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Chapter 7 Lesson 2 Solving Equations with Grouping Symbols pgs. 334-338

Chapter 7 Lesson 2 Solving Equations with Grouping Symbols pgs. 334-338. What you will learn: Solve equations that involve grouping symbols Identify equations that have no solution or an infinite number of solutions. Vocabulary.

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Chapter 7 Lesson 2 Solving Equations with Grouping Symbols pgs. 334-338

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  1. Chapter 7 Lesson 2Solving Equations with Grouping Symbolspgs. 334-338 What you will learn: Solve equations that involve grouping symbols Identify equations that have no solution or an infinite number of solutions

  2. Vocabulary • Null/empty set (336): equations that have no solution. No value of the variable results in a true sentence. Represented by  or { } • Identity (336): an equation that is true for every value of the variable

  3. Josh starts walking at a rate of 2 mph. One hour later, his sister Maria starts on the same path on her bike, riding at 10mph What does t represent? Why is Maria’s time shown as t-1? Write an equation that represents the time when Maria catches up to Josh. The time Josh travels She left 1 hour later than Josh 2t = 10(t-1)

  4. Example 1: Solve Equations with Parentheses Solve the equation from the previous chart: 2t = 10(t-1) Write the problem: 2t = 10(t -1) Distributive Property: 2t = 10(t) - 10(1) Simplify: 2t = 10t -10 Subtract 10t from each side: 2t -10t = 10t - 10t -10 Simplify: -8t = -10 Divide each side by -8: -8t = -10 -8 -8 Simplify/Solve: t = 5 or 1 1/4 4

  5. Now check the previous problem. Josh traveled 2 miles 5 hour or 2.5 miles hour 4 Maria traveled 1 hour less than Josh. She traveled: 10 miles  1 hour hour 4 or 2.5 miles Therefore, Maria caught up to Josh in 1/4 hour or 15 minutes.

  6. Another Example 1: Solve Equations with Parentheses Solve: 6(n-3) = 4(n + 2.1) Distributive Property on both sides: 6(n) - 6(3) = 4(n) + 4(2.1) Simplify: 6n - 18 = 4n + 8.4 Subtract 4n from each side: 6n -4n -18 = 4n -4n + 8.4 Simplify: 2n - 18 = 8.4 Add 18 to both sides: 2n - 18 +18 = 8.4 + 18 Simplify: 2n = 26.4 Simplify/Solve: n = 13.2 Divide both sides by 2: 2n = 26.4 2 2 Check your solution!

  7. Example 2: Use an Equation to Solve a Problem The perimeter of a rectangle is 20 feet. The width is 4 feet less than the length. Find the dimensions of the rectangle. Then find its area. Words: The width is 4 feet less than the length. The perimeter is 20 feet. Symbols: Let A = area Let L-4 = width 2length + 2width = perimeter Equation: 2length + 2(L-4) = 20 2L + 2L - 8 = 20 4L - 8 = 20 4L = 28 L = 7

  8. Formula: 2L + 2W = Perimeter 2(7) + 2W = 20 14 + 2W = 20 2W = 6 W = 3 So the width is 3 feet • Since we know the length is 7 ft, now we need to find the width. Check: 2(7) + 2(3) = 20 14 + 6 = 20 20 = 20  Now find the area Of the rectangle. A = LW A = 73 A = 21 ft2

  9. Example 3: No Solution Solve: 12 - h = -h + 3 Add an h to both sides: 12 - h + h = -h + h + 3 Simplify: 12 = 3 The sentence 12 = 3 is never true. So the Solution set is 

  10. Example 4: All Numbers as solutionsRemember, an equation that is true for every value of the variable is called an identity. Solve: 3(2g + 4) = 6(g+2) Distributive Property: 3(2g) + 3(4) = 6(g) + 6(2) Simplify: 6g + 12 = 6g + 12 Subtract 12 from each side: 6g + 12 - 12 = 6g + 12 - 12 Simplify: 6g = 6g Mentally divide each side by 6: g = g The sentence g = g is always true, the solution set is all numbers.

  11. Your Turn!Solve each equation. Check your solution a = 11 Check: 3(11-5) = 18 3(6) = 18 18 = 18  • 3(a-5) = 18 • 3(s+22) = 4(s+12) • 4(f+3) + 5 = 17 + 4f • 8y - 3 = 5(y - 1) +3y S = 18 Check: 3(18+22) = 4(18+12) 3(40) = 4(30) 120 = 120 f = f The solution set is all numbers The solution set is 

  12. One More! Find the dimension of the rectangle. P = 460ft 2(w) + 2(w+30)=460 2w + 2w + 60 = 460 4w + 60 = 460 4w + 60-60 = 460 -60 4w = 400 w =100 w w + 30 w + 30 =Length 100 + 30 = Length 130 = L So the the rectangle is 130ft by 100ft

  13. PRACTICE IS BY THE DOOR ON YOUR WAY OUT! • QUIZ TOMORROW OVER 7-1 & 7-2

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