of $1456 in
interest from two investments. He invested a total of
$28,000; part of it he invested at 4.8% and the rest at 5.5%.
How much did he invest at each rate?
Amount
Invested
Interest
Rate (%)
Time (in
years)
Interest Earned
$28,000 mixed 1 yr $1456
x 4.8 1 yr x.0481
28,000 – x 5.5 1 yr 28000–x.0551

14. Investment Problems (cont.)
The partial interest amounts have to add up to the total
interest:
So, Owen invested $12,000 at 4.8% and $16,000 at 5.5%.
      
.048 1 28000 .055 1 1456
x x
  
.048 1540 .055 1456
x x
  
.007 84
x
  
12000
x 

15. Literal Equations
 A literal equation is an equation that relates two or
more variables. Formulas are examples of literal
equations.
 When using literal equations, sometimes the variable is
not the one that is isolated, and so we have to solve for a
different variable.
 In this case we use some of the same methods used to
solve other equations – we treat the specified
variable as if it were the only variable, and the other
variables we treat as constants (although we
generally can’t combine them).

16. Literal Equations (cont.)
Examples: Solve for the specified variable.
(1) , for t (2) , for k
I Prt
 2
S kr krm
 

17. Literal Equations (cont.)
Examples: Solve for the specified variable.
(1) , for t (2) , for k
I Pr
Pr P
t
r
I
t
Pr


I Prt
 2
S kr krm
 

18. Literal Equations (cont.)
Examples: Solve for the specified variable.
(1) , for t (2) , for k
I Pr
Pr P
t
r
I
t
Pr


I Prt
 2
S kr krm
 
 
 
2
2
2 2
2
S r rm
k r rm
S
r rm r rm
S
k
r rm
k
 


 



19. Classwork
 College Algebra 2e
 2.3: 6-16 (even); 2.2: 22-34 (even); 2.1: 36-46 (even)
 2.3 Classwork Check
 Quiz 2.2