1. Week 1 Real Numbers and Their Properties (Section 1.6, 1.7, 1.8)

2. Week 1 Objectives This week students will: Utilize proper math terminology in written explanations. Compute basic operations with signed numbers. Exemplify properties of real numbers. Execute simplification techniques on expressions and equations.

3. Solving equation Given x + 3 = 5 Step 1: Assume x represents a number Step 2: Through trial and error, find a number such that 3 added to that number gives you 5. Answer: The solution is x = 2 because 2 + 3 = 5 x + 3 = 5 is what is known as algebraic equation. x is the variable; 3 and 5 are the constants. Solving an algebraic equation means find a value for the variable such that the equation remains true.

4. Solving equations If x is a number, then x – 10 = 3. Step 1: through trial and error come up with a number such that the number minus ten is equal to 3. Solution: the number is 13 because 13 – 10 = 3 X -10 = 3 is an expression that involves difference. X + 3 = 5 is an expression that involves sum. is an expression that involves quotient 3x 2 y is an expression that involves product. It is important to realize what sort of operation is involved in an equation because, often times, based on that we need to apply appropriate operations to solve the equation.

5. Examples 1.X + 4 = 9 2.X – 5 = 10 3.X + 2 = -3 (hint: think about what negative numbers that you need to add to 2 to give you -3) 4.X – 3 = -9 (hint: again think about what negative numbers that you need to add to -3 to give you -9). Answers: (1). 5, (2). 15, (3). -5, (4). -6

6. Equations An equation is a mathematical expression that involves an unknown term. So, given x – 2 = 3 you try to find a number such that the number minus two gives you a three. From trial and error, you know the number is 5. But trial and error methods cannot be used all the time. If I give you an equation like x – 1986754 = 354680891, trial and error method will not work here. So, we need to learn about various methods, tricks that we apply to solve equations. The following few slides will teach you about all those. The slides will contain definitions, properties of real numbers (section 1.7) and using the properties to solve equations and simplify expressions.

7. Properties of addition Addition property of 0: adding zero to any number leaves that number unchanged. Thus, 13 + 0 = 13 Commutative property of addition: changing the order of numbers in an addition does not affect the result: 26 + 37 = 37 + 26 = 63 Associative property of addition: changing the grouping of numbers in an addition does not affect the result: (2 + 3) + 5 = 2 + (3 + 5) = 10 Distributive Property of addition and multiplication: 2(3 + 5) = 2*3 + 2*5 = 6 + 10 = 16; it is known as the distributive property because we are distributing 2 to 3 and 5 (multiplying 3 and 5 by 2).

8. Properties of multiplication 1.Distributive Property: the sum of two numbers times a third number is equal to the sum of each number multiplied to the third number: 2(3 + 4) = 2*3 + 2* 4 = 6 + 8 = 14 2.Multiplication Property of zero: any number multiplied to zero gives us a zero: 2 * 0 = 0, 7 * 0 = 0, 10 * 0 = 0 3.Multiplication property of 1: any number multiplied to one gives us back the number: 2 * 1 = 2, 4 * 1= 4 4.Commutative property: the order of the numbers in the product does not affect the result. Thus, 2*3 = 3*2 = 6 5.Associative property: Grouping does not affect the result: (2*3)*1 = 2(3*1) = 6

9. Simplifying expressions with variables The first step in solving any equation is simplifying the given expression. Such simplifications can be done by applying associative, distributive and commutative properties of addition and multiplication. Example 1: 3(2x) can be written as (3*2)x by applying the associative property of multiplication. Thus, 3(2x) = (3*2)x = 6x is the simpler form. Example 2: -4(8x) = (-4*8)x = -32x, by associative property. Example 3: 2(x + 1) = 2x + 2, by the distributive property – distributing 2 to both x and 1

10. Simplifying expression with variables continues Example 4: Simplify 4(x – 1) + 2 Step 1: Distribute 4 to x and -1 to get 4x – 4 + 2 Step 2: Combine -4 and 2 to get 4x +(-4 + 2) = 4x + (-2) = 4x – 2 Example 5: Simplify -2(3x – 5y) + 9 Step 1: Rewrite the expression as -2(3x + (-5y)) + 9 Step 2: Distribute -2 to get (-2*3)x + (-2*-5)y + 9 = -6x + 10y + 9 Example 6: Simplify 18x + 3x Step 1: 18x + 3x = (18 + 3)x, by distributive property Step 2: Add 18 + 3 to get 21x as the solution. Example 7: Simplify 20x – 5x Step 1: 20x – 5x = (20 – 5) x, by distributive property Step 2: Subtract to get 15x as the solution

11. Quiz Simplify the following: 1.-9(x + 2) 2.11(5x – 13) 3.5(2x + 3y) – 10 4.14x – 10x 5.3(2x + 6) - 10 Answers: (1). -9x -18, (2). 55x – 143, (3). 10x + 15y – 10, (4). 4x, (5). 6x + 18

