Copyright © 2013, 2009, 2006 Pearson Education, Inc. 1 1 Section 2.3 Solving Linear Equations Copyright © 2013, 2009, 2006 Pearson Education, Inc. 1

2 2

3 1.Simplify the algebraic expression on each side. 2.Collect all the variable terms on one side and all the constant terms on the other side. 3.Isolate the variable and solve. 4.Check the proposed solution in the original equation. Linear Equations A step-by-step procedure for Solving Linear Equations:

Copyright © 2013, 2009, 2006 Pearson Education, Inc. 4 Solve and check: 7x = 3(x + 8) 7x = 3(x + 8) 7x = 3x + 24 Use the Distributive Property. 7x – 3x = 3x + 24 – 3x Subtract 3x from both sides. 4x = 24 Simplify. x = 6 Solve. 7(6) =3(6 + 8) Check the solution in the original equation. 42 = 3(14) 42 = 42 The true statement 42 = 42 verifies 6 is the solution. Solving Linear EquationsEXAMPLE

Copyright © 2013, 2009, 2006 Pearson Education, Inc. 5 Solve and check: 7x = 3(x + 8) 7x = 3(x + 8) 7x = 3x + 24 Use the Distributive Property. 7x – 3x = 3x + 24 – 3x Subtract 3x from both sides. 4x = 24 Simplify. x = 6 Solve. 7(6) =3(6 + 8) Check the solution in the original equation. 42 = 3(14) 42 = 42 The true statement 42 = 42 verifies 6 is the solution. Solving Linear EquationsEXAMPLE

Copyright © 2013, 2009, 2006 Pearson Education, Inc. 6 6 Objective #1: Examples

Copyright © 2013, 2009, 2006 Pearson Education, Inc. 7 7 Objective #1: Examples

Copyright © 2013, 2009, 2006 Pearson Education, Inc. 8 8 Objective #1: Examples

Copyright © 2013, 2009, 2006 Pearson Education, Inc. 9 9 Objective #1: Examples

Copyright © 2013, 2009, 2006 Pearson Education, Inc. 10 Copyright © 2013, 2009, 2006 Pearson Education, Inc. 10

Copyright © 2013, 2009, 2006 Pearson Education, Inc. 11 Solving a linear equation involving fractionsEXAMPLE SOLUTION Solve and check: 1) Simplify the algebraic expressions on each side. Multiply both sides by the LCD: 30 Distributive Property

Copyright © 2013, 2009, 2006 Pearson Education, Inc. 12 Solving a linear equation involving fractions Cancel CONTINUED Multiply Distribute

Copyright © 2013, 2009, 2006 Pearson Education, Inc. 13 Solving a linear equation involving fractionsCONTINUED Combine like terms14x + 2 = 15x 2) Collect variable terms on one side and constant terms on the other side. 2 = x Subtract 14x from both sides 14x – 14x + 2 = 15x – 14x Simplify 3) Isolate the variable and solve.Already done.

Copyright © 2013, 2009, 2006 Pearson Education, Inc. 14 Solving a linear equation involving fractionsCONTINUED 4) Check the proposed solution in the original equation. Replace x with 2. Simplify. Original Equation Simplify. ? ? ?

Copyright © 2013, 2009, 2006 Pearson Education, Inc. 15 Solving a linear equation involving fractionsCONTINUED Simplify 1 = 1 Simplify Since the proposed x value of 2 makes the statement true, 2 is indeed the solution of the original equation.

Copyright © 2013, 2009, 2006 Pearson Education, Inc. 16 Copyright © 2013, 2009, 2006 Pearson Education, Inc. 16 Objective #2: Examples

Copyright © 2013, 2009, 2006 Pearson Education, Inc. 17 Copyright © 2013, 2009, 2006 Pearson Education, Inc. 17 Objective #2: Examples

Copyright © 2013, 2009, 2006 Pearson Education, Inc. 18 Copyright © 2013, 2009, 2006 Pearson Education, Inc. 18

Copyright © 2013, 2009, 2006 Pearson Education, Inc. 19 Copyright © 2013, 2009, 2006 Pearson Education, Inc. 19 Objective #3: Examples

Copyright © 2013, 2009, 2006 Pearson Education, Inc. 20 Copyright © 2013, 2009, 2006 Pearson Education, Inc. 20 Objective #3: Examples

Copyright © 2013, 2009, 2006 Pearson Education, Inc. 21 Copyright © 2013, 2009, 2006 Pearson Education, Inc. 21

