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Solving Equations with Integers - Course 2

Learn how to solve one-step equations containing integers. Use mental math to find each solution. Includes warm-up problems and helpful hints.

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Solving Equations with Integers - Course 2

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  1. 2-5 Solving Equations Containing Integers Course 2 Warm Up Problem of the Day Lesson Presentation

  2. 2-5 Solving Equations Containing Integers Course 2 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. 2-5 Solving Equations Containing Integers Course 2 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

  4. 2-5 Solving Equations Containing Integers Course 2 Learn to solve one-step equations with integers.

  5. 2-5 Solving Equations Containing Integers + – – + + – – – – – + – – – + + Course 2 To solve integer equations such as x – 2 = -3 you must 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. -2 + 2 = 0

  6. 2-5 Solving Equations Containing Integers Helpful Hint 3 + (–3) = 0 3 is the opposite of –3. Course 2 The Golden Rule: ALWAYS use inverse operations when solving for variables. For example: Inverse of addition is __________. Inverse of subtraction is ___________. Inverse of multiplication is _____________. Inverse of division is ________________. Think about opposite operations! What you do to one side you have to do to the other.

  7. 2-5 Solving Equations Containing Integers ? –6 + (–1) = –7 ? –7 = –7 Course 2 Additional Example 1A: Solving Addition and Subtraction Equations Solve each question. Check each answer. –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. True. –1 is the solution to –6 + x = –7. 

  8. 2-5 Solving Equations Containing Integers ? –8 + 5 = –3 ? – 3 = –3 Course 2 Additional Example 1B: Solving Addition and Subtraction Equations Solve each equation. Check each answer. 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.  True. –8 is the solution to p + 5 = –3.

  9. 2-5 Solving Equations Containing Integers ? –31 – 9 = –40 ? –40 = –40 Course 2 Additional Example 1C: Solving Addition and Subtraction Equations Solve each equation. Check each answer. 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.  True. –31 is the solution to y – 9 = –40.

  10. 2-5 Solving Equations Containing Integers ? –3 + (–6) = –9 ? –9 = –9 Course 2 Insert Lesson Title Here Check It Out: Example 1A Solve each equation. Check each answer. –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. True. –6 is the solution to –3 + x = –9. 

  11. 2-5 Solving Equations Containing Integers ? –8 + 2 = –6 ? –6 = –6 Course 2 Insert Lesson Title Here Check It Out: Example 1B Solve each equation. Check each answer. 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. True. –8 is the solution to q + 2 = –6. 

  12. 2-5 Solving Equations Containing Integers ? –27 – 7 = –34 ? –34 = –34 Course 2 Insert Lesson Title Here Check It Out: Example 1C Solve each equation. Check each answer. 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.  True. –27 is the solution to y – 7 = –34.

  13. 2-5 Solving Equations Containing Integers Course 2 Additional Example 2A: Solving Multiplication and Division Equations Solve each equation. Check each answer. b –5 = 6 b –5 = 6 b –5 Multiply both sides by –5 to isolate the variable. = (–5)6 (–5) b = –30

  14. 2-5 Solving Equations Containing Integers Course 2 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.

  15. 2-5 Solving Equations Containing Integers Course 2 Additional Example 2B: Solving Multiplication and Division Equations Solve each equation. Check each answer. –400 = 8y –400 = 8y Divide both sides by 8 to isolate the variable. –400 = 8y 8 8 –50 = y

  16. 2-5 Solving Equations Containing Integers Course 2 Additional Example 2B: Solving Multiplication and Division Equations Check Substitute –50 for y in the original equation. –400 = 8y ? –400 = 8(–50) True. –50 is the solution to –400 = 8y. –400 = –400 

  17. 2-5 Solving Equations Containing Integers Course 2 Insert Lesson Title Here Check It Out: Example 2A Solve each equation. Check each answer. c 4 = –24 c 4 = –24 c 4 Multiply both sides by 4 to isolate the variable. = 4(–24) 4 c = –96

  18. 2-5 Solving Equations Containing Integers Course 2 Insert Lesson Title Here Check It Out: 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.

  19. 2-5 Solving Equations Containing Integers Course 2 Insert Lesson Title Here Check It Out: Example 2B Solve each equation. Check each answer. –200 = 4x –200 = 4x Divide both sides by 4 to isolate the variable. –200 = 4x 4 4 –50 = x

  20. 2-5 Solving Equations Containing Integers Course 2 Insert Lesson Title Here Check It Out: Example 2B Continued Check. Substitute –50 for x in the original equation. –200 = 4x ? –200 = 4(–50) True. –50 is the solution to –200 = 4x. –200 = –200 

  21. 2-5 Solving Equations Containing Integers Course 2 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). is = More than + This year’s profit 300 100 million 100 Last year’s profit p 300 = 100 + p –100 –100 200 = p The profit was $200 million in 2002.

  22. 2-5 Solving Equations Containing Integers Course 2 Check It Out: 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 they made last year. is = More than + This year’s profit 243 100 million 125 Last year’s profit x 243 = 125 + x –125 –125 118 = x The class earned $118 last year.

  23. 2-5 Solving Equations Containing Integers Course 2 Insert Lesson Title Here Lesson Quiz Solve each equation. 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

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