1. Mr. Istik Grade 9

2.  The Student Will Be Able To Solve One- Variable Equations using addition, subtraction, multiplication, division, and a combination of them, each to an accuracy of 90%  The Student Will Be Able To Relate Algebra Problems to real-life by solving word problems to an accuracy of 90%

3.  Navigate the following PowerPoint by clicking on the links that look like this: This Link will take you from one page to the next.  Occasionally, question slides will pop up. On these slides, answer the question. If your answer is correct, you’ll be sent to a “correct” slide, and will be able to continue on. If your answer is incorrect, you will be sent to an “incorrect” slide, and will either be directed to go back to the beginning of the content, or just sent back to the question slide to give another answer.

4.  The following pages and links combine to form the next unit. This unit is called Solving One-Variable Equations. The lesson goes in conjunction with Chapter 3 in your Algebra 1 book.  This lesson is extremely important in understanding math from now until you graduate from high school and eventually college, so please take this seriously.

5.  Before getting into actually solving the equations, we will first discuss the importance of symbols and variables.  SYMBOL - something that is used to represent something else  Go to the next slide and take note of the symbols!

6.  Note what the following pictures represent: America ReligionPirates

7. Money STOP The Number 4

8. a VARIABLE is just a letter that represents a number! For instance, x = 4 makes x a symbol that represents 4 in this example! All letters can be used as a symbol in algebra.

9.  Identity Property Says that 1 a = a  Commutative Property Says that a b = b a  Distributive Property Says that a(b + c) = ab + ac  Associative Property a (b c) = (a b) c

10.  Coefficient The number multiplied by the variable in a term Examples The Coefficient Here is 4. The Coefficient Here is 3.

11.  Constant The number added to or subtracted to the variable in a term  This number has no variable attached to it (like a coefficient does). Examples The Constant Here is -1. The Constant Here is 7.

12.  Equivalent Equations with the same solutions as the original one. Example: x + 7 and y – 3 both equal 5 in this case, making the expressions (x + 7) and (y – 3) equivalent.

13.  Reciprocal You will get 1 if you multiply a number by its reciprocal (or all of the terms will cancel out. Ex. a/b has the reciprocal b/a Ex2:  Are reciprocals of one another because:  3/2 times 2/3 = 1  And because one is the “inverse” of the other

14.  By the Associative Property, x (y z) equals which of the following x (y z) (x y z) (x y) z ((x ((y) z)))

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17. Now, let’s move on to actually SOLVING these equations!

19. It is easiest to proceed by example. Try the following example: Solve x – 5 = -13 To solve this equation, you want to get x by itself. How might you do this?

20.  Solve x – 5 = -13  x – 5 = -13  Rewrite the original equation  x – 5 + 5 = -13 + 5  Add 5 to each side.  x = -8.  Simplify, and our solution is x = -8.

21.  Now Check Your Answer  x – 5 = -13  Rewrite the original equation  -8 - 5 = -13  Substitute -8 in for x.  -13 = -13  The Solution is Correct!

22.  Did you understand what was meant by the word “simplify?”  Definition: Simplify  To Take an equation down to the simplest terms.  Ex: we had x – 5 + 5 = -13 + 5 in the earlier example.  Obviously -5 + 5 = 0, so those terms essentially CANCEL.  Additionally, -13 + 5 = -8  Therefore you could write x – 5 + 5 = -13 + 5 more simply as x = -8.

23.  Now Try Another Example!  Solve r + 3 = 2  r + 3 = 2  Rewrite the original equation  r + 3 – 3 = 2 – 3  Subtract 3 from each side.  r = -1.  Simplify, and our solution is r = -1.

24.  Now Check Your Answer  r + 3 = 2  Rewrite the original equation  -1 + 3 = 2  Substitute -1 in for r.  2 = 2  The Solution is Correct!

25.  x + 5 = 29 x = 24 x = 34 x = 5.4 x = 22

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29. Again, it is best to proceed by example. Try the following example: Solve -4x = 16 You want to get x by itself. What would you do?

30.  Solve -4x = 16  -4x = 16  Rewrite the original equation  -4x = 16 -4  Divide each side by -4  x = -4.  Simplify, and our solution is x = -4.

31.  Now Check Your Answer  -4x = 16  Rewrite the original equation  -4 (-4) = 16  Substitute -4 in for x.  -16 = -16  The Solution is Correct!

