2. Solving Algebraic Equations Equation: 2 + 3 = 5

3. Solving Algebraic Equations Equation: 2 + 3 = 5 -3 -3 2 + 0 = 2 2 = 2

4. Solving Algebraic Equations Equation: 2 + 3 = 5 +11 +11 2 + 14 = 16 16 = 16

5. Solving Algebraic Equations 5 (3) =15 ÷3 ÷3 5 (1) = 5 5 = 5

6. Solving Algebraic Equations

7. x + 13 = 20

8. Solving Algebraic Equations x + 13 = 20

9. Solving Algebraic Equations x + 13 = 20 -13 -13

10. Solving Algebraic Equations x + 13 = 20 -13 -13 x + 0 = 17 x = 17

11. Solving Algebraic Equations x + 13 = 20 -13 -13 x + 0 = 7 x = 7 Check: 7 + 13 = 20 20 = 20

12. Solving Algebraic Equations Opposites: + - x ÷ √ x 2 * When you see a number right next to a variable, with no operation, it means multiply. If you see a number right next to a parentheses (), it means multiply. 4x is the same as 4xx, but that would be confusing to read.

13. Solving Algebraic Equations 3t=24

14. Solving Algebraic Equations 3t=24

15. Solving Algebraic Equations 3t=24 ÷3

16. Solving Algebraic Equations 3t=24 ÷3 1t = 8 t = 8

17. Solving Algebraic Equations 3t=24 ÷3 1t = 8 t = 8 Check 3(8) = 24 24 = 24

18. Solving Algebraic Equations r – 7 = 22 8 + y = 17

19. Solving Algebraic Equations r – 7 = 22 8 + y = 17

20. Solving Algebraic Equations r – 7 = 22 8 + y = 17 + 7 + 7 -8 -8

21. Solving Algebraic Equations r – 7 = 22 8 + y = 17 + 7 + 7 -8 -8 r = 29 y = 9 Check 29 – 7 = 22 8 + 9 = 17 22 = 22 17 = 17

22. Solving Algebraic Equations 9s= 27 v/4= 6

23. Solving Algebraic Equations 9s= 27 v/4= 6

24. Solving Algebraic Equations 9s= 27 v/4= 6 ÷ 9 ÷9 ×4 ×4

25. Solving Algebraic Equations 9s= 27 v/4= 6 ÷ 9 ÷9 ×4 ×4 s = 3 v = 24 Check: 9(3) = 27 24/4 = 6 27 = 27 6 = 6

26. Solving Algebraic Equations = 9 d 2 = 64

27. Solving Algebraic Equations = 9 d 2 = 64

28. Solving Algebraic Equations = 9 d 2 = 64 2 2√ √

29. Solving Algebraic Equations = 9 d 2 = 64 2 2 √ √

30. Solving Algebraic Equations = 9 d 2 = 64 2 2 √ √ p = 9 2 d = p = 81 d = 8 Check: 8 2 = 64 64 = 64

31. When you have an equation that has multiplication or division and addition or subtraction, you have to move the addition or subtraction first. say you have 4 x 2 - 2 = 6 if you get rid of the multiplication first by dividing both sides by 2, what happens? 4-2=3 2=3??!?! It just doesn’t work.

32. Solving Algebraic Equations 4 x 2 – 2 = 6 get rid of the subtraction first by adding 2 to both sides, 4 x 2 – 2 = 6 + 2 +2 4 x 2 = 8 8 = 8 That’s more like it.

33. Algebraic Equations Multiplication with Subtraction 2x – 4 = 8 +4 2x=12 22 x=6

34. Algebraic Equations Multiplication with Addition 5x + 10 = 80 -10 5x=70 55 x=14

35. Algebraic Equations Multiplication with Subtraction -3x – 4 = -82 +4 -3x=-78 -3 x=26

36. Algebraic Equations Division with Addition x/5 + 2 = 8 -2 =6 x=30 x5x5 (5)

37. Algebraic Equations C.L.T. -20 10x = 60 10 4x +6x + 20 = 80 ( ) 10x + 20 = 80 10 x = 6

38. Algebraic Equations C.L.T. -12 5x = 80 5 3x + 2x + 20 - 8 = 92 5x + 12 = 92 5 x = 16 ( ) ( )

39. Common Formulas SIMPLE INTEREST: Interest = Principle x Rate x Time DISTANCE: Distance = Rate x Time TOTAL COST: Total Cost = (number of units)x (price per unit)

40. Substituting in Formulas If Sarah drove 390 miles at an average speed of 60 miles per hour, how long did it take her to complete her trip?

41. Substituting in Formulas If Sarah drove 390 miles at an average speed of 60 miles per hour, how long did it take her to complete her trip? 1) Write the formula you need: Distance = Rate x Time D = RT

42. Substituting in Formulas If Sarah drove 390 miles at an average speed of 60 miles per hour, how long did it take her to complete her trip? D = RT 2) Substitute numbers for variables where possible: 390 = 60T

43. Substituting in Formulas 1) Write the formula you need: D = RT 2) Substitute numbers for variables where possible: 390 = 60T 3) Solve the equation using your algebra skills 390 = 60T ÷60 6.5 = T It took Sarah 6.5 hours to complete her trip.

44. Substituting in Formulas 1) If the total cost of a shipment of pet food cans was $350 and there were 875 cans, how much did each can cost? 2) Write the formula you need: Cost = # of Units x Price per unit C = UP 2) Substitute numbers for variables where possible: 350 = 875P 3) Solve the equation using your algebra skills 350 = 875P ÷875 ÷875.4 = P Each can costs.4 dollars – or 40 cents.

45. Solving Algebraic Equations Try the practice problems on pages 144 and 145 and check your answers online. Try pages 24 and 25 in the GED Math Practice booklet and enter your answers on Blackboard.