Copyright © 2013, 2009, 2006 Pearson Education, Inc. 22 Some equations are not true for any real number. Such equations are called inconsistent equations or contradictions. Example: Solve: 2x = 2(x +3) 2x = 2x + 6 Distributive Property. 2x – 2x = 2x + 6 – 2x Subtract 2x from both sides. 0 = 6 Simplify. False, 0 ≠ 6. Inconsistent, no solution. Linear equations having no solutions

Copyright © 2013, 2009, 2006 Pearson Education, Inc. 23 Categorizing an Equations Type of EquationsDefinitions IdentityAn equation that is true for all real numbers ConditionalAn equation that is not an identity but is true for at least one real number Inconsistent (contradiction) An equation that is not true for any real number

Copyright © 2013, 2009, 2006 Pearson Education, Inc. 24 Categorizing an EquationEXAMPLE SOLUTION Solve and determine whether the equation is an identity, a conditional equation or an inconsistent equation x = 9x + 5 4x = 9x 4x – 4x = 9x – 4x Subtract 5 from both sides Simplify Subtract 4x from both sides

Copyright © 2013, 2009, 2006 Pearson Education, Inc. 25 Categorizing an Equation 0 = x Divide both sides by 5 Simplify The original equation is only true when x = 0. Therefore, it is a conditional equation. CONTINUED 0 = 5x Simplify

Copyright © 2013, 2009, 2006 Pearson Education, Inc. 26 Categorizing an EquationEXAMPLE SOLUTION Solve and determine whether the equation is an identity, a conditional equation or an inconsistent equation. 5 – (2x – 4) = 4(x +1) – 2x 5 – 2x + 4 = 4x + 4 – 2x 9 – 2x = 4 – 2x Distribute the – 1 and the 4. Simplify. Since after simplification we see a contradiction, we know that the original equation is inconsistent and can never be true for any x. 9 = 4Add 2x to both sides.

Copyright © 2013, 2009, 2006 Pearson Education, Inc. 27 Categorizing an EquationEXAMPLE SOLUTION Solve and determine whether the equation is an identity, a conditional equation or an inconsistent equation x = 3(x +1) – x 3 + 2x = 3x + 3 – x 3 + 2x = 2x + 3 Distribute the 3. Simplify. Since after simplification we can see that the left hand side (LHS) is equal to the RHS of the equation, this is an identity and is always true for all x.

Copyright © 2013, 2009, 2006 Pearson Education, Inc. 28 Copyright © 2013, 2009, 2006 Pearson Education, Inc. 28 Objective #4: Examples

Copyright © 2013, 2009, 2006 Pearson Education, Inc. 29 Copyright © 2013, 2009, 2006 Pearson Education, Inc. 29 Objective #4: Examples

Copyright © 2013, 2009, 2006 Pearson Education, Inc. 30 Copyright © 2013, 2009, 2006 Pearson Education, Inc. 30 Objective #4: Examples

Copyright © 2013, 2009, 2006 Pearson Education, Inc. 31 Copyright © 2013, 2009, 2006 Pearson Education, Inc. 31 Objective #4: Examples

Copyright © 2013, 2009, 2006 Pearson Education, Inc. 32 Copyright © 2013, 2009, 2006 Pearson Education, Inc. 32

Copyright © 2013, 2009, 2006 Pearson Education, Inc. 33 The formula describes the pressure of sea water, p, in pounds per square foot, at a depth d feet below the surface. At what depth is the pressure 30 pounds per square foot? Substitute the given pressure into the equation for p, p = 35. The equation becomes: Linear Equations – An application

Copyright © 2013, 2009, 2006 Pearson Education, Inc. 34 The depth where the pressure is 30 pounds per square foot is 33 feet. Linear EquationsCONTINUED

Copyright © 2013, 2009, 2006 Pearson Education, Inc. 35 The depth where the pressure is 30 pounds per square foot is 33 feet. Linear EquationsCONTINUED

Copyright © 2013, 2009, 2006 Pearson Education, Inc. 36 Copyright © 2013, 2009, 2006 Pearson Education, Inc. 36 Objective #5: Examples Where x represents the intensity of a negative life event (from a low of 1 to a high of 10) and D is the level of depression in response to that event.

Copyright © 2013, 2009, 2006 Pearson Education, Inc. 37 Copyright © 2013, 2009, 2006 Pearson Education, Inc. 37 Objective #5: ExamplesCONTINUED

Copyright © 2013, 2009, 2006 Pearson Education, Inc. 38 Copyright © 2013, 2009, 2006 Pearson Education, Inc. 38 Objective #5: ExamplesCONTINUED