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Chapter 1: Tools of Algebra 1-3: Solving Equations

Chapter 1: Tools of Algebra 1-3: Solving Equations. Essential Question: What is the procedure to solve an equation for a variable?. 1-3: Solving Equations. The solution of an equation is a number that can be used in place of the variable that makes the equation true.

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Chapter 1: Tools of Algebra 1-3: Solving Equations

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  1. Chapter 1: Tools of Algebra1-3: Solving Equations Essential Question: What is the procedure to solve an equation for a variable?

  2. 1-3: Solving Equations • The solution of an equation is a number that can be used in place of the variable that makes the equation true. • You can manipulate equations to help find a solution, so long as you do the same thing to both sides of the equation. • Addition Property If a = b, then a + c = b + c • Subtraction Property If a = b, then a – c = b – c • Multiplication Property If a = b, then ac = bc • Division Property If a = b, then a/c = b/c

  3. 1-3: Solving Equations • Solve 13y + 48 = 8y – 47

  4. 1-3: Solving Equations • Solve 13y + 48 = 8y – 47 • 13y + 48= 8y – 47– 48 – 48 (subtract 48 from both sides) • 13y = 8y – 95

  5. 1-3: Solving Equations • Solve 13y + 48 = 8y – 47 • 13y + 48 = 8y – 47 – 48 – 48 (subtract 48 from both sides) • 13y = 8y – 95 –8y –8y (subtract 8y from both sides) • 5y = – 95

  6. 1-3: Solving Equations • Solve 13y + 48 = 8y – 47 • 13y + 48 = 8y – 47 – 48 – 48 (subtract 48 from both sides) • 13y = 8y – 95 –8y –8y (subtract 8y from both sides) • 5y = – 9555 (divide both sides by 5) • y = -19

  7. 1-3: Solving Equations • Solve 3x – 7(2x – 13) = 3(-2x + 9)

  8. 1-3: Solving Equations • Solve 3x – 7(2x – 13) = 3(-2x + 9) • 3x – 7(2x – 13) = 3(-2x + 9) • 3x – 14x + 91 = -6x + 27 (distribute)

  9. 1-3: Solving Equations • Solve 3x – 7(2x – 13) = 3(-2x + 9) • 3x – 7(2x – 13) = 3(-2x + 9) • 3x – 14x + 91 = -6x + 27 (distribute) • -11x + 91 = -6x + 27 (combine like terms)

  10. 1-3: Solving Equations • Solve 3x – 7(2x – 13) = 3(-2x + 9) • 3x – 7(2x – 13) = 3(-2x + 9) • 3x – 14x + 91 = -6x + 27 (distribute) • -11x + 91 = -6x + 27 (combine like terms)- 91- 91 (subtract 91 from both sides) • -11x = -6x – 64

  11. 1-3: Solving Equations • Solve 3x – 7(2x – 13) = 3(-2x + 9) • 3x – 7(2x – 13) = 3(-2x + 9) • 3x – 14x + 91 = -6x + 27 (distribute) • -11x + 91 = -6x + 27 (combine like terms) - 91 - 91 (subtract 91 from both sides) • -11x = -6x – 64+6x +6x (add 6x to both sides) • -5x = -64

  12. 1-3: Solving Equations • Solve 3x – 7(2x – 13) = 3(-2x + 9) • 3x – 7(2x – 13) = 3(-2x + 9) • 3x – 14x + 91 = -6x + 27 (distribute) • -11x + 91 = -6x + 27 (combine like terms) - 91 - 91 (subtract 91 from both sides) • -11x = -6x – 64+6x +6x (add 6x to both sides) • -5x = -64-5 -5 (divide both sides by -5) • x = 12.8

  13. 1-3: Solving Equations • Solving a Formula for One of Its Variables • The formula for the area of a trapezoid isA = ½ h(b1 + b2). Solve the formula for h. • The goal is to use PEMDAS (in reverse) toget the variable in question alone. • A = ½ h(b1 + b2)

  14. 1-3: Solving Equations • Solving a Formula for One of Its Variables • The formula for the area of a trapezoid isA = ½ h(b1 + b2). Solve the formula for h. • The goal is to use PEMDAS (in reverse) toget the variable in question alone. • A = ½ h(b1+ b2)x2 x2 (multiply each side by 2, the reciprocal of ½) • 2A = h(b1 + b2)

  15. 1-3: Solving Equations • Solving a Formula for One of Its Variables • The formula for the area of a trapezoid isA = ½ h(b1 + b2). Solve the formula for h. • The goal is to use PEMDAS (in reverse) toget the variable in question alone. • A = ½ h(b1 + b2)x2 x2 (multiply each side by 2, the reciprocal of ½) • 2A = h(b1 + b2)(b1 + b2) (b1 + b2) (divide each side by b1 + b2)

  16. 1-3: Solving Equations • Solving a Formula for One of Its Variables • The formula for the area of a trapezoid isA = ½ h(b1 + b2). • Solve the formula for b1

  17. 1-3: Solving Equations • Solve for x

  18. 1-3: Solving Equations • Solve for x • (multiply both sides by a, to clear the first denominator)

  19. 1-3: Solving Equations • Solve for x • (multiply both sides by a, to clear the first denominator) • (multiply both sides by b, to clear the second denominator)

  20. 1-3: Solving Equations • Solve for x • (multiply both sides by a, to clear the first denominator) • (multiply both sides by b, to clear the second denominator) • (subtract bx on both sides, to get the x terms together)

  21. 1-3: Solving Equations • Solve for x • (multiply both sides by a, to clear the first denominator) • (multiply both sides by b, to clear the second denominator) • (subtract bx on both sides, to get the x terms together) • (distributive property, backwards)

  22. 1-3: Solving Equations • Solve for x • (multiply both sides by a, to clear the first denominator) • (multiply both sides by b, to clear the second denominator) • (subtract bx on both sides, to get the x terms together) • (distributive property, backwards) • (divide both sides by “a – b”)

  23. 1-3: Solving Equations • Solve for x • Are there any restrictions on the variables? Is there any number we couldn’t use in place of a, b, or x?

