1. Solving Linear Equations in One Variable Digital Lesson

2. Equations and inequalities compare algebraic expressions. An equation is a statement that two algebraic expressions are equal. An equation always contains an equals symbol, while an expression does not. 3x – 7 = 23x – 7 Left side Right side Equation (to solve) Expression (to simplify or evaluate ) Slide 2.1- 2 Distinguish between expressions and equations.

3. Simplify algebraic expressions, combine like terms Simplify means to combine as many terms as possible in an expression… Be sure to combine only like terms by addition or subtraction of the coefficients (same variables with identical exponents) Example: Simplify 4m 2 + 3m – 2m 2 + 5m. subtract coefficients of m with power 2 (4 - 2) (m 2 ) add coefficients of m with power 1 (3 + 5) (m) 2m 2 + 8m Example: Simplify ( s – 3t) + (2t – 3s). stack or line up coefficients of s and coefficients of t s – 3t -3s + 2t add coefficients of s and t vertically -2s + -t

4. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 4 A linear equation in one variable is an equation which can be written in the form: ax + b = c for a, b, and c real numbers with a  0. Linear equations in one variable: 2x + 3 = 11 2(x  1) = 8 Not linear equations in one variable: 2x + 3y = 11 Two variables can be rewritten 2x + (  2) = 8. x is squared.Variable in the denominator (x  1) 2 = 8 can be rewritten x + 5 =  7. Linear Equation in One Variable

5. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 5 A solution of a linear equation in one variable is a real number which, when substituted for the variable in the equation, makes the equation true. Example: Is 3 a solution of 2x + 3 = 11? 2x + 3 = 11 2(3) + 3 = 11 6 + 3 = 11 Original equation Substitute 3 for x. False equation 3 is not a solution of 2x + 3 = 11. Example: Is 4 a solution of 2x + 3 = 11? 2x + 3 = 11 2(4) + 3 = 11 8 + 3 = 11 Original equation Substitute 4 for x. True equation 4 is a solution of 2x + 3 = 11. Example: Solution

6. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 6 Addition Property of Equations Addition and Subtraction Property of Equations If a = b, then a + c = b + c and a  c = b  c. Use these properties to solve linear equations. Example: Solve x  5 = 12. x  5 = 12 x  5 + 5 = 12 + 5 x = 17 17  5 = 12 Original equation The solution is preserved when 5 is added to both sides of the equation. Check the answer. 17 is the solution. That is, the same number can be added to or subtracted from each side of an equation without changing the solution of the equation.

7. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 7 Multiplication Property of Equations Multiplication and Division Property of Equations If a = b and c  0, then ac = bc and. That is, an equation can be multiplied or divided by the same nonzero real number without changing the solution of the equation. Example: Solve 2x + 7 = 19. 2x + 7 = 19 2x + 7  7 = 19  7 2x = 12 x = 6 2(6) + 7 = 12 + 7 = 19 Original equation The solution is preserved when 7 is subtracted from both sides. Simplify both sides. 6 is the solution. Check the answer. The solution is preserved when each side is multiplied by.

8. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 8 Solving Linear Equations To solve a linear equation in one variable: 1. Simplify both sides of the equation. 2. Use the addition and subtraction properties to get all variable terms on the left-hand side and all constant terms on the right-hand side. 3. Simplify both sides of the equation. 4. Divide both sides of the equation by the coefficient of the variable. Example: Solve x + 1 = 3(x  5). x + 1 = 3(x  5) x + 1 = 3x  15 x = 3x  16  2x =  16 x = 8 The solution is 8. Check the solution: Original equation Simplify right-hand side. Subtract 1 from both sides. Subtract 3x from both sides. Divide both sides by  2. True (8) + 1 = 3((8)  5)  9 = 3(3)

9. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 9 Solving Linear Equations To solve a linear equation in one variable: 1. Simplify both sides of the equation. 2. Use the guess and check method by using a common value for both variables and the substitution property. 3. Simplify both sides of the equation. 4. Check to see if solutions match; If not try values higher or lower. Example: Solve x + 1 = 3(x  5). x + 1 = 3(x  5) 5 + 1 = 3(5  5) 5 + 1 = 6 : 3 (0) = 0 8 + 1 = 3(8  5) 8 + 1 = 9 : 3 (3) = 9 The solution is 8. Check the solution: Original equation Substitute a value; when x = 5. Check to see if solutions match. Try another value for x = 8. Check – Solutions match True (8) + 1 = 3((8)  5)  9 = 3(3)

10. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 10 Solving Linear Equations To solve a linear equation in one variable: 1. Simplify both sides of the equation. 2. Count the number of constant value in the equation. Draw that many boxes. Label the first box with your variable; last box is solution. 3. Use order of operations to construct flow chart from left to right. 4. Use inverse operation to construct flow chart from right to left. Example: Solve x + 1 = 3(x  5). x + 1 = 3(x  5) x + 1 = 3x  15  2x =  16 x = The solution is 8. Original equation Simplify right-hand side. Subtract 1 from both sides. Subtract 3x from both sides. Construct two boxes ( -2 and -16) -16 Fill in arrows with operations; solve x = x = 8 -16  -2  -2

11. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 11 Example: Solve Example: Solve 3(x + 5) + 4 = 1 – 2(x + 6). 3(x + 5) + 4 = 1 – 2(x + 6) 3x + 15 + 4 = 1 – 2x – 12 3x + 19 = – 2x – 11 3x = – 2x – 30 5x = – 30 x =  6 The solution is  6. Original equation Simplify. Subtract 19. Add 2x. Divide by 5. Check. True  3 + 4 = 1 3(– 1) + 4 = 1 – 2(0) 3(– 6 + 5) + 4 = 1 – 2(– 6 + 6)

12. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 12 Using a Common Denominator Equations with fractions can be simplified by multiplying both sides by a common denominator. The lowest common denominator of all fractions in the equation is 6. 3x + 4 = 2x + 8 3x = 2x + 4 x = 4 Multiply by 6. Simplify. Subtract 4. Subtract 2x. Check. True Example: Solve. 4 4 66

13. Solve. 4x + 8x = –9 + 17x – 1 CLASSROOM EXAMPLE 2 Using the Properties of Equality to Solve a Linear Equation Solve. 6 – (4 + x) = 8x – 2(3x + 5) Solve. 0.02(60) + 0.04x = 0.03(50 + x)

14. Type of Linear Equation Number of Solutions Indication when Solving ConditionalOneFinal line is x = a number. IdentityInfinite; solution set {all real numbers} Final line is true, such as 0 = 0. ContradictionNone; solution set  Final line is false, such as –15 = –20. Slide 2.1- 14 Identify conditional equations, contradictions, and identities.

15. Solve each equation. Decide whether it is a conditional equation, an identity, or a contradiction. 5(x + 2) – 2(x + 1) = 3x + 1 CLASSROOM EXAMPLE 6 Recognizing Conditional Equations, Identities, and Contradictions 5(3x + 1) = x + 5

16. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 16 Example: Word Problem Alice has a coin purse containing $5.40 in dimes and quarters. There are 24 coins all together. How many dimes are in the coin purse? Let the number of dimes in the coin purse = d. Then the number of quarters = 24  d. 10d + 25(24  d) = 540 Linear equation 10d + 600  25d = 540 10d  25d =  60  15d =  60 d = 4 Simplify left-hand side. Subtract 600. Simplify right-hand side. Divide by  15. There are 4 dimes in Alice’s coin purse.

17. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 17 Example: Word Problem The sum of three consecutive integers is 54. What are the three integers? Three consecutive integers can be represented as n, n + 1, n + 2. n + (n + 1) + (n + 2) = 54 3n + 3 = 54 3n = 51 n = 17 Simplify left-hand side. Subtract 3. Divide by 3. The three consecutive integers are 17, 18, and 19. 17 + 18 + 19 = 54. Check. Linear equation

18. Modeling Algebraic Expressions Using operation words: +, , *, /, , , ,  Examples: 8 less than the product of 5 times a number. 5n - 8 The quotient of 11 and the sum of 7 and the value of x. 11 / (x + 7) The total cost of notebooks if each notebook costs $2.25 2.25n Teaching Concept #1 Total, Sum, Difference, Between, Product, per, Quotient, average, Split equally, each, Give or take.

