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

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2-4

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  1. Solving Equations with Variables on Both Sides 2-4 Warm Up Lesson Presentation Lesson Quiz Holt Algebra 1 Holt McDougal Algebra 1

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

  4. Vocabulary identity

  5. 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. To solve an equation with variables on both sides, use inverse operations to "collect" variable terms on one side of the equation.

  6. –5n –5n + 2 + 2 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 2n = 8 Since n is multiplied by 2, divide both sides by 2 to undo the multiplication. n = 4

  7. –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 b = –2

  8. –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

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

  10. +36 +36 + 2a +2a 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

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

  12. 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

  13. –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 3x + 6 = 2x + 4 Combine like terms. To collect the variable terms on one side, subtract 2x from both sides. x + 6= 4 Since 6 is added to x, subtract 6 from both sides to undo the addition. x = –2

  14. An identity is an equation that is true for all values of the variable. An equation that is an identity has infinitely many solutions. Some equations are always false. These equations have no solutions.

  15. Identities and False Equations

  16. Identities and False Equations WORDS False Equations When solving an equation, if you get a false equation, the original equation has no solutions. NUMBERS 1 = 1 + 2 1 = 3  ALGEBRA x = x + 3 –x–x 0 = 3 

  17. + 5x + 5x 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. True statement. 11 = 11  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.

  18. –13x –13x 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 false equation. There is no value of x that will make the equation true. There are no solutions.

  19. –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 false equation. There is no value of y that will make the equation true. There are no solutions.

  20. –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.

  21. 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?

  22. 32 bulbs each hour 44 bulbs each hour the same as 60 bulbs 96 bulbs ? When is plus plus Example 4: Application 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

  23. Example 4: Application 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

  24. Example 4: Application 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.

  25. 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

  26. –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.

  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

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