1. Application Problems- Day 2 Digit and Coin Problems Goal: To use a system of equations to solve coin and digit problems

2. A vending machine only takes nickels and dimes. There are three times as many dimes as nickels in the machine. The face value of the coins is $5.25. How many of each coin are in the machine? Let n = the number of nickels Let d = the number of dimes If 3 nickels, how many dimes? Example 1

3. A vending machine only takes nickels and dimes. There are three times as many dimes as nickels in the machine. The face value of the coins is $5.25. How many of each coin are in the machine? Example 1

4. A jar of dimes and nickels contains $2.55. There are 30 coins in all. How many of each are there? Example 2 -5 ( ) =

5. The sum of the digits of a two digit number is 10. If the digits are reversed, the new number is 36 less than the original number. Find the original number. Example 3 Let x = the 10s digit and y = units digit Think about it: The first number is xy and the second number is yx Is there another way to represent this using addition? (Hint 53 = 5(?) + 3)

6. The sum of the digits of a two digit number is 10. If the digits are reversed, the new number is 36 less than the original number. Find the original number. Example 3 Yikes! Simplify this! You can divide everything by 9

7. The sum of the digits of a two digit number is 10. If the digits are reversed, the new number is 36 less than the original number. Find the original number. Example 3 The original number is 73 and the new number is 37.

8. The sum of the digits of a two digit number is 11. If the digits are reversed, the new number is 9 more than the original number. Find the original number. Try This The first number is 56 and the new number is 65.

9. Pick 5 of 7 notecards Check your solution upfront

10. Notecard #1 Find the value of two numbers if their sum is 12 and their difference is 4.

11. Notecard #2 The difference of two numbers is 3. Their sum is 13. Find the numbers.

12. Notecard #3 The sum of the digits of a certain two-digit number is 7. Reversing its digits increases the number by 9. What is the number?

13. Notecard #4 The sum of the digits of a two digit number is 13. If the digits are reversed, the new number is 45 less than the original number. Find the original number.

14. Notecard #5 A jar of quarters and nickels contains $3.00. There are 6 more nickels than quarters. How many of each is there?

15. Notecard #6 The sum of the digits of a two digit number is 12. If the digits are reversed, the new number is 18 less than the original number. Find the original number.

16. Notecard #7 A jar of dimes and quarters contains $15.25. There are 103 coins in all. How many of each are there?

17. Homework 4 Application Problems Coin and Number Problems

19. Application Problems- Day 3

20. Current Problems When traveling in water, going with the current will add to the rate in still water (no current). Downstream rate rate in still water (r) + rate of current (c) Traveling against the current (upstream) will be slower. The rate will be the still water rate minus the rate of the current. Upstream rate rate in still water (r) - rate of current (c)

21. A motorboat travels 531 km in 9 hours going upstream and 428 km in 4 hours going downstream. What is the speed of the boat in still water, and what is the speed of the current? Example 1 Let r = the speed of motorboat in still water Let c = the speed of current DirectionRateTimeDistance Against Upstream 9531 Downstream With current 4428

22. A motorboat travels 531 km in 9 hours going upstream and 428 km in 4 hours going downstream. What is the speed of the boat in still water, and what is the speed of the current? Example 1 The motorboat’s speed in still water is 83 km/h and the speed of the current is 24 km/h

23. Flying with the wind a plane went 183 km/h. Flying against the same wind the plane went 141 km/h. Find the speed of the plane in still air and the speed of the wind. Example 2 Let x = the speed of plane in still air Let y = the speed of wind The plane’s speed in still air is 162 km/h and the wind speed is 21 km/h They give you the rate!! Easier!!

24. A child in an airport is able to cover 348 meters in 3 minutes running at a steady speed down a moving sidewalk in the direction of the sidewalk's motion. Running at the same speed in the direction opposite to the sidewalk's movement, the child is able to cover 312 meters in 4 minutes. What is the child's running speed on a still sidewalk, and what is the speed of the moving sidewalk? Example 3

25. Directio n RateTimeDistanc e With Sidewal k 3348 Against Sidewal k 4312 Confused? Make a table The child can run 97 m/min and the sidewalk is moving at a rate of 19 m/min

26. Classwork/Homework Pick any 4 problems for homework

28. Application Problems- Day 4 Types of Problems –Mixture Problems –Geometry Applications

29. You have 50 ounces of a 25% saline solution ( a mixture of water and salt). How many ounces of a 10% saline solution must you add to make a new solution that is 15% saline? Example 1 Let’s find out… https://www.khanacademy.org/math/algebra/sy stems-of-eq-and-ineq/systems-word- problems/v/mixture-problems-2

30. A chemist has 70 mL of a 50% methane solution. How much of a 80% solution must she add so the final solution is 60% methane? Example 2 Total amount of solution % of solution # of methane Starting Solution Solution Added Resulting Solution

31. A chemist has 70 mL of a 50% methane solution. How much of a 80% solution must she add so the final solution is 60% methane? Example 2 Total # of solution % of solution # of methane Starting Solution 70 ml50% or 0.535mL Solution Added Resulting Solution

32. A chemist has 70 mL of a 50% methane solution. How much of a 80% solution must she add so the final solution is 60% methane? Example 2 Total # of solution % of solution # of methane Starting Solution 70 ml50% or 0.535mL Solution Added x80% or 0.80.8x Resultin g Solution

33. A chemist has 70 mL of a 50% methane solution. How much of a 80% solution must she add so the final solution is 60% methane? Example 2 Total # of solution % of solution # of methane Starting Solution 70 ml50% or 0.535mL Solution Added x80% or 0.80.8x Resultin g Solution 70+x60% or 0.6

34. A chemist has 70 mL of a 50% methane solution. How much of a 80% solution must she add so the final solution is 60% methane? Example 2 Total # of solution % of solution # of methane Starting Solution 700.535 Solution Added x0.80.8x Resultin g Solution 70+x0.6 0.6(70+x) = 35 + 0.8x

35. A chemist has 70 mL of a 50% methane solution. How much of a 80% solution must she add so the final solution is 60% methane? Example 2 So there is 35 mL of solution added to the mixture. So you have 63 mL of methane in a mixture of 105 mL

36. Two angles are supplementary. The larger angle is 48 degrees more than 10 times the smaller angle. Find the measure of each angle. Example 3 Let x = the measure of the smaller angle y = the measure of the larger angle The larger angle is 114° and the smaller angle is 66°

37. Classwork/Homework Mixture Problems Solve at least 3 Geometry Problems Solve at least 5