Steps for Solving an Equations: Topic: Solving Complex Equations Essential Question: How do we solve Complex equations? Standard: 8.EE.7 Question / Key Words Notes or Solutions to Problems Simplify each side by distributing and/or combining like terms Steps for Solving an Equations: 2. Collect variables terms on the side where the coefficient is greater. Use inverse operations to isolate the variable. 4. Check your solution in the original equation.

Solve 3(2 – x) – x = -5(x+ 1) 6 – 3x – x = – 5x – 5 – 4x + 6 = – 5x Question / Key Words Notes or Solutions to Problems Solve 3(2 – x) – x = -5(x+ 1) 1) 3(2 – x) – x = -(5x + 1) 6 – 3x – x = – 5x – 5 – 4x + 6 = – 5x – 5 + 5x + 5x x + 6 = – 5 – 6 – 6 x = – 11

Solve 3(2 – 4x) + 3x = 9(x + 2) – 6x 6 – 12x +3x 9x + 18 – 6x = – 9x Question / Key Words Notes or Solutions to Problems Solve 3(2 – 4x) + 3x = 9(x + 2) – 6x 2) 3(2 – 4x) + 3x = 9(x +2) –6x 6 – 12x +3x 9x + 18 – 6x = – 9x + 6 = 3x + 18 + 9x +9x 6 = 12x + 18 – 18 – 18 1 – 12 = 12x 12 12 – 1 = x

Solve 3) (15x + 20) = 6 – 2(x – 4) 3x + 4 = 6 – 2x + 8 3x + 4 = – 2x Question / Key Words Notes or Solutions to Problems Solve 1 5 3) (15x + 20) = 6 – 2(x – 4) 1 5 (15x + 20) = 6 – 2(x – 4) 3x + 4 = 6 – 2x + 8 3x + 4 = – 2x + 14 + 2x + 2x 5x + 4 = 14 – 4 – 4 1 5x = 10 5 5 x = 2

2(4x + 5) = 8x + 10 8x + 10 = 8x + 10 –8x –8x 10 = 10 Identity 6) Question / Key Words Notes or Solutions to Problems Solve the equation if possible. Determine whether it has one solution, no solution, or is an identity. 2(4x + 5) = 8x + 10 8x + 10 = 8x + 10 6) 2(4x + 5) = 8x + 10 –8x –8x 10 = 10 Answer: The equation 10 = 10 is always true, so all values of x are solutions. The original equation is an identity. Identity

x – 1 = x + 7 –x –x -1 = 7 No solution. 7) x – 1 = x + 7 Question / Key Words Notes or Solutions to Problems Solve the equation if possible. Determine whether it has one solution, no solution, or is an identity. x – 1 = x + 7 –x –x 7) x – 1 = x + 7 -1 = 7 Answer: The equation -1 = 7 is never true no matter what the value of x. The original equation has no solution. No solution.

Question / Key Words Notes or Solutions to Problems 4) A rental car agency charges $20 per day plus 11 cents per mile to rent a certain car. Another agency charges $26 per day plus 5 cents per mile to rent the same car. How many miles per day will have to be driven for the cost of a car from the first agency to equal the cost of a car from the second agency? Cost of first rental Agency C = 0.11x + 20 Cost of second rental Agency C = 0.05x + 26 0.11x + 20 = 0.05x + 26 – 0.05x – 0.05x 0.06x + 20 = 26 – 20 – 20 0.06x = 6

4) 0.06x = 6 0.06 0.06 x = 100 miles Question / Key Words Notes or Solutions to Problems 4) A rental car agency charges $20 per day plus 11 cents per mile to rent a certain Car. Another agency Charges $26 per day plus 5 cents per mile to rent the same car. How many miles per day will have to be driven for the cost of a car from the first agency to equal the cost of a car from the second agency? 1 0.06x = 6 0.06 0.06 x = 100 miles