1. 1 So far you have followed these steps to solve equations with fractions: Undo any addition or subtraction in order to get the variable term alone on one side of the equation. Multiply both sides of the equation by the multiplicative inverse of the coefficient of the variable term. Another way to solve an equation with fractions is to clear fractions by multiplying each side of the equation by the LCD of the fractions. The resulting equation is equivalent to the original equation. Equations and Inequalities with Rational Numbers 5.7 LESSON

2. 2 –10x = 3 – x + = 5 6 1 2 3 4 12 ( – x + ) = 12 ( ) 5656 1212 3434 Solving an Equation by Clearing Fractions EXAMPLE 1 Original equation Use distributive property. Multiply each side by LCD of fractions. Simplify. 5 6 – x + = 1 2 3 4 –10x + 6 = 9 12 ( – x ) + 12 ( ) = 12 ( ) 5 6 1 2 3 4 Subtract 6 from each side. –10x + 6 = 9 – 6 Simplify. –10x = 3 Divide each side by –10. Simplify. 3 10 x = – –10 = Equations and Inequalities with Rational Numbers 5.7 LESSON

3. 3 Solving Equations with Decimals As we are about to see, you can clear decimals from an equation. Equations and Inequalities with Rational Numbers 5.7 LESSON

4. 4 2.3 = 5.14 + 0.8m Solving an Equation by Clearing Decimals EXAMPLE 2 Multiply each side by 100. Write original equation. Use distributive property. Simplify. 230 = 514 + 80m (2.3) = (5.14 + 0.8m) Subtract 514 from each side. 230 = 514 + 80m– 514 Simplify. –284 = 80m Divide each side by 80. –284 = 80m Simplify. –3.55 = m 80 = Solve the equation 2.3 = 5.14 + 0.8m. Because the greatest number of decimal places in any of the terms with decimals is 2, multiply each side of the equation by 10 2, or 100. 100 Equations and Inequalities with Rational Numbers 5.7 LESSON

5. 5 Solving Inequalities You can use the methods you have learned for solving equations with fractional coefficients to solve inequalities. Equations and Inequalities with Rational Numbers 5.7 LESSON

6. 6 Solving an Inequality with Fractions EXAMPLE 3 Substitute. Combine like terms. Simplify. Write a verbal model. Let x represent the original prices of the shirts you can afford to buy. SOLUTION Shopping A sign in a clothing store says to take off the marked price of a shirt. You have $20 in cash and a $5 gift certificate. What are the original prices of the shirts you can afford to buy? 3 1 Original price – – Gift certificate amount Cash on hand of original price 3 1 ≤ ( 1 – ) x – 5 ≤ 20 3 1 x – 5 ≤ 20 3 2 x – x – 5 ≤ 20 3 1 Equations and Inequalities with Rational Numbers 5.7 LESSON

7. 7 Solving an Inequality with Fractions EXAMPLE 3 Simplify. SOLUTION x ≤ 37.50 Simplify. x – 5 ≤ 20 3 2 Add 5 to each side. x – 5 ≤ 20 3 2 + 5 ( x ) ≤ (25) 3 2 2 3 2 3 Multiply each side by multiplicative inverse of. 3 2 ANSWER You can afford a shirt whose original price is $37.50 or less. Equations and Inequalities with Rational Numbers 5.7 LESSON Shopping A sign in a clothing store says to take off the marked price of a shirt. You have $20 in cash and a $5 gift certificate. What are the original prices of the shirts you can afford to buy? 3 1

8. 8 Solving an Inequality by Clearing Fractions EXAMPLE 4 Original inequality Distributive property Multiply each side by LCD of fractions. Simplify. 3 4 – m – ≤ – 1 8 1 4 –6m – 1 ≤ –2 8 ( – m ) – 8 ( ) ≤ 8 ( – ) 3 4 1 8 1 4 Add 1 to each side. –6m – 1 ≤ –2 + 1 Simplify. –6m ≤ –1 Divide each side by –6. Reverse inequality symbol. –6m ≤ –1 Simplify. 1 6 m ≥ –6 ≥ ( – m – ) ≤ ( – ) 3 4 1 8 1 4 8 Equations and Inequalities with Rational Numbers 5.7 LESSON