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2-4 Solving Equations with Variables on Both Sides Warm Up Warm Up Lesson Quiz Lesson Quiz Lesson Presentation Lesson Presentation

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2-4 Solving Equations with Variables on Both Sides Warm Up Simplify. 1. 4x – 10x 2. –7(x – 3) 3. 4. 15 – (x – 2) Solve. 5. 3x + 2 = 8 6. –6x –7x + 21 17 – x 2x + 3 2 28

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2-4 Solving Equations with Variables on Both Sides MA.912.A.3.1 Solve linear equations in one variable that include simplifying algebraic expressions. Also MA.912.A.3.5. Sunshine State Standards

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2-4 Solving Equations with Variables on Both Sides Solve equations in one variable that contain variable terms on both sides. Objective

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2-4 Solving Equations with Variables on Both Sides Vocabulary identity

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2-4 Solving Equations with Variables on Both Sides 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.

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

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2-4 Solving Equations with Variables on Both Sides Solve 4b + 2 = 3b. Check It Out! Example 1a To collect the variable terms on one side, subtract 3b from both sides. 4b + 2 = 3b –3b b + 2 = 0 b = –2 – 2 – 2 To isolate b, subtract 2 from both sides.

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2-4 Solving Equations with Variables on Both Sides Solve 0.5 + 0.3y = 0.7y – 0.3. Check It Out! Example 1b To collect the variable terms on one side, subtract 0.3y from both sides. 0.5 + 0.3y = 0.7y – 0.3 –0.3y 0.5 = 0.4y – 0.3 0.8 = 0.4y +0.3 + 0.3 2 = y Since 0.3 is subtracted from 0.4y, add 0.3 to both sides to undo the subtraction. Since y is multiplied by 0.4, divide both sides by 0.4 to undo the multiplication.

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2-4 Solving Equations with Variables on Both Sides To solve more complicated equations, you may need to first simplify by using the Distributive Property or combining like terms.

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2-4 Solving Equations with Variables on Both Sides Solve 4 – 6a + 4a = –1 – 5(7 – 2a). Additional Example 2: Simplifying Each Side Before Solving Equations Combine like terms. 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 4 – 2a = –36 + 10a +36 40 – 2a = 10a + 2a +2a 40 = 12a Since –36 is added to 10a, add 36 to both sides. To collect the variable terms on one side, add 2a to both sides.

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2-4 Solving Equations with Variables on Both Sides Solve 4 – 6a + 4a = –1 – 5(7 – 2a). Additional Example 2: Continued 40 = 12a Since a is multiplied by 12, divide both sides by 12.

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

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2-4 Solving Equations with Variables on Both Sides Solve 3x + 15 – 9 = 2(x + 2). Check It Out! Example 2b 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 3x + 6 = 2x + 4 –2x x + 6 = 4 – 6 – 6 x = –2 To collect the variable terms on one side, subtract 2x from both sides. Since 6 is added to x, subtract 6 from both sides to undo the addition. Subtract.

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2-4 Solving Equations with Variables on Both Sides 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.

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2-4 Solving Equations with Variables on Both Sides Solve 10 – 5x + 1 = 7x + 11 – 12x. Additional Example 3A: Infinitely Many Solutions or No Solutions Add 5x to both sides. Identify like terms. 10 – 5x + 1 = 7x + 11 – 12x 11 – 5x = 11 – 5x 11 = 11 + 5x + 5x True statement. Combine like terms on the left and the right. 10 – 5x + 1 = 7x + 11 – 12x 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.

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2-4 Solving Equations with Variables on Both Sides Solve 12x – 3 + x = 5x – 4 + 8x. Additional Example 3B: Infinitely Many Solutions or No Solutions Subtract 13x from both sides. Identify like terms. 12x – 3 + x = 5x – 4 + 8x 13x – 3 = 13x – 4 –3 = –4 –13x False statement. Combine like terms on the left and the right. 12x – 3 + x = 5x – 4 + 8x 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.

