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1-9. Simplifying Algebraic Expressions. Course 2. Warm Up. Problem of the Day. Lesson Presentation. Warm Up Evaluate each expression for y = 3. 1. 3 y + y 2. 7 y 3. 10 y – 4 y 4. 9 y 5. y + 5y + 6 y 6. 10 y. 12. 21. 18. 27. 36. 30. Problem of the Day

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1-9

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  1. 1-9 Simplifying Algebraic Expressions Course 2 Warm Up Problem of the Day Lesson Presentation

  2. Warm Up Evaluate each expression for y = 3. 1.3y + y 2. 7y 3. 10y – 4y 4. 9y 5.y + 5y + 6y 6. 10y 12 21 18 27 36 30

  3. Problem of the Day Emilia saved nickels, dimes, and quarters in a jar. She had as many quarters as dimes, but twice as many nickels as dimes. If the jar had 844 coins, how much money had she saved? $94.95

  4. Learn to simplify algebraic expressions.

  5. Vocabulary term coefficient

  6. In the expression 7x + 9y + 15, 7x, 9y, and 15 are called terms. A term can be a number, a variable, or a product of numbers and variables. Terms in an expression are separated by + and –. x 3 7x + 5 – 3y2 + y + term term term term term In the term 7x, 7 is called the coefficient. A coefficient is a number that is multiplied by a variable in an algebraic expression. A variable by itself, like y, has a coefficient of 1. So y = 1y. Coefficient Variable 7 x

  7. Like terms are terms with the same variable raised to the same power. The coefficients do not have to be the same. Constants, like 5, , and 3.2, are also like terms. 1 2 w 7 3x and 2x w and 5 and 1.8 5x2 and 2x 6a and 6b 3.2 and n Only one term contains a variable The exponents are different. The variables are different

  8. Helpful Hint Use different shapes or colors to indicate sets of like terms. Additional Example 1: Identifying Like Terms Identify like terms in the list. 3t 5w2 7t 9v 4w2 8v Look for like variables with like powers. 3t 5w2 7t 9v 4w2 8v Like terms: 3t and 7t 5w2 and 4w2 9v and 8v

  9. Check It Out: Example 1 Identify like terms in the list. 2x 4y3 8x 5z 5y3 8z Look for like variables with like powers. 2x 4y3 8x 5z 5y3 8z Like terms: 2x and 8x 4y3 and 5y3 5z and 8z

  10. Combining like terms is like grouping similar objects. x x x x x x x x + = x x x x x x x x x x = 9x 4x + 5x To combine like terms that have variables, add or subtract the coefficients.

  11. Additional Example 2: Simplifying Algebraic Expressions Simplify. Justify your steps using the Commutative, Associative, and Distributive Properties when necessary. A. 6t – 4t 6t and 4t are like terms. 6t– 4t 2t Subtract the coefficients. B. 45x – 37y + 87 In this expression, there are no like terms to combine.

  12. Additional Example 2: Simplifying Algebraic Expressions Simplify. Justify your steps using the Commutative, Associative, and Distributive Properties when necessary. C. 3a2 + 5b + 11b2 – 4b + 2a2 – 6 Identify like terms. 3a2 + 5b+ 11b2 – 4b + 2a2 – 6 Group like terms. (3a2 + 2a2) + (5b – 4b)+ 11b2 – 6 5a2 + b + 11b2 – 6 Add or subtract the coefficients.

  13. 2 2x – 26x + 6 Check It Out: Example 2 Simplify. Justify your steps using the Commutative, Associative, and Distributive Properties when necessary. A. 5y + 3y 5y and 3y are like terms. 5y+ 3y 8y Add the coefficients. B. 2(x2 – 13x) + 6 Distributive Property. There are no like terms to combine.

  14. Check It Out: Example 2 Simplify. Justify your steps using the Commutative, Associative, and Distributive Properties when necessary. C. 4x2 + 4y + 3x2 – 4y + 2x2 + 5 4x2 + 4y + 3x2 – 4y + 2x2 + 5 Identify like terms. (4x2 + 3x2 +2x2)+ (4y – 4y) + 5 Group like terms. Add or subtract the coefficients. 9x2 + 5

  15. Additional Example 3: Geometry Application Write an expression for the perimeter of the triangle. Then simplify the expression. 2x + 3 3x + 2 x Write an expression using the side lengths. 2x + 3 + 3x + 2 + x Identify and group like terms. (x+ 3x + 2x) + (2 + 3) 6x + 5 Add the coefficients.

