1. Copyright © 2017, 2013, 2009 Pearson Education, Inc.
Equations and Inequalities
Copyright © 2017, 2013, 2009 Pearson Education, Inc. 1

2. 1.2 Applications and Modeling with Linear Equations
Solving Applied Problems
Geometry Problems
Motion Problems
Mixture Problems
Modeling with Linear Equations

3. SOLVING AN APPLIED PROBLEM
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. Write down what the variable represents. If necessary, express any other unknown values in terms of the variable.

4. SOLVING AN APPLIED PROBLEM
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 the answer in the words of the original problem.

5. EXAMPLE 1
FINDING THE DIMENSIONS OF A SQUARE
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 the original square. Find the dimensions of the original square.
Solution
Step 1 Read the problem. We must find the length of the original square.

6. FINDING THE DIMENSIONS OF A SQUARE
EXAMPLE 1
FINDING THE DIMENSIONS OF A SQUARE
Step 2 Assign a variable. Since the length of a side of the original square is to be found, let the variable represent this length. x
Original square x
The length of a side of the new square is 3 cm more than the length of a side of the old square.
Side is increased by 3.
x + 3
x + 3

7. EXAMPLE 1
FINDING THE DIMENSIONS OF A SQUARE
Step 2 Assign a variable.
Write a variable expression for the perimeter of the new square. The perimeter of a square is 4 times the length of a side.

8. EXAMPLE 1
FINDING THE DIMENSIONS OF A SQUARE
Step 3 Write an equation. Translate the English sentence that follows into its equivalent algebraic expression.
The new perimeter
more than
twice the length of a side of the original square. is 40

9. Step 4 Solve the equation.
EXAMPLE 1
FINDING THE DIMENSIONS OF A SQUARE
Step 4 Solve the equation.
Distributive property
Subtract 2x and 12.
Divide by 2.

10. EXAMPLE 1
FINDING THE DIMENSIONS OF A SQUARE
Step 5 State the answer. Each side of the original square measures 14 cm.

11. EXAMPLE 1
FINDING THE DIMENSIONS OF A SQUARE
Step 6 Check. Go back to the words of the original problem to see that all necessary conditions are satisfied.
The length of a side of the new square would be = 17 cm.
The perimeter of the new square would be 4(17)= 68 cm.
Twice the length of a side of the original square would be 2(14) = 28 cm.
Since = 68, the answer checks.

12. Motion Problems and its related forms and
Problem-Solving Hint In a motion problem, the components distance, rate, and time are denoted by the letters d, r, and t, respectively. (The rate is also called the speed or velocity. Here, rate is understood to be constant.)
and its related forms
and

13. Example 2
SOLVING A MOTION PROBLEM
Maria and Eduardo are traveling to a business conference. The trip takes 2 hr for Maria and 2.5 hr for Eduardo, since he lives 40 mi farther away. Eduardo travels 5 mph faster than Maria. Find their average rates.

14. Example 2
SOLVING A MOTION PROBLEM
Solution
Step 1 Read the problem. We must find Maria’s and Eduardo’s average rates.

15. Example 2
SOLVING A MOTION PROBLEM
Step 2 Assign a variable. Since average rates are to be found, we let the variables represent one of these rates.
Let x = Maria’s rate. Then x + 5 = Eduardo’s rate. Summarize the given information in a table. r t d
Maria x 2 2x
Eduardo
x + 5
2.5
2.5(x + 5)
Use
d = rt.

16. Example 2
SOLVING A MOTION PROBLEM
Step 3 Write an equation. Eduardo’s distance traveled exceeds Maria’s distance by 40 mi. Translate this relationship into its algebraic form.
Eduardo’s distance
40 more
than Maria’s. is

17. Step 4 Solve. SOLVING A MOTION PROBLEM Example 2 Distributive property
Subtract 2x and 12.5.
Divide by 0.5

18. Example 2
SOLVING A MOTION PROBLEM
Step 5 State the answer. Maria’s rate of travel is 55 mph, and Eduardo’s rate is
= 60 mph.
Step 6 Check. The diagram shows that the conditions of the problem are satisfied.
Distance traveled by Maria: (55) = 110 mi
Distance traveled by Eduardo: (60) = 150 mi
150 – 110 = 40

19. Mixture Problems
Problem-Solving Hint In mixture problems involving solutions, the rate (percent) of concentration is multiplied by the quantity to get the amount of pure substance present. The concentration of the final mixture must be between the concentrations of the two solutions making up the mixture.

20. Example 3
SOLVING A MIXTURE PROBLEM
A chemist needs a 20% solution of alcohol. She has a 15% solution on hand, as well as a 30% solution. How many liters of the 15% solution should she add to 3L of the 30% solution to obtain her 20% solution?

21. Example 3
SOLVING A MIXTURE PROBLEM
Solution
Step 1 Read the problem. We must find the required number of liters of 15% alcohol solution.

