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

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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.  You can manipulate equations to help find a solution, so long as you do the same thing to both sides of the equation.  Addition PropertyIfa = b, then a + c = b + c  Subtraction PropertyIf a = b, then a – c = b – c  Multiplication PropertyIf a = b, then ac = bc  Division PropertyIf a = b, then a / c = b / c

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

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

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

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

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

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)

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

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

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

1-3: Solving Equations  Solving a Formula for One of Its Variables  The formula for the area of a trapezoid is A = ½ h(b 1 + b 2 ). Solve the formula for h.  The goal is to use PEMDAS (in reverse) to get the variable in question alone.  A = ½ h(b 1 + b 2 )

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

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

1-3: Solving Equations  Solving a Formula for One of Its Variables  The formula for the area of a trapezoid is A = ½ h(b 1 + b 2 ).  Solve the formula for b 1

1-3: Solving Equations  Solve for x

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

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)

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)

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)

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”)

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?

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)

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)

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

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

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

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

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

1-3: Solving Equations  Perimeter = 2 width + 2 length  100 = 2 x + 2 5x

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

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

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

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

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

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 = s 1 + s 2 + s 3  Determine the variables

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 = s 1 + s 2 + s 3  Determine the variables  Let perimeter = 18  Let s 1 (shortest side) = 3x  Let s 2 (second side) = 4x  Let s 3 (third side) = 5x  Use variable in the equation, and solve

1-3: Solving Equations  Perimeter = s 1 + s 2 + s 3  18 = 3x + 4x + 5x

1-3: Solving Equations  Perimeter = s 1 + s 2 + s 3  18 = 3x + 4x + 5x  18 = 12x

1-3: Solving Equations  Perimeter = s 1 + s 2 + s 3  18 = 3x + 4x + 5x  18 = 12x  12  12  1.5 = x  Determine both of your variables (unknowns) from the beginning of the problem

1-3: Solving Equations  Perimeter = s 1 + s 2 + s 3  18 = 3x + 4x + 5x  18 = 12x  12  12  1.5 = x  Determine both of your variables (unknowns) from the beginning of the problem  s 1 = 3x = 3 1.5 = 4.5 in  s 2 = 4x = 4 1.5 = 6 in  s 3 = 5x = 5 1.5 = 7.5 in

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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