1. Section 4.4 Problem Solving Using Systems of Equations

2. 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

3. 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

4. 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. Objective 1: Assign Variables to Two Unknowns

5. EXAMPLE 1 Each year, Academy Award winners are presented with Oscars. The 13.5-inch statuette has a base on which a gold- plated 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. Motion Pictures 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.

6. EXAMPLE 1 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. Motion Pictures

7. EXAMPLE 1 Motion Pictures Solution Since the second equation is solved for x, we will use substitution to solve the system.

8. 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.

9. 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°. Objective 2: Use Systems to Solve Geometry Problems

10. EXAMPLE 3 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. History 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.

11. 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

12. EXAMPLE 3 History Solution Since the second equation is solved for l, we will use substitution to solve the system.

13. 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.

14. Solve Example 4 on the next few slides. Objective 3: Use Systems to Solve Number-Value Problems

15. EXAMPLE 4 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. Photography 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.

16. EXAMPLE 4 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. Photography 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:

17. 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.

18. 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.

19. Solve Example 6 on the next few slides. Objective 4: Use Systems to Solve Interest, Uniform Motion, and Mixture Problems

20. EXAMPLE 6 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. Boating 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.

21. EXAMPLE 6 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. Boating

22. EXAMPLE 6 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. Boating

23. 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.

24. 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.