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Solving Equations with Variables on Both Sides - Warm Up Lesson

This lesson presentation provides a warm-up activity, examples, and practice problems to help students solve equations with variables on both sides. The lesson also includes an explanation of helpful hints and vocabulary.

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Solving Equations with Variables on Both Sides - Warm Up Lesson

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  1. Warm Up Lesson Presentation Lesson Quiz

  2. Warm Up Simplify. 1. 4x – 10x 2. –7(x – 3) 3. 4. 15 – (x – 2) Solve. 5. 3x + 2 = 8 6. –6x –7x + 21 2x + 3 17 – x 2 28

  3. Sunshine State Standards MA.912.A.3.1 Solve linear equations in one variable that include simplifying algebraic expressions. AlsoMA.912.A.3.5.

  4. Objective Solve equations in one variable that contain variable terms on both sides.

  5. Vocabulary identity

  6. Helpful Hint Equations are often easier to solve when the variable has a positive coefficient. Keep this in mind when deciding on which side to "collect" variable terms.

  7. –5n –5n + 2 + 2 Additional Example 1: Solving Equations with Variables on Both Sides Solve 7n – 2 = 5n + 6. 7n – 2 = 5n + 6 To collect the variable terms on one side, subtract 5n from both sides. 2n – 2 = 6 Since 2 is subtracted from 2n, add 2 to undo the subtraction. 2n = 8 Since n is multiplied by 2, divide both sides by 2 to undo the multiplication. n = 4

  8. –3b –3b – 2 – 2 Check It Out! Example 1a Solve 4b + 2 = 3b. 4b + 2 = 3b To collect the variable terms on one side, subtract 3b from both sides. b + 2 = 0 To isolate b, subtract 2 from both sides. b = –2

  9. –0.3y –0.3y +0.3 + 0.3 Check It Out! Example 1b Solve 0.5 + 0.3y = 0.7y – 0.3. To collect the variable terms on one side, subtract 0.3y from both sides. 0.5 + 0.3y = 0.7y – 0.3 0.5 = 0.4y – 0.3 Since 0.3 is subtracted from 0.4y, add 0.3 to both sides to undo the subtraction. 0.8 = 0.4y Since y is multiplied by 0.4, divide both sides by 0.4 to undo the multiplication. 2 = y

  10. To solve more complicated equations, you may need to first simplify by using the Distributive Property or combining like terms.

  11. +36 +36 + 2a +2a Additional Example 2: Simplifying Each Side BeforeSolving Equations Solve 4 – 6a + 4a = –1 – 5(7 – 2a). Distribute –5 to the expression in parentheses. 4 – 6a + 4a = –1 –5(7 – 2a) 4 – 6a + 4a = –1 –5(7) –5(–2a) 4 – 6a + 4a = –1 – 35 + 10a Combine like terms. 4 – 2a = –36 + 10a Since –36 is added to 10a, add 36 to both sides. 40 – 2a = 10a To collect the variable terms on one side, add 2a to both sides. 40 = 12a

  12. Additional Example 2: Continued Solve 4 – 6a + 4a = –1 – 5(7 – 2a). 40 = 12a Since a is multiplied by 12, divide both sides by 12.

  13. Distribute to the expression in parentheses. 1 2 To collect the variable terms on one side, subtract b from both sides. 1 2 + 1 + 1 Check It Out! Example 2a Solve . 3 = b – 1 Since 1 is subtracted from b, add 1 to both sides. 4 = b

  14. –2x –2x – 6 – 6 Check It Out! Example 2b Solve 3x + 15 – 9 = 2(x + 2). Distribute 2 to the expression in parentheses. 3x + 15 – 9 = 2(x + 2) 3x + 15 – 9 = 2(x) + 2(2) 3x + 15 – 9 = 2x + 4 Subtract. To collect the variable terms on one side, subtract 2x from both sides. 3x + 6 = 2x + 4 x + 6= 4 Since 6 is added to x, subtract 6 from both sides to undo the addition. x = –2

  15. An identity is an equation that is always true, no matter what value is substituted for the variable. The solutions of an identity are all real numbers. Some equations are always false. They have no solutions.

  16. + 5x + 5x Additional Example 3A: Infinitely Many Solutions or No Solutions Solve 10 – 5x + 1 = 7x + 11 – 12x. 10 – 5x + 1 = 7x + 11 – 12x 10– 5x+ 1 = 7x+ 11– 12x Identify like terms. 11 – 5x = 11 – 5x Combine like terms on the left and the right. Add 5x to both sides. 11 = 11 True statement.  The equation 10 – 5x + 1 = 7x + 11 – 12x is an identity. All values of x will make the equation true. All real numbers are solutions.