12. Similar terms and simplification Given 2x + 4x, 2x and 4x are known as similar terms because the variable terms are same. Similarly in 2x + 3 + 9, 3 and 9 are similar terms both of them are some constant numbers and do not involve any variables. Using the concept of similar terms, we can simplify more complicated equations like 2x + 4x + 3 + 9. Step 1: Combine (group together) the like terms (2x + 4x) + (3 + 9) Step 2: Simplify each grouped terms. Thus, 2x + 4x = (2 + 4)x = 6x. And, 3 + 9 = 12. Step 3: Write down the final solution as 6x + 12

13. Similar terms and simplification -- examples Simplify 6x – 3 + 4x + 8 Step 1 Grouping: (6x + 4x) + (-3 + 8) Step 2 Simplify each group: 6x + 4x = (6 + 4)x = 10x and -3 + 8 = 5 Step 3 Write down the answer: 10x + 5 Simplify 3(2x + 1) + 5(3x – 6) Step 1: Distribute: 3*2x + 3*1 + 5*3x + 5*-6 = 6x + 3 + 15x – 30 Step 2: Combine like terms (6x + 15x) + (3 – 30) Step 3: Simplify each group (6 + 15)x + (3 + -30) = 21x + -27 = 21x – 27

14. Quiz Simplify the following 1.7x + 5x + 2 – 9 2.-8a + 12 + 3a + 1 3.2(x + 3) + 4(x + 2) 4.7(2a + 2) + 4(5a – 1) Answers: (1). 12x – 7, (2). -5a + 13, (3). 6x + 14, (4). 34a + 10

15. Value of algebraic expression An equation of the form 3x + 5 is known as an algebraic expression because it involves a variable (an unknown term) x. An algebraic expression will take on different values depending upon the value of the unknown term (variable). Thus, for 3x + 5, it can take various values : X = 2  3(2) + 5 = 6 + 5 = 11 X = 0  3(0) + 5 = 0 + 5 = 5 X = -1  3(-1) + 5 = -3 + 5 = 2

16. Value of algebraic expression -- example Find the value of the expression, 9x + 21, when x = -3 Step 1: Substitute x = -3 in the expression: 9(-3) + 21 Step 2: Compute the value: -27 + 21 = -6 Find the value: 5a + 5, a = 1 Step 1: Substitute a = 1 in the expression: 5(1) + 5 Step 2: Compute the value: 5 + 5 = 10 Find the value 11t – 9, t = 2 Step 1: Substitute t = 2 in the expression: 11(2) – 9 Step 2: Compute the value: 22 – 9 = 13

17. quiz Compute the values of the following: 1.8x – 9, x = -3 2.-9 - 3x, x = 1 3.4x + 10, x = 7 Answers: (1). -33, (2). -12, (3). 38

18. Solutions of equations A solution for an equation is a number that when used in place of the variable makes the equation a true statement. Thus, for 2x + 3 = 7, the solution is x = 2 because: Step 1: substitute x = 2: 2(2) + 3 = 7 Step 2: compute: 4 + 3 = 7 Step 3: simplify: 7 = 7 Step 4: reach conclusion: because 7 = 7 is a true statement, thus x = 2 is the solution for 2x + 3 = 7 We found the solution through trial-and-error method. Next we will learn how to find the solution by actually solving an equation and not using trial and error method.

19. quiz Is 2 a solution of 5x + 5 = 15? Answer: It is a solution because 5(2) + 5 = 15  10 + 5 = 15  15 = 15 and it is true.

20. Addition property of equality The property says that if A = B, then A + C = B + C, that is adding same quantity to both sides of an equation does not change the solution to an equation. We apply this property all the time in solving any equation. So, it is very important that memorize this property. Example: We are going to solve the equation x - 3 = 10 using the above property. Step 1: Since we have a negative 3 on the left side, so we add positive 3 on both sides: x – 3 + 3 = 10 + 3 Step 2: Cancel the negative and positive 3 to get x = 13 and that is our solution.

21. Subtraction property of equality Just like the addition property, we also have a subtraction property of equality which says that if A = B, then A – C = B – C Application of the property example: Solve x + 3 = 10 Step 1: We have positive 3 on the left side. So, add negative 3 on both sides to get x + 3 + (-3) = 10 + (-3) Step 2: Simplify: x + 3 – 3 = 10 – 3 Step 3: Cancel the positive and number in the left to get x = 10 – 3 Or x = 7 and that is our solution.

22. Rules to remember and follow for solving equations The following rules/steps are usually applied in solving equations. Thus, you should really try to remember them: 1.Apply the distributive, associative, commutative properties of addition and multiplication to simplify expressions. 2.Apply the addition/subtraction property of equality 3.Combine like terms 4.In an equation, all the terms involving variables should always be on one side of the equation; all the constant numbers should be on the other side of the equation.

