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Algebra 3.3 Solving Multi-Step Equations Solve for x ► 1) 1 x 5 10 2 Solve for x ► 1) 1 x 5 10 2 5 5 Solve for x ► 1) 1 x 5 10 2 5 5 1 x 15 2 Solve for x ► 1) 1 x 5 10 2 5 5 1 x 15 2 2 1 15 2 x 12 1 1 Solve for x ► 1) 1 x 5 10 2 5 5 1 x 15 2 2 1 15 2 x 12 1 1 x 30 Combining Like Terms First 2) 2x – 9x + 17 = -4 -7x + 17 = -4 - 17 - 17 -7x = -21 x=3 Using the Distributive Property 3) 5x + 3(x + 4) = 28 5x + 3x + 12 = 28 8x + 12 = 28 - 12 - 12 8x = 16 x=2 Distributing a Negative…Careful! 4) 2x – 5(x – 9) = 27 2x – 5x + 45 = 27 -3x + 45 = 27 - 45 - 45 -3x = -18 x=6 Multiplying by a Reciprocal First 5) Multiplying by a Reciprocal First 3 5) 12 x 2 10 Save time…do not distribute! Multiplying by a Reciprocal First 3 5) 12 x 2 10 Save time…do not distribute! 10 3 10 12 x 2 3 10 3 Multiply by the reciprocal. Multiplying by a Reciprocal First 3 5) 12 x 2 10 Save time…do not distribute! 10 3 10 12 x 2 3 10 3 40 x 2 Multiply by the reciprocal. Multiplying by a Reciprocal First 3 5) 12 x 2 10 Save time…do not distribute! 10 3 10 12 x 2 3 10 3 40 x 2 2 2 Multiply by the reciprocal. Subtract 2 from each side. Multiplying by a Reciprocal First 3 5) 12 x 2 10 Save time…do not distribute! 10 3 10 12 x 2 3 10 3 40 x 2 2 2 38 x Multiply by the reciprocal. Subtract 2 from each side. Three numbers add to 101. The first number is 12 more than the second number. The third number is nine more than twice the second number. Find the numbers. Three numbers add to 101. a + b + c = 101 The first number is 12 more than the second number. a = b + 12 The third number is nine more than twice the second number. c = 2b + 9 Substitute and solve. (b + 12) + b + (2b + 9) = 101 4b + 21 = 101 4b = 80 b = 20 a = b +12 = 20 + 12 = 32 c = 2b + 9 = 2(20) + 9 = 40 + 9 = 49 a = 32, b = 20, c = 49 They add to 101!! Homework P. 148 (#11-43 Odds) 51, 54