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

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. 2. 28. Objective. Solve equations in one variable that contain variable terms on both sides. Vocabulary. identity contradiction.

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

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  1. 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

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

  3. Vocabulary identity contradiction

  4. 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.

  5. The Goal The goal when solving equations with variables on both sides is to… isolate the variable

  6. To isolate the variable we will use … …inverse operations

  7. Steps to solving equations with variables on both sides. 1. Simplify both sides of the equation - Distributive Property - Combine Like Terms

  8. 2. Move the variable to one side of the equation using INVERSE OPERATIONS. (Hint: You will only use addition or subtraction to move the variable)

  9. 3. Isolate the variable using inverse operations. (Remember the first move will be addition or subtraction and the second move will be to multiply or divide.)

  10. Directions: Solve

  11. –5n –5n + 2 + 2 Example 1 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

  12. –3b –3b – 2 – 2 Example 2 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

  13. –0.3y –0.3y +0.3 + 0.3 Example 3 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

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

  15. +36 +36 + 2a +2a Example 4 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

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

  17. 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 Example 5 Solve . 3 = b – 1 Since 1 is subtracted from b, add 1 to both sides. 4 = b

  18. –2x –2x – 6 – 6 Example 6 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

  19. An identity is an equation that is true for all values of the variable. An equation that is an identity has infinitely many solutions. A contradiction is an equation that is not true for any value of the variable. It has no solutions.

  20. Identities and Contradictions

  21. Identities and Contradictions WORDS Contradiction When solving an equation, if you get a false equation, the original equation is a contradiction, and it has no solutions. NUMBERS 1 = 1 + 2 1 = 3  ALGEBRA x = x + 3 –x–x 0 = 3 

  22. + 5x + 5x Example 7 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 – 12xis an identity. All values of x will make the equation true. All real numbers are solutions.

  23. –13x–13x Example 8 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 + 8xis a contradiction. There is no value of x that will make the equation true. There are no solutions.

  24. –3y–3y Example 9 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.

  25. –3c –3c Example 10 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.

  26. Example 11: 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?

  27. 32 bulbs each hour 44 bulbs each hour the same as 60 bulbs 96 bulbs ? When is plus plus Example 11: 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

  28. Example 11: 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

  29. Example 11: 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.

  30. 7 . Example 12 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

  31. –3g –3g + 3+ 3 Example 12Continued 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.

  32. Lesson Summary 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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