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3-6 Solving Equations Containing Integers Course 2 Warm Up Warm Up Problem of the Day Problem of the Day Lesson Presentation Lesson Presentation

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Warm Up Use mental math to find each solution y = x ÷ 9 = x = x – 12 = 30 y = 8 x = 81 x = 4 Course Solving Equations Containing Integers x = 42

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Problem of the Day Zelda sold her wet suit to a friend for $156. She sold her tank, mask, and snorkel for $85 less than she sold her wet suit. She bought a used wet suit for $80 and a used tank, mask, and snorkel for $36. If she started with $0, how much money does she have left? $111 Course Solving Equations Containing Integers

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Learn to solve one-step equations with integers. Course Solving Equations Containing Integers

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Course Solving Equations Containing Integers When you are solving equations with integers, the goal is the same as with whole numbers isolate the variable on one side of the equation. One way to isolate the variable is to add opposites. Recall that the sum of a number and its opposite is –– – 3 + (–3) = 0 a + (–a) = 0

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Course Solving Equations Containing Integers 3 + (–3) = 0 3 is the opposite of –3. Helpful Hint

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Solve. Check each answer. Additional Example 1A: Solving Addition and Subtraction Equations Course Solving Equations Containing Integers A. –6 + x = –7 –6 + x = –7 + 6 x = –1 Check –6 + x = –7 –6 + (–1) = –7 ? Add 6 to both sides to isolate the variable. Substitute –1 for x in the original equation. True. –1 is the solution to –6 + x = –7. –7 = –7 ?

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Additional Example 1B: Solving Addition and Subtraction Equations Course Solving Equations Containing Integers Solve. Check the answer. B. p + 5 = –3 p + 5 = –3 + (–5) +(–5) p = –8 Check –8 + 5 = –3 ? Add –5 to both sides to isolate the variable. Substitute –8 for p in the original equation. True. –8 is the solution to p + 5 = –3. p + 5 = –3 – 3 = –3 ?

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Additional Example 1C: Solving Addition and Subtraction Equations Course Solving Equations Containing Integers Solve. Check the answer. C. y – 9 = –40 y – 9 = – y = –31 Check –31 – 9 = –40 ? Add 9 to both sides to isolate the variable. Substitute –31 for y in the original equation. True. –31 is the solution to y – 9 = –40. y – 9 = –40 –40 = –40 ?

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Try This: Example 1A Insert Lesson Title Here Course Solving Equations Containing Integers Solve. Check each answer. A. –3 + x = –9 –3 + x = – x = –6 Check –3 + x = –9 –3 + (–6) = –9 ? Add 3 to both sides to isolate the variable. Substitute –6 for x in the original equation. True. –6 is the solution to –3 + x = –9. –9 = –9 ?

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Try This: Example 1B Insert Lesson Title Here Course Solving Equations Containing Integers Solve. Check the answer. B. q + 2 = –6 q + 2 = –6 +(–2) q = –8 Check –8 + 2 = –6 ? Add –2 to both sides to isolate the variable. Substitute –8 for q in the original equation. True. –8 is the solution to q + 2 = –6. q + 2 = –6 –6 = –6 ?

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Try This: Example 1C Insert Lesson Title Here Course Solving Equations Containing Integers Solve. Check the answer. C. y – 7 = –34 y – 7 = –34 +7 y = –27 Check –27 – 7 = –34 ? Add 7 to both sides to isolate the variable. Substitute –27 for y in the original equation. True. –27 is the solution to y – 7 = –34. y – 7 = –34 –34 = –34 ?

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Solve. Check each answer. Additional Example 2A: Multiplication and Division Equations Course Solving Equations Containing Integers A. b –5 = 6 b –5 = 6 b –5 = (–5)6 (–5) b = –30 Multiply both sides by –5 to isolate the variable.

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Additional Example 2A Continued Course Solving Equations Containing Integers Check b –5 = 6 –30 – 5 = 6 ? 6 = 6 Substitute –30 for b in the original equation. True. –30 is the solution to b –5 = 6.

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Additional Example 2B: Solving Multiplication and Division Equations Course Solving Equations Containing Integers Solve. Check each answer. B. –400 = 8y –400 = 8y 8 –50 = y Divide both sides by 8 to isolate the variable.

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Check. Additional Example 2B: Solving Multiplication and Division Equations Course Solving Equations Containing Integers –400 = 8y –400 = 8(–50) ? –400 = –400 Substitute –50 for y in the original equation. True. –50 is the solution to –400 = 8y.

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Try This: Example 2A Insert Lesson Title Here Course Solving Equations Containing Integers Solve. Check each answer. A. c4c4 = –24 c4c4 c4c4 = 4(–24) 4 c = –96 Multiply both sides by 4 to isolate the variable.

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Try This: Example 2A Continued Insert Lesson Title Here Course Solving Equations Containing Integers Check c4c4 = –24 –96 4 = –24 ? –24 = –24 Substitute –96 for c in the original equation. True. –96 is the solution to c4c4 = –24.

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Try This: Example 2B Insert Lesson Title Here Course Solving Equations Containing Integers Solve. Check each answer. B. –200 = 4x –200 = 4x 4 –50 = x Divide both sides by 4 to isolate the variable.

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Try This: Example 2B Continued Insert Lesson Title Here Course Solving Equations Containing Integers Check. –200 = 4x –200 = 4(–50) ? –200 = –200 Substitute –50 for x in the original equation. True. –50 is the solution to –200 = 4x.

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In 2003, a manufacturer made a profit of $300 million. This amount was $100 million more than the profit in What was the profit in 2002? Additional Example 3: Business Application Course Solving Equations Containing Integers Let p represent the profit in 2002 (in millions of dollars). Profit in 2003 = p Profit in 2003 = $300 million p = 300 –100 p = 200 The profit in 2002 was $200 million.

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Try This: Example 3 This year the class bake sale made a profit of $243. This was an increase of $125 over last year. How much did they make last year? Insert Lesson Title Here Course Solving Equations Containing Integers This year = x This year = $243 x = 243 –125 x = 118 The class earned $118 last year. Let x represent the money made last year.

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Lesson Quiz Solve. Check your answer. 1. –8y = – x – 22 = –18 3. – = 7 4. w + 72 = –21 5. Last year a phone company had a loss of $25 million. This year the loss is $14 million more last year. What is this years loss? Insert Lesson Title Here –49 –93 Course Solving Equations Containing Integers y7y7 $39 million

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