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Bell work 1. Solve. 0.5a + 0.75a +1.2 = 1.45 0.5a + 0.75a +1.2 = 1.45 2. Solve. (x/2) + 3/16 = (5/8) 3. Solve. 5(3 + 2c) = 12 – 4(c – 6) 4. Solve. (2y.

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Presentation on theme: "Bell work 1. Solve. 0.5a + 0.75a +1.2 = 1.45 0.5a + 0.75a +1.2 = 1.45 2. Solve. (x/2) + 3/16 = (5/8) 3. Solve. 5(3 + 2c) = 12 – 4(c – 6) 4. Solve. (2y."— Presentation transcript:

1 Bell work 1. Solve. 0.5a + 0.75a +1.2 = 1.45 0.5a + 0.75a +1.2 = 1.45 2. Solve. (x/2) + 3/16 = (5/8) 3. Solve. 5(3 + 2c) = 12 – 4(c – 6) 4. Solve. (2y – 5)(y + 8) = 0

2 Bell work answers 1. (25/125) = (1/5) = 0.2 2. x = (7/8) = 0.875 3. c = (3/2) = 1½ = 1.5 4. y = { 5/2, - 8 }

3 Algebra 3 Chapter 2 Section 2 Objective: Students will: 1. Solve word problems by translating words to equations that can be used to solve the problems and by using the guidelines specific to this section.

4 Guidelines for Solving Word Problems Using Equations (p.68) 1. If two #’s are consecutive, Call one x and the other (x + 1). 2. If two numbers are consecutive odd or consecutive even,call one number x and the other (x + 2). 3. If a number is increased by n%, the new number is (x + (n/100 ) x).

5 Word problems Example 1. Solve. A carpenter works one third as long after lunch as she worked before lunch. If she works a total of 8 hours, how long did she work before lunch?

6 Word problems solution Example 1. Solution. x ⅓x x ⅓x|-------------|-------| 0 hours Lunch 8 hours let x = time (hours worked before lunch) let x = time (hours worked before lunch) Equation: x + ⅓x = 8 3 (x + ⅓x = 8) 3 3 (x + ⅓x = 8) 3 3x + 1x = 24 4x = 24 4x = 24 x = 6 hours x = 6 hours

7 Word problems Example 2. Solve. A certain amount of money was deposited in a bank. The value of the money increased by 25% to a final value of $100.00. How much was originally invested?

8 Word problems solution Example 2. Solution. let x = the original amount invested (deposited) let x = the original amount invested (deposited) 25% =.25 (means the money earned 25 cents/dollar) Equation: x +.25 x = 100 1.25 x = 100 1.25 x = 100 x = 100/1.25 x = 100/1.25 x = $80.00 x = $80.00

9 Word problems Example 3. Solve. The sum of two consecutive odd integers is 36. What are the two integers?

10 Word problems solution Example 3. Solution. let x = the first odd integer & let x = the first odd integer & x+2 = the next odd integer Equation: x + (x + 2 ) = 36 2x +2 = 36 2x +2 = 36 2x = 34 2x = 34 x = 34/2 = 17 ; (x +2 ) = 17+2 =19 Thus the two odd consecutive integers are 17,19

11 Homework ALGEBRA 3 P. 69 (2- 28) even

12 Grade Homework p. 64- 65 (2- 46) even&41 # correct/ 28 2. 2 4. (49/9) 6. (-502/100)8. 310. 1 12. 7 14. 716. (39/14) 18. 620. (-37/5)22.{ - 4, 8 } 24. { 3, 7}26. { 4/3, ¼ }28. { 0, 5 } 30.{0, 4, - 2} 32. (-16/3)34. { (c – 3)/ 8}36. {(5a -3h)/ c } 38. { 12/(a – b)}40. solve without clearing the decimals due to use of a calculator 41. (x -7) (x + 8) = 0 42. {1, -1 } 44. { 0, 2 } 46. a. {whole numbers}b. {pos. integers} c. {neg. integers} d.{ even whole numbers} e. { whole # multiple s of 10} f. {integers} d.{ even whole numbers} e. { whole # multiple s of 10} f. {integers}

13 Journal Topic Write what the problems solving guidelines are for word problems from (p. 66 in your book) in the blue box in the upper left corner of the page.


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