32.  Try Another! Solve ½x = 5  ½x = 5  Rewrite the original equation  ½x (2) = 5 (2)  Multiply each side by 2.  x = 10.  Simplify, and our solution is x = 10.

33.  Now Check Your Answer  ½x = 5  Rewrite the original equation  ½(10) = 5  Substitute 10 in for x.  5 = 5  The Solution is Correct!

34.  3x= 12 x = 36 x = 9 x = 15 x = 4

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38. Multi-Step Equations are equations that include solving by addition and/or subtraction as well as by multiplication and/or division. Once again, we will proceed by example. Try the following: Solve 7x + 6 = -8

39.  We will start manipulating the equation by doing the addition and subtraction part.  7x + 6 = -8  Rewrite the original equation  7x + 6 – 6 = -8 – 6  Subtract 6 from each side.  7x = -14  Simplify.

40.  Now we’ll do the multiplication part. Remember, from the last slide we were left with 7x = -14  7x = -14 7 7  Divide each side by 7  x = -2.  Simplify once more, and our solution is x = -2.

41.  Now Check Your Answer  7x +6 = -8  Rewrite the original equation  7(-2) +6 = -8  Substitute -2 in for x.  -14 +6 = -8  Distribute (multiply the 7 and -2).  -8 = -8  The Solution is Correct!

42.  Now Try Another Example  Solve 7x – 3x – 8 = 24  7x – 3x – 8 = 24  Rewrite the original equation  Notice that there are 2 “x” terms.  4x – 8 = 24  Combine Like Terms  4x – 8 + 8 = 24 + 8  Add 8 to each side.

43.  4x = 32  Simplify.  4x = 32 4  Divide each side by 4.  x = 8.  Simplify once more to get x = 8.

44.  Now Check Your Answer.  7x – 3x – 8 = 24  Rewrite the original equation  7(8) – 3(8) – 8 = 24  Substitute 8 in for EACH x.  56 – 24 – 8 = 24  Distribute.  24 = 24  The Solution is Correct!

45.  5 – 4x = 8 – x x = 5 x = -1 x = 3 x = 8

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49.  A word problem is just a paragraph version of an equation.  The tricky part is to determine what the word problem means in equation terms.  Once you figure out how to do that part, it becomes either an addition/subtraction, multiplication/division problem or a multi-step problem.

50.  In order to solve word problems, the best way, again, is to proceed by example.  Try this one: Jason sells chocolate bars for his baseball team. Each chocolate bar costs 2 dollars. If at the end of the day he had 38 dollars, how many chocolate bars did Jason sell?

51. WORD PROBLEMHOW TO SOLVE  Jason sells chocolate bars for his baseball team. Each chocolate bar costs 2 dollars. If at the end of the day he had 38 dollars, how many chocolate bars did Jason sell?  First, identify all of the important facts/numbers in the problem.  We see that Jason sold each chocolate bar for 2 dollars and ended up with 38 dollars

52. WORD PROBLEMHOW TO SOLVE  Jason sells chocolate bars for his baseball team. Each chocolate bar costs 2 dollars. If at the end of the day he had 38 dollars, how many chocolate bars did Jason sell?  Now that we have these facts, we can derive the equation  Since each bar costs 2 dollars, then we know 2 will be in the problem somehow

53. WORD PROBLEMHOW TO SOLVE  Jason sells chocolate bars for his baseball team. Each chocolate bar costs 2 dollars. If at the end of the day he had 38 dollars, how many chocolate bars did Jason sell?  We want to MULTIPLY the 2 by our variable (call it x).  Therefore, a 2x will appear somewhere in the equation that we want to solve

54. WORD PROBLEMHOW TO SOLVE  Jason sells chocolate bars for his baseball team. Each chocolate bar costs 2 dollars. If at the end of the day he had 38 dollars, how many chocolate bars did Jason sell?  Finally, 38 is the result of our equation.  Combing the two together, we can derive the following equation: 2x = 38

55.  Now that we have our equation, we can solve it just like we did earlier.  2x = 38  Rewrite the original equation  2x = 38 2  Divide each side by 2  x = 19.  Simplify, and our solution is x = 19.

56.  Now Check Your Answer  2x = 38  Rewrite the original equation  2 (19) = 38  Substitute 19 in for x.  38 = 38  The Solution is Correct!