  16. Check It Out: Example 3 Write an expression for the perimeter of the triangle. Then simplify the expression. 2x + 1 2x + 1 x Write an expression using the side lengths. x + 2x + 1 + 2x + 1 (x + 2x + 2x) + (1 + 1) Identify and group like terms. 5x + 2 Add the coefficients.

  17. Lesson Quiz: Part I Identify like terms in the list. 1. 3n2 5n 2n3 8n 2.a5 2a2a3 3a 4a2 Simplify. Justify your steps using the Commutative, Associative, and Distributive Properties when necessary. 3. 4a + 3b + 2a 4.x2 + 2y + 8x2 5n, 8n 2a2, 4a2 6a + 3b 9x2 + 2y

  18. Lesson Quiz: Part II 5. Write an expression for the perimeter of the given figure. 2x + 3y x + y x + y 2x + 3y 6x + 8y

  19. Equations and Their Solutions 1-10 Course 2 Warm Up Problem of the Day Lesson Presentation

  20. Equations and Their Solutions 1-10 Course 2 Warm Up Evaluate each expression for x = 12. 1. x + 2 2. 3. x – 8 4. 10x – 4 5. 2x + 12 6. 5x + 7 14 x 4 3 4 116 36 67

  21. Equations and Their Solutions 1-10 Course 2 Problem of the Day Alicia buys buttons at a cost of 8 for $20. She resells them for $5 each. How many buttons does Alicia need to sell for a profit of $120? 48 buttons

  22. Equations and Their Solutions 1-10 Course 2 Learn to determine whether a number is a solution of an equation.

  23. Equations and Their Solutions 1-10 Course 2 Insert Lesson Title Here Vocabulary equation solution

  24. Equations and Their Solutions 1-10 Course 2 Ella has 22 CDs. This is 9 more than her friend Kay has. This situation can be written as an equation. An equation is a mathematical statement that two expressions are equal in value. An equation is like a balanced scale. Number of CDs Ella has 22 is equal to = 9 more than Kay has j + 9 Left expression Right expression

  25. Equations and Their Solutions 1-10 Course 2 Just as the weights on both sides of a balanced scale are exactly the same, the expressions on both sides of an equation represent exactly the same value. When an equation contains a variable, a value of the variable that makes the statement true is called a solution of the equation. 22 = j + 9 j = 13 is a solution because 22 = 13 + 9. 22 = j + 9 j = 15 is not a solution because 22  15 + 9. Reading Math The symbol ≠ means “is not equal to.”

  26. Equations and Their Solutions 1-10 ? 26 + 9 = 17 ? 35 = 17 Course 2 Additional Example 1A: Determining Whether a Number is a Solution of an Equation Determine whether the given value of the variable is a solution of t + 9= 17. 26 t + 9= 17 Substitute 26 for t.  26 is not a solution of t + 9 = 17.

  27. Equations and Their Solutions 1-10 ? 8 + 9 = 17 ? 17 = 17 Course 2 Additional Example 1B: Determining Whether a Number is a Solution of an Equation Determine whether the given value of the variable is a solution of t + 9= 17. 8 t + 9= 17 Substitute 8 for t.  8 is a solution of t + 9 = 17.

  28. Equations and Their Solutions 1-10 ? 22– 5 = 12 ? 17 = 12 ? 8– 5 = 12 ? 3 = 12 Course 2 Insert Lesson Title Here Check It Out: Example 1 Determine whether each number is a solution of x– 5= 12. A. 22 x– 5= 12 Substitute 22 for x.  22 is not a solution of x– 5 = 12. B. 8 x – 5= 12 Substitute 8 for x.  8 is not a solution of x– 5 = 12.