23. Solving equations -- examples Solve 4x – 3 - 3x = 5 – 10 Step 1 combine like terms (4x – 3x) – 3 = 5 – 10 Step 2 simplify: (4-3)x – 3 = -5  x – 3 = -5 Step 3 add positive 3 on both sides: x – 3 + 3 = -5 + 3 Step 4 simplify by cancelling the positive and negative 3 to get: x = -2 and that is our solution Check (optional): You can always check if you have the correct solution or not. Substitute your solution in the given equation and see if you get a correct statement: 4(-2) – 3 – 3(-2) = 5 – 10  -8 – 3 + 6 = -5  -11 + 6 = -5  -5 = 5 and a true statement.

24. Example continues Solve 5x + 9 = 4x + 3 Step 1: Add negative 9 to both sides of the equation and simplify: 5x + 9 + (-9) = 4x + 3 + (-9)  5x + 9 – 9 = 4x +(3 -9)  5x = 4x – 6 Step 2: Since there is a positive 4x on the right side of the equation, we need to bring it back to the left side of the equation because we need to have all the variable terms on one side and numbers on the other. So, add negative 4x on both sides of the equation and simplify: 5x + (-4x) = 4x – 6 + (-4x)  5x – 4x = 4x – 6 – 4x  (5 -4)x = (4x – 4x ) – 6  x = 0 – 6  x = 6 and that is our solution.

25. Quiz Solve for x: 1.7x + 5 = 19 2.7a – 6 – 6a = -3 + 1 3.2x + 3 = x + 9 Solutions: (1). x = 2, (2). a = 4, (3). x = 6

26. Reciprocal Reciprocal of a number is 1 divided by that number. Thus: Example 1: Reciprocal of 2 is ½ Example 2: Reciprocal of 8 is 1/8 Example 3: Reciprocal of 1 is 1/1 which is nothing but 1 Example 4: Reciprocal of -4 is 1/-4 which is written as -1/4 Example 5: Reciprocal of -10 is 1/-10 or -1/10 NOTE: There is no reciprocal of 0 because 1/0 is not acceptable in mathematics. Reciprocal of -1 is 1/-1 which is nothing but -1

27. Multiplication property of equality If A = B, then AC = BC, that is, multiplying both sides of an equation with a non-zero quantity does not change the solution of the equation. The property is used in solving equations Example 1: Solve 2x = 8 Step 1: 2 is attached to the variable x. Reciprocal of 2 is ½. So, multiply both sides of the equation with ½: Step 2: simplify by cancelling out the 2’s to get: That is, x = 4 is the solution.

28. Example continues Solve 5x – 8x + 3 = 4 – 10 Step 1: (5 – 8)x + 3 = 4 + (-10)  -3x + 3 = -6 Step 2: Add negative 3 to both sides to get -3x + 3 + (-3) = -6 + ( 3)  -3x + (3 – 3) = -9  -3x = -9 Step 3: Multiply both sides by the reciprocal of -3 which is -1/3: Step 4: Simplify by cancelling out the 3’s to get: x = 3 as the solution

29. Quiz Solve for x: 1.-7x = -42 2.x/3 = 7 3.-2y – 8 = 2 4.4x – 7 = 2x + 11 Solutions: (1). x = 6, (2). x = 21, (3). y = -5, (4). x = 9

30. Another example Solve 6(3x – 6) + 10 = 5(2x + 8) – 10 Step 1 Distribute: 6*3x + 6*-6 +10 = 5*2x + 5*8 – 10 Step 2 Simplify: 18x + (-36) + 10 = 10x + 40 – 10  18x + (-36 + 10) = 10x + 30  18x -26 = 10x + 30 Step 3 Add positive 26 to both sides: 18x – 26 + 26 = 10x + 30 + 26  18x = 10x + 56

31. Example continues Step 4 Add negative 10x from both sides: 18x + (-10x) = 10x + 56 + (-10x) Step 5 Simplify: (18 – 10)x = (10 – 10)x + 56  8x = 56 Step 6 Multiply the reciprocal of 8 to both sides: Step 7 Simplify by cancelling the 8’s to get x = 7 and that is our solution

32. Distributive property of fraction multiplication Multiply Step 1: Add the fractions in the parenthesis: LCD of 5 and 7 is 35. Thus, multiply top and bottom of 2/5 with 7; multiply top and bottom of 1/7 with 5. Add the results: Step 2: Multiply the fraction outside the parenthesis to the fraction obtained in step 1:

33. Distribution property – another method Multiplication of can also be done by applying the distributive property a(b + c) = ab + ac: Factors of 15 are 3*5; factors of 21 are 3*7. Thus, the LCD is 3*5*7 = 105 Multiply the top and bottom of 4/15 with 7; multiply the top and bottom of 2/21 with 5 and add the result: Note: we obtain the same result