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

3-6 Solving Equations Containing Integers Warm Up Course 2 3-6 Solving Equations Containing Integers Warm Up Use mental math to find each solution. 1. 7 + y = 15 2. x ÷ 9 = 9 3. 6x = 24 4. x – 12 = 30 y = 8 x = 81 x = 4 x = 42

3-6 Solving Equations Containing Integers Problem of the Day Course 2 3-6 Solving Equations Containing Integers 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

3-6 Learn to solve one-step equations with integers. Course 2 3-6 Solving Equations Containing Integers Learn to solve one-step equations with integers.

3-6 Solving Equations Containing Integers Course 2 3-6 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 0. + + + – – – 3 + (–3) = 0 a + (–a) = 0

3-6 Solving Equations Containing Integers Helpful Hint 3 + (–3) = 0 Course 2 3-6 Solving Equations Containing Integers 3 + (–3) = 0 3 is the opposite of –3. Helpful Hint

Additional Example 1A: Solving Addition and Subtraction Equations Course 2 3-6 Solving Equations Containing Integers Additional Example 1A: Solving Addition and Subtraction Equations Solve. Check each answer. A. –6 + x = –7 –6 + x = –7 + 6 + 6 Add 6 to both sides to isolate the variable. x = –1 Check –6 + x = –7 Substitute –1 for x in the original equation. –6 + (–1) = –7 ? –7 = –7 ?  True. –1 is the solution to –6 + x = –7.

Additional Example 1B: Solving Addition and Subtraction Equations Course 2 3-6 Solving Equations Containing Integers Additional Example 1B: Solving Addition and Subtraction Equations Solve. Check the answer. B. p + 5 = –3 p + 5 = –3 + (–5) +(–5) Add –5 to both sides to isolate the variable. p = –8 Check p + 5 = –3 Substitute –8 for p in the original equation. –8 + 5 = –3 ? – 3 = –3 ?  True. –8 is the solution to p + 5 = –3.

Additional Example 1C: Solving Addition and Subtraction Equations Course 2 3-6 Solving Equations Containing Integers Additional Example 1C: Solving Addition and Subtraction Equations Solve. Check the answer. C. y – 9 = –40 y – 9 = –40 + 9 + 9 Add 9 to both sides to isolate the variable. y = –31 Check y – 9 = –40 Substitute –31 for y in the original equation. –31 – 9 = –40 ? –40 = –40 ?  True. –31 is the solution to y – 9 = –40.

Insert Lesson Title Here Course 2 3-6 Solving Equations Containing Integers Insert Lesson Title Here Try This: Example 1A Solve. Check each answer. A. –3 + x = –9 –3 + x = – 9 + 3 + 3 Add 3 to both sides to isolate the variable. x = –6 Check –3 + x = –9 Substitute –6 for x in the original equation. –3 + (–6) = –9 ? –9 = –9 ?  True. –6 is the solution to –3 + x = –9.

Insert Lesson Title Here Course 2 3-6 Solving Equations Containing Integers Insert Lesson Title Here Try This: Example 1B Solve. Check the answer. B. q + 2 = –6 q + 2 = –6 +(–2) +(–2) Add –2 to both sides to isolate the variable. q = –8 Check q + 2 = –6 Substitute –8 for q in the original equation. –8 + 2 = –6 ? –6 = –6 ?  True. –8 is the solution to q + 2 = –6.

Insert Lesson Title Here Course 2 3-6 Solving Equations Containing Integers Insert Lesson Title Here Try This: Example 1C Solve. Check the answer. C. y – 7 = –34 y – 7 = –34 +7 +7 Add 7 to both sides to isolate the variable. y = –27 Check y – 7 = –34 Substitute –27 for y in the original equation. –27 – 7 = –34 ? –34 = –34 ?  True. –27 is the solution to y – 7 = –34.

Additional Example 2A: Multiplication and Division Equations Course 2 3-6 Solving Equations Containing Integers Additional Example 2A: Multiplication and Division Equations Solve. Check each answer. b –5 A. = 6 b –5 = 6 b –5 (–5) = (–5)6 Multiply both sides by –5 to isolate the variable. b = –30

Additional Example 2A Continued Course 2 3-6 Solving Equations Containing Integers Additional Example 2A Continued Check b –5 = 6 Substitute –30 for b in the original equation. –30 ? = 6 – 5 True. –30 is the solution to 6 = 6  b –5 = 6.

Additional Example 2B: Solving Multiplication and Division Equations Course 2 3-6 Solving Equations Containing Integers Additional Example 2B: Solving Multiplication and Division Equations Solve. Check each answer. B. –400 = 8y –400 = 8y –400 = 8y Divide both sides by 8 to isolate the variable. 8 8 –50 = y

Additional Example 2B: Solving Multiplication and Division Equations Course 2 3-6 Solving Equations Containing Integers Additional Example 2B: Solving Multiplication and Division Equations Check. –400 = 8y Substitute –50 for y in the original equation. ? –400 = 8(–50) True. –50 is the solution to –400 = 8y. –400 = –400 

Insert Lesson Title Here Course 2 3-6 Solving Equations Containing Integers Insert Lesson Title Here Try This: Example 2A Solve. Check each answer. c 4 A. = –24 c 4 = –24 c 4 4 = 4(–24) Multiply both sides by 4 to isolate the variable. c = –96

Insert Lesson Title Here Course 2 3-6 Solving Equations Containing Integers Insert Lesson Title Here Try This: Example 2A Continued Check c 4 = –24 Substitute –96 for c in the original equation. –96 ? = –24 4 True. –96 is the solution to –24 = –24  c 4 = –24.

Insert Lesson Title Here Course 2 3-6 Solving Equations Containing Integers Insert Lesson Title Here Try This: Example 2B Solve. Check each answer. B. –200 = 4x –200 = 4x –200 = 4x Divide both sides by 4 to isolate the variable. 4 4 –50 = x

Insert Lesson Title Here Course 2 3-6 Solving Equations Containing Integers Insert Lesson Title Here Try This: Example 2B Continued Check. –200 = 4x Substitute –50 for x in the original equation. ? –200 = 4(–50) True. –50 is the solution to –200 = 4x. –200 = –200 

Additional Example 3: Business Application Course 2 3-6 Solving Equations Containing Integers Additional Example 3: Business Application In 2003, a manufacturer made a profit of $300 million. This amount was $100 million more than the profit in 2002. What was the profit in 2002? Let p represent the profit in 2002 (in millions of dollars). Profit in 2003 = p + 100 Profit in 2003 = $300 million p + 100 = 300 –100 –100 p = 200 The profit in 2002 was $200 million.

Insert Lesson Title Here Course 2 3-6 Solving Equations Containing Integers Insert Lesson Title Here 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? Let x represent the money made last year. This year = x + 125 This year = $243 x + 125 = 243 –125 –125 x = 118 The class earned $118 last year.

Insert Lesson Title Here Course 2 3-6 Solving Equations Containing Integers Insert Lesson Title Here Lesson Quiz Solve. Check your answer. 1. –8y = –800 2. 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? 100 4 y 7 –49 –93 $39 million