  24. 1-3: Solving Equations • Solve for x • Are there any restrictions on the variables? Is there any number we couldn’t use in place of a, b, or x? • Looking at the beginning problem • a ≠ 0 and b ≠ 0 (can’t have a denominator of 0)

  25. 1-3: Solving Equations • Solve for x • Are there any restrictions on the variables? Is there any number we couldn’t use in place of a, b, or x? • Looking at the beginning problem • a ≠ 0 and b ≠ 0 (can’t have a denominator of 0) • Looking at the solution • a – b ≠ 0 (again, denominator can’t be 0) • a ≠ b (add b to both sides)

  26. 1-3: Solving Equations • Assignment • Page 21 • 1 – 27, odd problems • Show your work • Tomorrow: Word problems

  27. Chapter 1: Tools of Algebra1-3: Solving Equations (Day 2) Essential Question: What is the procedure to solve an equation for a variable?

  28. 1-3: Solving Equations • Writing Equations to Solve Problems • Ex 5: A dog kennel owner has 100 ft of fencing to enclose a rectangular dog run. She wants it to be 5 times as long as it is wide. Find the dimensions of the dog run. • (Optional) Draw a diagram • Determine the formula to use

  29. 1-3: Solving Equations • Writing Equations to Solve Problems • Ex 5: A dog kennel owner has 100 ft of fencing to enclose a rectangular dog run. She wants it to be 5 times as long as it is wide. Find the dimensions of the dog run. • (Optional) Draw a diagram • Determine the formula to use • Perimeter = 2 • width + 2 • length • Determine the unknowns

  30. 1-3: Solving Equations • Writing Equations to Solve Problems • Ex 5: A dog kennel owner has 100 ft of fencing to enclose a rectangular dog run. She wants it to be 5 times as long as it is wide. Find the dimensions of the dog run. • (Optional) Draw a diagram • Determine the formula to use • Perimeter = 2 • width + 2 • length • Determine the unknowns • Let perimeter = 100 • Let width = x • Let length= 5x • Use variable in the equation, and solve

  31. 1-3: Solving Equations • Perimeter = 2 • width + 2 • length • 100 = 2 • x + 2 • 5x

  32. 1-3: Solving Equations • Perimeter = 2 • width + 2 • length • 100 = 2 • x + 2 • 5x • 100 = 2x + 10x

  33. 1-3: Solving Equations • Perimeter = 2 • width + 2 • length • 100 = 2 • x + 2 • 5x • 100 = 2x + 10x • 100 = 12x

  34. 1-3: Solving Equations • Perimeter = 2 • width + 2 • length • 100 = 2 • x + 2 • 5x • 100 = 2x + 10x • 100 = 12x12 12 • 8 1/3 = x • Determine both of your unknowns from the beginning of the problem

  35. 1-3: Solving Equations • Perimeter = 2 • width + 2 • length • 100 = 2 • x + 2 • 5x • 100 = 2x + 10x • 100 = 12x12 12 • 8 1/3 = x • Determine both of your unknowns from the beginning of the problem • Width = x = 8 1/3 ft • Length = 5x = 5 • 8 1/3 = 41 2/3 ft

  36. 1-3: Solving Equations • Writing Equations to Solve Problems • Ex 6: The lengths of the sides of a triangle are in the ratio 3:4:5. The perimeter of the triangle is 18 in. Find the length of the sides. • (Optional) Draw a diagram • Determine the formula to use

  37. 1-3: Solving Equations • Writing Equations to Solve Problems • Ex 6: The lengths of the sides of a triangle are in the ratio 3:4:5. The perimeter of the triangle is 18 in. Find the length of the sides. • (Optional) Draw a diagram • Determine the formula to use • Perimeter = s1 + s2 + s3 • Determine the variables

  38. 1-3: Solving Equations • Writing Equations to Solve Problems • Ex 6: The lengths of the sides of a triangle are in the ratio 3:4:5. The perimeter of the triangle is 18 in. Find the length of the sides. • (Optional) Draw a diagram • Determine the formula to use • Perimeter = s1 + s2 + s3 • Determine the variables • Let perimeter = 18 • Let s1 (shortest side) = 3x • Let s2 (second side) = 4x • Let s3 (third side) = 5x • Use variable in the equation, and solve

  39. 1-3: Solving Equations • Perimeter = s1 + s2 + s3 • 18 = 3x + 4x + 5x

  40. 1-3: Solving Equations • Perimeter = s1 + s2 + s3 • 18 = 3x + 4x + 5x • 18 = 12x

  41. 1-3: Solving Equations • Perimeter = s1 + s2 + s3 • 18 = 3x + 4x + 5x • 18 = 12x12 12 • 1.5 = x • Determine both of your variables (unknowns) from the beginning of the problem

  42. 1-3: Solving Equations • Perimeter = s1 + s2 + s3 • 18 = 3x + 4x + 5x • 18 = 12x12 12 • 1.5 = x • Determine both of your variables (unknowns) from the beginning of the problem • s1 = 3x = 3 • 1.5 = 4.5 in • s2 = 4x = 4 • 1.5 = 6 in • s3 = 5x = 5 • 1.5 = 7.5 in

  43. 1-3: Solving Equations • Assignment • Page 22 • 29 – 35, all problems • Skip 35b • Show your work • What equation you used to solve the problem • Some of the steps you took to find your solution

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