# 3-6 Solving Equations Containing Integers Warm Up Problem of the Day

## Presentation on theme: "3-6 Solving Equations Containing Integers Warm Up Problem of the Day"— Presentation transcript:

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. 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

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 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.

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.

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 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 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 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

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. –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. –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