19. Modeling Algebraic Expressions Using Graphs: A number greater than or equal to 1 Positive numbers Values between -2 and 2 The cost of x notebooks is each notebook costs $2.25 Teaching Concept #2

20. Modeling Algebraic Expressions Using Tables: The cost of notebooks purchased if each notebook costs $2.25. Teaching Concept #5 The data plan for your cell phone with has a flat rate of $30 for 10MB and $10 for each additional 2MB. 012345 02.254.506.759.0011.25 0510121420 30.00 40.0050.00130.00

21. What’s an inequality? Is a range of values, rather than ONE set number An algebraic relation showing that a quantity is greater than or less than another quantity. Speed limit:

22. Symbols Less than Greater than Less than OR EQUAL TO Greater than OR EQUAL TO

23. Solutions…. You can have a range of answers…… -5 -4 -3 -2 -1 0 1 2 3 4 5 All real numbers less than 2 x< 2

24. Solutions continued… -5 -4 -3 -2 -1 0 1 2 3 4 5 All real numbers greater than -2 x > -2

25. Solutions continued…. -5 -4 -3 -2 -1 0 1 2 3 4 5 All real numbers less than or equal to 1

26. Solutions continued… -5 -4 -3 -2 -1 0 1 2 3 4 5 All real numbers greater than or equal to -3

27. Did you notice, Some of the dots were solid and some were open? -5 -4 -3 -2 -1 0 1 2 3 4 5 Why do you think that is? If the symbol is > or < then dot is open because it can not be equal. If the symbol is  or  then the dot is solid, because it can be that point too.

28. Write and Graph a Linear Inequality Sue ran a 2-K race in 8 minutes. Write an inequality to describe the average speeds of runners who were faster than Sue. Graph the inequality. Faster average speed > Distance Sue’s Time -5 -4 -3 -2 -1 0 1 2 3 4 5

29. Solving an Inequality Solving a linear inequality in one variable is much like solving a linear equation in one variable. Isolate the variable on one side using inverse operations. Add the same number to EACH side. x – 3 < 5 Solve using addition: +3 x < 8

30. Solving Using Subtraction Subtract the same number from EACH side. -6

31. Using Subtraction… Graph the solution. -5 -4 -3 -2 -1 0 1 2 3 4 5

32. Using Addition… -5 -4 -3 -2 -1 0 1 2 3 4 5 Graph the solution.

33. THE TRAP….. When you multiply or divide each side of an inequality by a negative number, you must reverse the inequality symbol to maintain a true statement.

34. Solving using Multiplication Multiply each side by the same positive number. (2)

35. Solving Using Division Divide each side by the same positive number. 33

36. Solving by multiplication of a negative # Multiply each side by the same negative number and REVERSE the inequality symbol. Multiply by (-1). (-1) See the switch

37. Solving by dividing by a negative # Divide each side by the same negative number and reverse the inequality symbol. -2

38. Finding Absolute Value Absolute value is used to describe how to operate with positive and negative numbers. The absolute value of 3 is +3 because 3 is 3 units from 0 on the number line. Geometric Meaning of Absolute Value The absolute value of a real number a, denoted is the distance from 0 to a on the number line. This distance is always nonnegative.

39. Exploration Determine the solution for each equation. – 4,4 4, -4 9,9 9, -9 No Solution

40. What did you notice? Summarize what you noticed from the previous solutions. When the absolute value is equal to zero. When a variable is inside an absolute value, there are two solutions. When a variable is inside an absolute value, there are two solutions. When an absolute value is set equal to a negative number, there is no solution. (this is important to remember) When an absolute value is set equal to a negative number, there is no solution. (this is important to remember) Can you think of a situation where there would be one solution? Can you think of a situation where there would be one solution?

41. Steps for solving absolute value equations. **Need to isolate the absolute value expression** 1)Undo addition or subtraction outside of absolute value. 2)Undo multiplication or division outside of absolute value. 3)Set expression inside absolute value equal to the given value and its opposite. 4)Solve for variable using steps for solving equations. 1.Distribute 2.Combine Like Terms 3.Move Variable to One Side 4.Undo + or – 5.Undo × or ÷

42. Examples Solving basic absolute value equations

43. Examples continued -24, 8 1, 5

44. More Examples Solving absolute value equations when there are terms outside the symbols

45. Even More Examples -2, 6 0, 8/3