  29. Equations and Their Solutions 1-10 ? 35 = 32 ? 52 - 17 = 32 Course 2 Additional Example 2: Writing an Equation to Determine Whether a Number is a Solution Mrs. Jenkins had $32 when she returned home from the supermarket. If she spent $17 at the supermarket, did she have $52 or $49 before she went shopping? You can write an equation to find the amount of money Mrs. Jenkins had before she went shopping. If m represents the amount of money she had before she went shopping, then m - 17 = 32. $52 m – 17 = 32 Substitute 52 for m. 

  30. Equations and Their Solutions 1-10 ? 32 = 32 ? 49 - 17 = 32 Course 2 Additional Example 2 Continued Mrs. Jenkins had $32 when she returned home from the supermarket. If she spent $17 at the supermarket, did she have $52 or $49 before she went shopping? You can write an equation to find the amount of money Mrs. Jenkins had before she went shopping. If m represents the amount of money she had before she went shopping, then m - 17 = 32. $49 m – 17 = 32 Substitute 49 for m.  Mrs. Jenkins had $49 before she went shopping.

  31. Equations and Their Solutions 1-10 ? 14 = 12 ? 61 - 47 = 12 Course 2 Check it Out: Additional Example 2 Mr. Rorke had $12 when he returned home from buying a hat. If he spent $47 at the hat store, did he have $61 or $59 before he bought the hat? You can write an equation to find the amount of money Mr. Rorke had before he purchased a hat. If m represents the amount of money he had before he purchased a hat, then m – 47 = 12. $61 m – 47 = 12 Substitute 61 for h. 

  32. Equations and Their Solutions 1-10 ? 12 = 12 ? 59 - 47 = 12 Course 2 Check it Out: Additional Example 2 Continued Mr. Rorke had $12 when he returned home from buying a hat. If he spent $47 at the hat store, did he have $59 or $61 before he bought the hat? You can write an equation to find the amount of money Mr. Rorke had before he purchased a hat. If m represents the amount of money he had before he purchased a hat, then m – 47 = 12. $59 m – 47 = 12 Substitute 59 for h.  Mr. Rorke had $59 before he purchased a hat.

  33. Equations and Their Solutions 1-10 $5 for admission 5 + $2 per ride 2x Course 2 Additional Example 3: Deriving a Real-World Situation from an Equation Which problem situation best matches the equation 5 + 2x = 13? Situation A: Admission to the county fair costs $5 and rides cost $2 each. Mike spent a total of $13. How many rides did he go on? Mike spent $13 in all, so 5 + 2x = 13. Situation A matches the equation.

  34. Equations and Their Solutions 1-10 $5 per ride 5x Course 2 Additional Example 3 Continued Which problem situation best matches the equation 5 + 2x = 13? Situation B: Admission to the county fair costs $2 and rides cost $5 each. Mike spent a total of $13. How many rides did he go on? The variable x represents the number of rides that Mike bought. Since 5x is not a term in the given equation, Situation B does not match the equation.

  35. Equations and Their Solutions 1-10 $13 per souvenir hat 13x Course 2 Check It Out: Additional Example 3 Which problem situation best matches the equation 13 + 4x = 25? Situation A: Admission to the baseball game costs $4 and souvenir hats cost $13 each. Trina spent a total of $25. How many souvenir hats did she buy? The variable x represents the number of souvenir hats Trina bought. Since 13x is not a term in the given equation, Situation A does not match the equation.

  36. Equations and Their Solutions 1-10 $13 for admission 13 + $4 per souvenir hat 4x Course 2 Check It Out: Additional Example 3 Continued Which problem situation best matches the equation 13 + 4x = 25? Situation B: Admission to the baseball game costs $13 and souvenir hats cost $4 each. Trina spent a total of $25. How many souvenir hats did she buy? Trina spent $25 in all, so 13 + 4x = 25. Situation B matches the equation.

  37. Equations and Their Solutions 1-10 Course 2 Insert Lesson Title Here Lesson Quiz Determine whether the given value of the variable is a solution of 5 + x = 47. 1. x = 42 2. x = 52 Determine whether the given value of the variable is a solution of 57 – y = 18. 3. y = 75 4. y = 39 5. Kwan has 14 marbles. This is 7 more than Drue has. Does Drue have 21 or 7 marbles? yes no no yes 7

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