Chapter 2: Equations and Inequalities 2.3: Applications of Equations

1. Chapter 2: Equations and Inequalities2.3: Applications of Equations Essential Question: What are the different methods to solve a quadratic equation?

2. 2.3 Applications of Equations • Guidelines for solving applied problems • Read the problem carefully, and determine what is asked for. • Label the unknown quantities with variables • Draw a picture of the situation, if appropriate • Translate the verbal statements in the problem and the relationships between the known and unknown quantities into mathematical language. • Consolidate the mathematical information into an equation in one variable that can be solved or an equation in two variables that can be graphed. • Solve for at least one of the unknown quantities. • Find all remaining unknown quantities by using the relationships given in the problem. • Check and interpret all quantities found in the original problem.

3. 2.3 Applications of Equations • Example 1: Number Relations • The average of two real numbers is 41.125, and their product is 1683. Find the two numbers. • Solution • Two equations: • Solve one equation for one variable, and then substitute.

4. 2.3 Applications of Equations • Example 1 (Continued) • a2-82.25a+1683=0 • Option 1 → Graph • Answers are where the graph crosses the x-axis • Option 2 → Ye old Quadratic Equation

5. 2.3 Applications of Equations • Example 1 (Finishing) • Check the answers • If a = 44 • Plug into the ab=1683 equation to find that b=38.25 • If a = 38.25 • Plug into the same function to find that b=44 • Check • The average of 44 & 38.25 is 41.125 • The two numbers are 44 and 38.25

6. 2.3 Applications of Equations • Example 2: Dimensions of a Rectangle • A rectangle is twice as wide as it is high. If it has an area of 24.5 square inches, what are its dimensions? • Two equations • Substitute and solve

7. 2.3 Applications of Equations • Example 2 (Continued) • It’s not possible to have a negative height, so the only value we have to check is h=3.5 inches • If the width is twice the height, then the width = 2(3.5) = 7 inches • Check your answer: (3.5 in)(7 in) = 24.5 in2 • The width is 7 inches and the height is 3.5 inches

8. 2.3 Applications of Equations • Example 4: Interest Applications • I = Prt • I = Interest • P = Principal (initial invested amount) • r = rate (percentage written as a decimal) • t = time (in years) • A high-risk stock pays dividends at a rate of 12% per year, and a savings account pays 6% interest per year. How much of a $9000 investment should be put in the stock and how much should be put in savings to obtain a return of 8% per year on the total investment?

9. 2.3 Application of Equations • Example 4: Interest Applications (continued) • Let s be the amount invested in stock. The rest of the money ($9000 – s) is the amount invested in savings • Translate English into math: • (stock at 12%) + (savings at 6%) = 8% of $9000 • 0.12s + 0.06(9000 – s) = 0.08(9000) [Distribute] • 0.12s + 540 – 0.06s = 720 [Combine like terms] • 0.06s + 540 - 540 = 720 - 540 • 0.06s ÷ 0.06 = 180÷ 0.06 • s = 3000

10. Application of Equations • Example 5: Distance Applications • d = rt • d = distance • r = rate • t = time • You can convert the equation if necessary • r = d/t • t = d/r • A pilot wants to make an 840-mile round trip from Cleveland to Peoria and back in 5 hours flying time. There will be a headwind of 30mph going to Peoria, and it is estimated that there will be a 40mph tailwind returning to Cleveland. At what constant engine speed should the plane be flown?

11. 2.3 Application of Equations • Example 5: Distance Applications (continued) • Let r be the engine speed of the plane • Headwind slows the velocity by 30 mph • Tailwind increases the velocity by 40 mph • Cleveland to Peoria • Distance = 420 • Actual velocity = r – 30 • Time = 420/(r – 30) • Peoria to Cleveland • Distance = 420 • Actual velocity = r + 40 • Time = 420/(r + 40)

12. Application of Equations • Example 5: Distance Applications (continued) • The total time is going to be 5 hours, so

13. 2.3 Application of Equations • Example 5: Distance Applications (concluded) • r2 - 158r - 2040 = 0 • This can be factored (you can use the quadratic equation as well of course) • (r – 170)(r + 12) = 0 • r – 170 = 0 or r + 12 = 0 • r = 170 or r = -12 • Because you can’t fly at a negative rate, the plane must fly at a constant rate of 170 mph

14. 2.3 Application of Equations • Example 8: Mixture Problem • A car radiator contains 12 quarts of fluid, 20% of which is antifreeze. How much fluid should be drained and replaced with pure antifreeze so that the resulting mixture is 50% antifreeze? • Let x be the number of quarts of fluid to be replaced by pure antifreeze. • When x quarts are drained, there are 12 – x quarts of fluid left in the radiator, 20% of which is antifreeze.

15. 2.3 Application of Equations • Example 8: Mixture Problem (continued) • Translate English into math • 20% of (12 – x) + x = 50% of 12 • 0.2(12 – x) + x = 0.5(12) [Distribute] • 2.4 – 0.2x + x = 6 [Combine like terms] • 0.8x + 2.4 – 2.4 = 6 – 2.4 • 0.8x ÷ 0.8 = 3.6 ÷ 0.8 • x = 4.5 • 4.5 quarts should be drained and replaced with pure antifreeze

16. 2.3 Applications of Equations • Assignment • Pages 105-106 • Tuesday: 9, 15, 17, 25 • Wednesday: 11, 13, 19, 21, 23 • For #25, you’re going to want to solve it by graphing and finding the x-intercept(s). Make sure you alter your window so that you can see a solution.