1. Copyright © Cengage Learning. All rights reserved. 1 Equations, Inequalities, and Mathematical Modeling

2. Copyright © Cengage Learning. All rights reserved. Modeling with Linear Equations 1.3

3. 3  Write and use mathematical models to solve real-life problems.  Solve mixture problems.  Use common formulas to solve real-life problems. Objectives

4. 4 Using Mathematical Models

5. 5 In this section, you will use algebra to solve problems that occur in real-life situations. The process of translating phrases or sentences into algebraic expressions or equations is called mathematical modeling. A good approach to mathematical modeling is to use two stages. Begin by using the verbal description of the problem to form a verbal model.

6. 6 Using Mathematical Models Then, after assigning labels to the quantities in the verbal model, form a mathematical model or algebraic equation. When you are constructing a verbal model, it is helpful to look for a hidden equality—a statement that two algebraic expressions are equal.

7. 7 Example 1 – Using a Verbal Model You accept a job with an annual income of $32,300. This includes your salary and a year-end bonus of $500. You are paid twice a month. What is your gross pay (pay before taxes) for each paycheck? Solution: Because there are 12 months in a year and you will be paid twice a month, it follows that you will receive 24 paychecks during the year.

8. 8 Example 1 – Solution Verbal Model: Labels: Income for year = 32,300 (dollars) Amount of each paycheck = x (dollars) Bonus = 500 (dollars) Equation: 32,300 = 24x + 500 The algebraic equation for this problem is a linear equation in the variable x, which you can solve as follows. 32,300 = 24x + 500 cont’d Write original equation.

9. 9 Example 1 – Solution 31,800 = 24x 1325 = x So, your gross pay for each paycheck will be $1325. Subtract 500 from each side. Divide each side by 24. cont’d

10. 10 Using Mathematical Models A fundamental step in writing a mathematical model to represent a real-life problem is translating key words and phrases into algebraic expressions and equations. The following list gives several examples.

11. 11 Example 2 – Finding the Percent of a Raise You accept a job that pays $10 per hour. You are told that after a two-month probationary period, your hourly wage will be increased to $11 per hour. What percent raise will you receive after the two-month period? Solution: Verbal Model: Labels: Old wage = 10 (dollars per hour)

12. 12 Example 2 – Solution Labels: Raise = 11 – 10 = 1 (dollars per hour) Percent = r (in decimal form) Equation: 1 = r  10 0.1 = r You will receive a raise of 0.1 or 10%. Divide each side by 10. Rewrite fraction as a decimal. cont’d

13. 13 Mixture Problems

14. 14 Mixture Problems Problems that involve two or more rates are called mixture problems.

15. 15 Example 7 – A Simple Interest Problem You invested a total of $10,000 at and simple interest. During one year, the two accounts earned $508.75. How much did you invest in each account? Solution: Let x represent the amount invested at. Then the amount invested at is 10,000 – x. Verbal Model:

16. 16 Example 7 – Solution Labels: Interest from (dollars) Interest from (dollars) Total interest = 508.75 (dollars) Equation: 0.045x + 0.055(10,000 – x) = 508.75 –0.01x = –41.25 x = 4125 So, $4125 was invested at and 10,000 – x = $5875 was invested at cont’d

17. 17 Common Formulas

18. 18 Common Formulas A literal equation is an equation that contains more than one variable. Many common types of geometric, scientific, and investment problems use ready-made literal equations called formulas. Knowing these formulas will help you translate and solve a wide variety of real-life applications.

19. 19 Common Formulas

20. 20 Common Formulas

21. 21 Common Formulas When working with applied problems, you often need to rewrite a literal equation in terms of another variable. You can use the methods for solving linear equations to solve some literal equations for a specified variable. For instance, the formula for the perimeter of a rectangle, P = 2l + 2w, can be rewritten or solved for w as w = (P – 2l).

22. 22 Example 9 – Using a Formula The cylindrical can has a volume of 200 cubic centimeters (cm 3 ). Find the height of the can.

23. 23 Example 9 – Solution The formula for the volume of a cylinder is V =  r 2 h. To find the height of the can, solve for h. Then, using V = 200 and r = 4, find the height. Substitute 200 for V and 4 for r. Simplify denominator.

24. 24 Example 9 – Solution ≈ 3.98 So, the height of the can is about 3.98 centimeters. You can use unit analysis to check that your answer is reasonable. Use a calculator. cont’d