© 2010 Pearson Prentice Hall. All rights reserved Solving Equations Using Two Properties § 10.5.

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© 2010 Pearson Prentice Hall. All rights reserved Solving Equations Using Two Properties § 10.5

Tobey & Slater, Basic College Mathematics, 6e2 Using Two Properties Example: Solve for c. 3c  9 =  24 3c  =  Add 9 to both sides of the equation. 3(  5)  9 =  24 3c =  15 Divide both sides of the equation by 3. c =  5  15  9 =  24  24 =  24 Check your answer in the original equation.

Tobey & Slater, Basic College Mathematics, 6e3 Variables on Both Sides of the Equation Example: Solve for w. 9w + 5 = 3w  13 9w (  5) = 3w  13 + (  5) Add  5 to both sides of the equation. 9(  3) + 5 = 3(  3)  13 9w = 3w  18 9w + (  3w) = 3w  18 + (  3w) Add (  3w) to both sides of the equation. 6w =  18 We need to keep the variables on one side of the equation and the numbers on the other side of the equation. Divide both sides of the equation by 6. w =  3 Check your answer in the original equation.  =  9  13  22 =  22

Tobey & Slater, Basic College Mathematics, 6e4 Variables on Both Sides of the Equation Example: Solve for a a  19 = 8a  6 + 3a 7a  14 = 11a  6 Add like terms on both sides of the equation. 7a  = 11a  a = 11a + 8 Add 14 to both sides of the equation. 7a  11a = 11a + 8  11a Divide both sides of the equation by  4. Add (  11a) to both sides of the equation.  4a = 8 Be sure to check your answer!

Tobey & Slater, Basic College Mathematics, 6e5 Procedure to Solve Equations 1.Remove any parentheses by using the distributive property. 2.Collect like terms on each side of the equation. 3.Add the appropriate term to both sides of the equation to get all numbers on one side. 4.Add the appropriate term to both sides of the equation to get all variable terms on the other side. 5.Divide both sides of the equation by the numerical coefficient of the variable term. 6.Check by substituting the solution back into the original equation.

Tobey & Slater, Basic College Mathematics, 6e6 Solving Equations with Parentheses Example: Solve for x. 5x  6(x + 1) = 2x  6 5x  6x + (  6) = 2x  6 Distribute to remove the parentheses.  x + (  6) = 2x  6  x + (  6) + 6 = 2x  Add like terms.  x = 2x Add (  2x) to both sides of the equation. Be sure to check your answer! Add 6 to both sides of the equation.  x  2x = 2x  2x x = 0