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Solve geometry, motion, mixture, and interest problems

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- 1. 1.2 Applications and Modeling with Linear Equations Chapter 1 Equations and Inequalities
- 2. Concepts & Objectives ⚫ Solving applied problems ⚫ Geometry problems ⚫ Motion problems ⚫ Mixture problems ⚫ Modeling with linear equations
- 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. 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 2x x
- 5. 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 2x x+ = + 4 12 40 2x x+ = + 2 28x = 14x =
- 6. Motion Problems ⚫ In a motion problem, the components distance, rate, and time are 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=
- 7. 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.
- 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. 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)
- 9. 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 40x x+ = + 2.5 12.5 2 40x x+ = + .5 27.5x = 55x =
- 10. Mixture Problems ⚫ Problems involving mixtures of two types of the same substance, salt solution, candy, and so on, often involve percent. ⚫ In mixture problems, 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.
- 11. Mixture Problems (cont.) Example: 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 3 L of the 30% solution to obtain her 20% solution?
- 12. Mixture Problems (cont.) Example: 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 3 L of the 30% solution to obtain her 20% solution? Strength Liters of Solution Liters of Pure Alcohol 15% x .15x 30% 3 .30(3) 20% x + 3 .20(x + 3)
- 13. Mixture Problems (cont.) Since the number of liters of pure alcohol in the 15% + 30% solution must equal the number of liters in the 20% solution, we can set them equal to each other: This means that the chemist needs 6 L of 15% solution to combine with 3 L of 30% solution to produce a solution which is 20% alcohol. ( ) ( ).15 .30 3 .20 3x x+ = + .15 .90 .20 .60x x+ = + .30 .05x= 6 x=
- 14. 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.
- 15. 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?
- 16. 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)
- 17. 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 1456x x+ − = .048 1540 .055 1456x x+ − = .007 84x− = − 12000x =
- 18. Classwork ⚫ 1.2 Assignment ⚫ Page 97: 10-16 (even); pg. 89: 30-46 (even); pg. 71: 94-106 (even) ⚫ 1.2 Classwork Check (due 10/2) ⚫ Quiz 1.1 (due 10/2)