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Simplify Algebraic Expressions: Coefficients, Like Terms, and Variables

Learn to simplify algebraic expressions by identifying coefficients, like terms, and variables. This course covers grouping similar terms, combining coefficients, and using properties like distributive and associative. Practice evaluating, identifying, and simplifying expressions. Engage in warm-up problems and explore real-world applications. Improve your algebra skills and master simplification techniques efficiently.

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Simplify Algebraic Expressions: Coefficients, Like Terms, and Variables

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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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