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Opener : Find three consecutive odd integers whose sum is -63

Opener : Find three consecutive odd integers whose sum is -63. Integer #1 = n Integer #2 = n + 2 Integer #3 = n + 4 (n) + (n + 2) + (n + 4) = -63 3n + 6 = -63 3n = -69 n = -23 Integer #1 n = -23 Integer #2 n + 2 = -23 + 2 = -21

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Opener : Find three consecutive odd integers whose sum is -63

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  1. Opener: Find three consecutive odd integers whose sum is -63 Integer #1 = n Integer #2 = n + 2 Integer #3 = n + 4 (n) + (n + 2) + (n + 4) = -63 3n + 6 = -63 3n = -69 n = -23 Integer #1 n = -23 Integer #2 n + 2 = -23 + 2 = -21 Integer #3 n + 4 = -23 + 4 = -19

  2. Algebra 1 ~ Chapter 3.5 Solving Equations with the Variable on Each Side

  3. * To solve equations with variables on both sides, we follow the same steps as solving any other equation. But first, we need to get all of the variables on one side. Example: 8 + 5x = 7x – 2 -5x -5x 8 = 2x – 2 +2 +2 10 = 2x 2 2 5 = x or x = 5 CHECK 8 + 5x = 7x – 2 8 + 5(5) = 7(5) – 2 8 + 25 = 35 – 2 33 = 33

  4. Example 1 – Solve and CHECK -2 + 10p = 8p – 1 -8p -8p -2 + 2p = -1 +2 +2 2p = 1 2 2 p = ½ CHECK -2 + 10p = 8p – 1 -2 + 10(½) = 8(½) – 1 -2 + 5 = 4 - 1 3 = 3

  5. Solving an Equation with Grouping Symbols You must use the Distributive Property to get rid of the parentheses!! ¼(16x – 4) = -3(x – 9) 4x - 1 = -3x + 27 +3x +3x 7x - 1 = 27 + 1 + 1 7x = 28 7 7 x = 4 CHECK ¼(16x – 4) = -3(x – 9) ¼(16∙4 – 4) = -3(4 – 9) ¼(64 – 4) = -3(-5) ¼(60) = 15 15 = 15

  6. Example 2 – Solve and Check 4(4 – w) = 3(2w + 2) 16 – 4w = 6w + 6 +4w +4w 16 = 10w + 6 -6 -6 10 = 10w 10 10 1 = w Or w = 1 CHECK 4(4 – w) = 3(2w + 2) 4(4 – 1) = 3(2∙1 + 2) 4(3) = 3(4) 12 = 12

  7. Solve 2m + 5 = 5(m – 7) – 3m 2m + 5 = 5m – 35 – 3m 2m + 5 = 2m – 35 -2m -2m 5 = -35 Since 5 = -35 is a false statement, this equation has no solution. *** Some equations with the variable on each side may have no solution. That is, there is no value of the variable that will result in a true equation.

  8. Solve 3( r + 1) – 5 = 3r – 2 3r + 3 – 5 = 3r – 2 3r – 2 = 3r – 2 Since the expressions on each side of the equation are the same, this equation is an identity. *** The statement 3(r + 1) – 5 = 3r – 2 is true for all values of r.

  9. Solving Equations Summary • Use the Distributive property to remove the grouping symbols. • Simplify the expressions on each side of the equals sign (combine like terms). • Collect variable terms on one side of the equation • Undo Addition or Subtraction • Undo Multiplication or Division • If the solution results in a false statement, there is no solution of the equation. • If the solution results in an identity, the solution is all numbers.

  10. Examples – Solve each equation 1.) 7 – 3r = r – 4(2 + r) 2.) 3.)

  11. Word Problem Example “The sum of one half of a number and six equals one third of the number. What is the number?”

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