57.  The last component of a word problem is a “summary sentence”  Definition: Summary Sentence  A sentence that goes with word problems that sums up your answer in sentence or word form.  So for our example, x = 19 would be answered as follows: Jason sold a total of 19 chocolate bars.

58.  Now You Try:  Kelly has 15 DVDs in her case. After buying more DVDs at The Exchange, she now has 23. How many more CDs did Kelly buy? Kelly Bought 4 CDsKelly Bought 23 CDs Kelly Bought 15 CDsKelly Bought 8 CDs

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61.  Now we’ll try a more difficult word problem, a multi-step problem.  Solve this: Dan and Amber are adding up their ages. Dan is 2 times Amber’s age then minus 5. Amber is 9 years old. How old is Dan?

62. WORD PROBLEMHOW TO SOLVE  Dan and Amber are adding up their ages. Dan is 2 times Amber’s age then minus 5. Amber is 9 years old. How old is Dan?  First, identify all of the important facts/numbers in the problem.  We see that Amber is 9 years old.  We see that Dan is 2 times Amber’s age then minus 5.

63. WORD PROBLEMHOW TO SOLVE  Dan and Amber are adding up their ages. Dan is 2 times Amber’s age then minus 5. Amber is 9 years old. How old is Dan?  Since Dan is 2 times Amber’s age then minus 5, and Amber is 9, our equation is then 2x – 5 = 9

64.  Now solve the equation we just got.  2x – 5 = 9 Rewrite the original equation  2x – 5 + 5 = 9 + 5 Add 5 to each side  2x = 14 2 2 Divide each side by 2  x = 7. Simplify, and our solution is x = 7.

65.  Now Check Your Answer  2x – 5 = 9  Rewrite the original equation  2 (7) – 5 = 9  Substitute 7 in for x.  14 – 5 = 9  Distribute (multiply the 7 and 2).  9 = 9  The Solution is Correct!

66.  Don’t forget! We have to write a summary sentence for this problem!  Our Summary Sentence is: Therefore, Amber is 9 years old and Dan is 7 years old.

67.  Now Try This problem: Mike sells cakes at a local bakery. He charges 10 dollars for each cake. At the end of the day, there was 180 dollars worth of cake sold. How many cakes did Mike sell? Mike Sold 18 CakesMike Sold 10 Cakes Mike Sold 170 CakesMike Sold 8 Cakes

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71.  Solving Equations with Fractions and Decimals in them is just the same as solving using addition, subtraction, multiplication, division or a combination, except instead of integers, fractions and decimals are used.

72.  What is a fraction? A fraction is a number that represents a part of a whole. These numbers are written with a forward slash or horizontal line.  Ex: ½, 2/3, 5/8,  NUMERATOR (the number on top)  DENOMENATOR (the number on bottom)

73.  What is a decimal? A decimal is also a number that represents part of a whole number. This time, it is represented with by a number after a period.  Ex:  1.4, 5.9,.08,.01763, etc  What is an integer? An integer is a whole number. It is every natural number, (0, 1, 2, 3,...) and every negative natural number (-1, -2, -3,...)

74.  Since you will solve these equations in the same way that you solved the other ones, we will proceed by example.  So, let’s try this example: Solve 0.2x + 5 = 15

75.  0.2x + 5 = 15  Rewrite the original equation  0.2x + 5 – 5 = 15 – 5  Subtract 5 from each side.  0.2x = 10  Simplify.  Keep GOING!!!

76.  0.2x = 10 0.2  Divide each side by 0.2  You Can Use Your Calculator if you need it!  x = 50.  Simplify once more, and our solution is x = 50.

77.  -1.2x - 3 = 9 x = 1 x = -1 x = 10 x = -10

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80.  Let’s try this example:  Solve ¼x + 6 = 7  ¼x + 6 = 7  Rewrite the original equation  ¼x + 6 – 6 = 7 – 6  Subtract 6 from each side.  ¼x = 1  Simplify.

81.  ¼x = 1  Rewrite the equation again  ¼x (4) = 1 (4)  Multiply each side by 4.  Note that 4 is the RECIPROCAL of ¼.  And remember two numbers that are reciprocals of one another when multiplied together leaves only 1, thus x is left here.  x = 4.  Simplify, and our solution is x = 4.

82.  ½x + 6 = 9 x = 1 x = 3 x = 6 x = 9

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