  17. –13x –13x Additional Example 3B: Infinitely Many Solutions or No Solutions Solve 12x – 3 + x = 5x – 4 + 8x. 12x – 3 + x = 5x – 4 + 8x 12x– 3+ x = 5x– 4+ 8x Identify like terms. 13x – 3 = 13x – 4 Combine like terms on the left and the right. Subtract 13x from both sides.  –3 = –4 False statement. The equation 12x – 3 + x = 5x – 4 + 8x is a contradiction. There is no value of x that will make the equation true. There are no solutions.

  18. –3y –3y Check It Out! Example 3a Solve 4y + 7 – y = 10 + 3y. 4y + 7 – y = 10 + 3y 4y+ 7– y = 10+ 3y Identify like terms. 3y + 7 = 3y + 10 Combine like terms on the left and the right. Subtract 3y from both sides.  7 = 10 False statement. The equation 4y + 7 – y = 10 + 3y is a contradiction. There is no value of y that will make the equation true. There are no solutions.

  19. –3c –3c Check It Out! Example 3b Solve 2c + 7 + c = –14 + 3c + 21. 2c + 7 + c = –14 + 3c + 21 2c+ 7+ c = –14+ 3c+ 21 Identify like terms. 3c + 7 = 3c + 7 Combine like terms on the left and the right. Subtract 3c both sides. 7 = 7 True statement.  The equation 2c + 7 + c = –14 + 3c + 21 is an identity. All values of c will make the equation true. All real numbers are solutions.

  20. Additional Example 4: Application Jon and Sara are planting tulip bulbs. Jon has planted 60 bulbs and is planting at a rate of 44 bulbs per hour. Sara has planted 96 bulbs and is planting at a rate of 32 bulbs per hour. In how many hours will Jon and Sara have planted the same number of bulbs? How many bulbs will that be?

  21. 32 bulbs each hour 44 bulbs each hour the same as 60 bulbs 96 bulbs ? When is plus plus Example 4 Continued Let b represent bulbs, and write expressions for the number of bulbs planted. 60 + 44b = 96 + 32b 60 + 44b = 96 + 32b To collect the variable terms on one side, subtract 32b from both sides. – 32b– 32b 60 + 12b = 96

  22. Additional Example 4 Continued 60 + 12b = 96 Since 60 is added to 12b, subtract 60 from both sides. –60 – 60 12b = 36 Since b is multiplied by 12, divide both sides by 12 to undo the multiplication. b = 3

  23. Additional Example 4 Continued After 3 hours, Jon and Sara will have planted the same number of bulbs. To find how many bulbs they will have planted in 3 hours, evaluate either expression for b = 3: 60 + 44b = 60 + 44(3) = 60 + 132 = 192 96 + 32b = 96 + 32(3) = 96 + 96 = 192 After 3 hours, Jon and Sara will each have planted 192 bulbs.

  24. 7 . Check It Out! Example 4 Four times Greg's age, decreased by 3 is equal to 3 times Greg's age increased by 7. How old is Greg? Let g represent Greg's age, and write expressions for his age. three times Greg's age four times Greg's age is equal to increased by decreased by 3 4g – 3 = 3g + 7

  25. –3g –3g + 3+ 3 Check It Out! Example 4Continued 4g – 3 = 3g + 7 To collect the variable terms on one side, subtract 3g from both sides. g – 3= 7 Since 3 is subtracted from g, add 3 to both sides. g = 10 Greg is 10 years old.

  26. Lesson Quizzes Standard Lesson Quiz Lesson Quiz for Student Response Systems

  27. Lesson Quiz Solve each equation. 1. 7x + 2 = 5x + 82. 4(2x – 5) = 5x + 4 3. 6 – 7(a + 1) = –3(2 – a) 4. 4(3x + 1) – 7x = 6 + 5x – 2 5. 6. A painting company charges $250 base plus $16 per hour. Another painting company charges $210 base plus $18 per hour. How long is a job for which the two companies costs are the same? 3 8 all real numbers 1 20 hours

  28. Lesson Quiz for Student Response Systems 1. Solve the equation. 9a + 3 = 6a + 6 C. a = 3 A. a = 1 B. a = 2 D. a = 6

  29. Lesson Quiz for Student Response Systems 2. Solve the equation. 7(3p − 4) = 4p + 6 C. p = 4 A. p = 2 B. p = 3 D. p = 5

  30. Lesson Quiz for Student Response Systems 3. Solve the equation. 5 − 2(3z + 2) = –4(1 − 2z) A. C. B. D.

  31. Lesson Quiz for Student Response Systems 4. Solve the equation. 6(s + 2) − 5s = 16 + s − 4 All real numbers A. C. (0,0) None of the above No solution B. D.

  32. Lesson Quiz for Student Response Systems 5. Solve the equation. a = 4 a = 2 A. C. a = 3 a = 5 B. D.

  33. Lesson Quiz for Student Response Systems 6. A moving company charges a $2,000 fee plus a $100 per hour. Another moving company charges an $1800 fee plus $150 per hour. How long is a job for which the costs of the two companies is the same? A. 2 h C. 4 h B. 3 h D. 5 h

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