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2-4 Solving Equations with Variables on Both Sides Solve 4y + 7 – y = 10 + 3y. Check It Out! Example 3a Subtract 3y from both sides. Identify like terms. 4y + 7 – y = 10 + 3y 3y + 7 = 3y + 10 7 = 10 –3y –3y False statement. Combine like terms on the left and the right. 4y + 7 – y = 10 + 3y 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.

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2-4 Solving Equations with Variables on Both Sides Solve 2c + 7 + c = –14 + 3c + 21. Check It Out! Example 3b Subtract 3c both sides. Identify like terms. 2c + 7 + c = –14 + 3c + 21 3c + 7 = 3c + 7 7 = 7 –3c –3c True statement. Combine like terms on the left and the right. 2c + 7 + c = –14 + 3c + 21 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.

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2-4 Solving Equations with Variables on Both Sides 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? Additional Example 4: Application PersonBulbs Jon60 bulbs plus 44 bulbs per hour Sara96 bulbs plus 32 bulbs per hour

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2-4 Solving Equations with Variables on Both Sides Example 4 Continued Let b represent bulbs, and write expressions for the number of bulbs planted. 60 bulbs plus 44 bulbs each hour the same as 96 bulbs plus 32 bulbs each hour When is ? 60 + 44b = 96 + 32b – 32b To collect the variable terms on one side, subtract 32b from both sides. 60 + 12b = 96

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2-4 Solving Equations with Variables on Both Sides Additional Example 4 Continued Since 60 is added to 12b, subtract 60 from both sides. 60 + 12b = 96 –60 – 60 12b = 36 Since b is multiplied by 12, divide both sides by 12 to undo the multiplication. b = 3

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2-4 Solving Equations with Variables on Both Sides 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.

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2-4 Solving Equations with Variables on Both Sides Four times Greg's age, decreased by 3 is equal to 3 times Greg's age increased by 7. How old is Greg? Check It Out! Example 4 Let g represent Greg's age, and write expressions for his age. four times Greg's age decreased by 3 is equal to three times Greg's age increased by 7. 4g – 3 = 3g + 7

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2-4 Solving Equations with Variables on Both Sides Check It Out! Example 4 Continued 4g – 3 = 3g + 7 To collect the variable terms on one side, subtract 3g from both sides. g – 3 = 7 –3g Since 3 is subtracted from g, add 3 to both sides. + 3 + 3 g = 10 Greg is 10 years old.

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2-4 Solving Equations with Variables on Both Sides Standard Lesson Quiz Lesson Quizzes Lesson Quiz for Student Response Systems

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2-4 Solving Equations with Variables on Both Sides Lesson Quiz Solve each equation. 1. 7x + 2 = 5x + 8 2. 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

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2-4 Solving Equations with Variables on Both Sides Lesson Quiz for Student Response Systems 1. Solve the equation. A. a = 1 D. a = 6 C. a = 3 B. a = 2 9a + 3 = 6a + 6

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2-4 Solving Equations with Variables on Both Sides Lesson Quiz for Student Response Systems 2. Solve the equation. A. p = 2 D. p = 5 C. p = 4 B. p = 3 7(3p − 4) = 4p + 6

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2-4 Solving Equations with Variables on Both Sides Lesson Quiz for Student Response Systems 3. Solve the equation. A. B. C. D. 5 − 2(3z + 2) = –4(1 − 2z)

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2-4 Solving Equations with Variables on Both Sides Lesson Quiz for Student Response Systems 4. Solve the equation. A. B. C. D. None of the above 6(s + 2) − 5s = 16 + s − 4 All real numbers No solution (0,0)

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2-4 Solving Equations with Variables on Both Sides Lesson Quiz for Student Response Systems 5. Solve the equation. A. B. C. D. a = 2 a = 5 a = 4 a = 3

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2-4 Solving Equations with Variables on Both Sides 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 B. 3 h C. 4 h D. 5 h

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