1. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. 9.3 Further Solving Linear Equations

2. Martin-Gay, Prealgebra & Introductory Algebra, 3ed 22 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. To Solve Linear Equations in One Variable Solving linear equations in one variable Step 1:If an equation contains fractions, multiply both sides by the LCD to clear the equation of fractions. Step 2: Use the distributive property to remove parentheses if they are present. Step 3:Simplify each side of the equation by combining like terms. Continued

3. Martin-Gay, Prealgebra & Introductory Algebra, 3ed 33 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. To Solve Linear Equations in One Variable Step 4:Get all variable terms on one side and all numbers on the other side by using the addition property of equality. Step 5: Get the variable alone by using the multiplication property of equality. Step 6:Check the solution by substituting it into the original equation.

4. Martin-Gay, Prealgebra & Introductory Algebra, 3ed 44 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Simplify. Divide both sides by 7. Simplify. Solving Linear Equations Example Multiply both sides by 5. Add –3y to both sides. Add –30 to both sides. Solve :

5. Martin-Gay, Prealgebra & Introductory Algebra, 3ed 55 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Solving Linear Equations Example 5x – 5 = 2(x + 1) + 3x – 7 5x – 5 = 2x + 2 + 3x – 7 Apply the distributive property. 5x – 5 = 5x – 5 Combine like terms. Both sides of the equation are identical. Since this equation will be true for every x that is substituted into the equation, the solution is “all real numbers.” Solve: 5x – 5 = 2(x + 1) + 3x – 7

6. Martin-Gay, Prealgebra & Introductory Algebra, 3ed 66 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Solving Linear Equations Example Since no value for the variable x can be substituted into this equation that will make this a true statement, there is “no solution.” 3x – 7 = 3(x + 1) 3x – 7 = 3x + 3 Apply the distributive property. – 7 = 3 Simplify. 3x + ( – 3x) – 7 = 3x + ( – 3x) + 3 Add –3x to both sides. Solve : 3x – 7 = 3(x + 1)