22. Let x = the number of liters of 15% solution to be added.
Example 3
SOLVING A MIXTURE PROBLEM
Step 2 Assign a variable.
Let x = the number of liters of 15% solution to be added.
Strength
Liters of Solution
Liters of Pure Alcohol
15% x
0.15x
30% 3
0.30(3)
20%
x + 3
0.20(x + 3)
Sum must equal

23. Example 3
SOLVING A MIXTURE PROBLEM
Step 3 Write an equation. The number of liters of pure alcohol in the 15% solution plus the number of liters in the 30% solution must equal the number of liters in the final 20% solution.
Liters of pure alcohol in 15%
Liters of pure alcohol in 30%
Liters of pure alcohol in 20% + =

24. Distributive property
Example 3
SOLVING A MIXTURE PROBLEM
Step 4 Solve.
Distributive property
Subtract 0.60 and 0.15x.
Divide by 0.05.

25. Example 3
SOLVING A MIXTURE PROBLEM
Step 5 State the answer. Thus, 6 L of 15% solution should be mixed with 3 L of 30% solution, giving = 9 L of 20% solution.
Step 6 Check. The answer checks since the amount of alcohol in the two solutions is equal to the amount of alcohol in the mixture.
Solutions
Mixture

26. Mixture Problems
Problem-Solving Hint In mixed investment problems, multiply each principal by the interest rate and the time in years to find the amount of interest earned.

27. Example 4
SOLVING AN INVESTMENT PROBLEM
An artist has sold a painting for $410,000. He invests a portion of the money for 6 months at 2.65% and the rest for a year at 2.91%. His broker tells him the two investments will earn a total of $ How much should be invested at each rate to obtain that amount of interest?

28. Example 4
SOLVING AN INVESTMENT PROBLEM
Solution
Step 1 Read the problem. We must find the amount to be invested at each rate.

29. Step 2 Assign a variable. SOLVING AN INVESTMENT PROBLEM Example 4
Let x = dollar amount invested for 6 months at 2.65%.
410,000 – x = dollar amount invested for 1 yr at 2.91%.
Invested Amount
Interest Rate (%)
Time
(in years)
Interest Earned x
2.65
0.5
x(0.0265)(0.5)
410,000 – x
2.91 1
(410,000 – x)(0.0291)(1)

30. Example 4
SOLVING AN INVESTMENT PROBLEM
Step 3 Write an equation. The sum of the two interest amounts must equal the total interest earned.
Interest from 2.65% investment
Interest from 2.91% investment
Total interest + =

31. Example 4
SOLVING AN INVESTMENT PROBLEM
Step 4 Solve.

32. Example 4
SOLVING AN INVESTMENT PROBLEM
Step 5 State the answer. The artist should invest $200,000 at 2.65% for 6 months and $410,000 – $200,000 = $210,000 at 2.91% for 1 yr to earn $8761 in interest.

33. Step 6 Check. The 6-month investment earns
Example 4
SOLVING AN INVESTMENT PROBLEM
Step 6 Check. The 6-month investment earns
while the 1-yr investment earns
The total amount of interest earned is

34. Modeling with Linear Equations
A mathematical model is an equation (or inequality) that describes the relationship between two quantities. A linear model is a linear equation.

35. Example 5
MODELING THE PREVENTION OF INDOOR POLLUTANTS
If a vented range hood removes contaminants such as carbon monoxide and nitrogen dioxide from the air at a rate of F liters of air per second, then the percent P of contaminants that are also removed from the surrounding air can be modeled by the linear equation
where
What flow F must a range hood have to remove 50% of the contaminants from the air? (Source: Proceedings of the Third International Conference on Indoor Air Quality and Climate.)

36. Solution Replace P with 50 in the linear model, and solve for F.
Example 5
MODELING THE PREVENTION OF INDOOR POLLUTANTS
Solution Replace P with 50 in the linear model, and solve for F.
Given model
Let P = 50.
Subtract 7.18.
Divide by 1.06.
Therefore, to remove 50% of the contaminants, the flow rate must be approximately L of air per second.

37. Example 6
MODELING HEALTH CARE COSTS
The projected per capita health care expenditures in the United States, where y is in dollars, and x is years after 2000, are given by the linear equation.
Linear model
(Source: Centers for Medicare and Medicaid Services.)
(a) What were the per capita health care expenditures in the year 2010?
(b) If this model continues to describe health care expenditures, when will the per capita expenditures reach $11,000?

38. (a) The year 2010 is 10 yr after the year 2000.
Example 6
MODELING HEALTH CARE COSTS
In part (a) we are given information to determine a value for x and asked to find the corresponding value of y, whereas in part (b) we are given a value for y and asked to find the corresponding value of x.
(a) The year 2010 is 10 yr after the year 2000.
Let x = 10 and find the value of y.
Given model
Let x = 10.
Multiply and then add.
In 2010, the estimated per capita health care expenditures were $8401.

39. (b) Let y = 11,000 in the given model, and find the value of x.
Example 6
MODELING HEALTH CARE COSTS
(b) Let y = 11,000 in the given model, and find the value of x.
Let y = 11,000.
Subtract 4512.
17 corresponds to = 2017.
Divide by 343.
The x-value of 17.9 indicates that per capita health care expenditures are projected to reach $11,000 during the 17th year after 2000—that is, 2017.