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CCSS Content Standards A.CED.1 Create equations and inequalities in one variable and use them to solve problems. Mathematical Practices 3 Construct viable arguments and critique the reasoning of others. 8 Look for and express regularity in repeated reasoning.
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Then/Now You used properties of real numbers to evaluate expressions. Translate verbal expressions into algebraic expressions and equations, and vice versa. Solve equations using the properties of equality.
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Vocabulary open sentence equation solution
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Example 1 Verbal to Algebraic Expression A. Write an algebraic expression to represent the verbal expression 7 less than a number. Answer:
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Example 1 Verbal to Algebraic Expression A. Write an algebraic expression to represent the verbal expression 7 less than a number. Answer: n – 7
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Example 1 Verbal to Algebraic Expression B. Write an algebraic expression to represent the verbal expression the square of a number decreased by the product of 5 and the number. Answer:
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Example 1 Verbal to Algebraic Expression B. Write an algebraic expression to represent the verbal expression the square of a number decreased by the product of 5 and the number. Answer: x 2 – 5x
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Example 1a A.6x B.x + 6 C.x 6 D.x – 6 A. Write an algebraic expression to represent the verbal expression 6 more than a number.
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Example 1a A.6x B.x + 6 C.x 6 D.x – 6 A. Write an algebraic expression to represent the verbal expression 6 more than a number.
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Example 1b A.x 3 – 2 B.2x 3 C.x 2 – 2 D.2 + x 3 B. Write an algebraic expression to represent the verbal expression 2 less than the cube of a number.
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Example 1b A.x 3 – 2 B.2x 3 C.x 2 – 2 D.2 + x 3 B. Write an algebraic expression to represent the verbal expression 2 less than the cube of a number.
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Example 2 Algebraic to Verbal Sentence A. Write a verbal sentence to represent 6 = –5 + x. Answer:
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Example 2 Algebraic to Verbal Sentence A. Write a verbal sentence to represent 6 = –5 + x. Answer: Six is equal to –5 plus a number.
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Example 2 Algebraic to Verbal Sentence B. Write a verbal sentence to represent 7y – 2 = 19. Answer:
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Example 2 Algebraic to Verbal Sentence B. Write a verbal sentence to represent 7y – 2 = 19. Answer: Seven times a number minus 2 is 19.
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Example 2a A.The difference of a number and 3 is 7. B.The sum of a number and 3 is 7. C.The difference of 3 and a number is 7. D.The difference of a number and 7 is 3. A. What is a verbal sentence that represents the equation n – 3 = 7?
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Example 2a A.The difference of a number and 3 is 7. B.The sum of a number and 3 is 7. C.The difference of 3 and a number is 7. D.The difference of a number and 7 is 3. A. What is a verbal sentence that represents the equation n – 3 = 7?
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Example 2b A.Five is equal to the difference of 2 and a number. B.Five is equal to twice a number. C.Five is equal to the quotient of 2 and a number. D.Five is equal to the sum of 2 and a number. B. What is a verbal sentence that represents the equation 5 = 2 + x?
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Example 2b A.Five is equal to the difference of 2 and a number. B.Five is equal to twice a number. C.Five is equal to the quotient of 2 and a number. D.Five is equal to the sum of 2 and a number. B. What is a verbal sentence that represents the equation 5 = 2 + x?
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Concept
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Example 4 Solve One-Step Equations A. Solve m – 5.48 = 0.02. Check your solution. m – 5.48=0.02Original equation m – 5.48 + 5.48=0.02 + 5.48Add 5.48 to each side. m=5.5Simplify. Checkm – 5.48=0.02Original equation Answer: 0.02=0.02Simplify. 5.5 – 5.48=0.02Substitute 5.5 for m. ?
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Example 4 Solve One-Step Equations Original equation Simplify.
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Example 4 Solve One-Step Equations Answer: Substitute 36 for t. Simplify. CheckOriginal equation ?
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Example 4a A.–8 B.–2 C.2 D.8 A. What is the solution to the equation x + 5 = 3?
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Example 4a A.–8 B.–2 C.2 D.8 A. What is the solution to the equation x + 5 = 3?
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Example 4b B. What is the solution to the equation A.5 B. C.15 D.30
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Example 4b B. What is the solution to the equation A.5 B. C.15 D.30
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Example 5 Solve a Multi-Step Equation Solve 53 = 3(y – 2) – 2(3y – 1). 53=3(y – 2) – 2(3y – 1)Original equation 53=3y – 6 – 6y + 2Apply the Distributive Property. 53=–3y – 4Simplify the right side. 57=–3yAdd 4 to each side. –19=yDivide each side by –3. Answer:
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Example 5 Solve a Multi-Step Equation Solve 53 = 3(y – 2) – 2(3y – 1). 53=3(y – 2) – 2(3y – 1)Original equation 53=3y – 6 – 6y + 2Apply the Distributive Property. 53=–3y – 4Simplify the right side. 57=–3yAdd 4 to each side. –19=yDivide each side by –3. Answer: The solution is –19.
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Example 5 What is the solution to 25 = 3(2x + 2) – 5(2x + 1)? A.–6 B. C. D.6
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Example 5 What is the solution to 25 = 3(2x + 2) – 5(2x + 1)? A.–6 B. C. D.6
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Example 7 ABCDABCD Read the Test Item You are asked to find the value of the expression 4g – 2. Your first thought might be to find the value of g and then evaluate the expression using this value. Notice that you are not required to find the value of g. Instead, you can use the Subtraction Property of Equality.
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Example 7 Solve the Test Item Answer: Original equation Subtract 7 from each side. Simplify.
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Example 7 Solve the Test Item Answer: C Original equation Subtract 7 from each side. Simplify.
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Example 7 A.12 B.6 C.–6 D.–12 If 2x + 6 = –3, what is the value of 2x – 3?
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Example 7 A.12 B.6 C.–6 D.–12 If 2x + 6 = –3, what is the value of 2x – 3?
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