Skill Check Lesson Presentation Lesson Quiz

Skill Check Solve the inequality. 1. 2x + 9 > 13 x > 2 2. 4x – 5 < –17 x < –3 3. 5x – 3 17 4. –7x + 1 –13

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.

1 – x + = 12(– x + ) = 12( ) – x + = 12(– x) + 12( ) = 12( ) EXAMPLE 1 Solving an Equation by Clearing Fractions 5 6 – x + = 1 2 3 4 Original equation 12(– x + ) = 12( ) 5 6 1 2 3 4 – x + = 5 6 1 2 3 4 Multiply each side by LCD of fractions. 12(– x) + 12( ) = 12( ) 5 6 1 2 3 4 Use distributive property. –10x + 6 = 9 Simplify. –10x + 6 = 9 – 6 – 6 Subtract 6 from each side. –10x = 3 Simplify. –10x = 3 –10 –10 = Divide each side by –10. 3 10 x = – Simplify.

Solving Equations with Decimals As we are about to see, you can clear decimals from an equation.

EXAMPLE 2 Solving an Equation by Clearing Decimals 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. 2.3 = 5.14 + 0.8m Write original equation. 100 100 (2.3) = (5.14 + 0.8m) Multiply each side by 100. 230 = 514 + 80m Use distributive property. Simplify. 230 = 514 + 80m – 514 – 514 Subtract 514 from each side. –284 = 80m Simplify. –284 = 80m 80 80 = Divide each side by 80. –3.55 = m Simplify.

Solving Inequalities You can use the methods you have learned for solving equations with fractional coefficients to solve inequalities.

3 (1 – )x – 5 < 20 x – x – 5 < 20 x – 5 < 20 EXAMPLE 3 Solving an Inequality with Fractions 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 SOLUTION Write a verbal model. Let x represent the original prices of the shirts you can afford to buy. Original price – Gift certificate amount Cash on hand of original price 3 1 < x – x – 5 < 20 3 1 Substitute. (1 – )x – 5 < 20 3 1 Combine like terms. x – 5 < 20 3 2 Simplify.

3 ( x) < (25) x – 5 < 20 x – 5 < 20 + 5 + 5 x < 37.50 EXAMPLE 3 Solving an Inequality with Fractions 1 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 SOLUTION Simplify. x – 5 < 20 3 2 x – 5 < 20 3 2 + 5 + 5 Add 5 to each side. 2 3 ( x) < (25) 3 2 Multiply each side by multiplicative inverse of . 3 2 x < 37.50 Simplify. ANSWER You can afford a shirt whose original price is $37.50 or less.

4 (– m – ) < (– ) – m – < – 8 8 8(– m) – 8( ) < 8(– ) EXAMPLE 4 Solving an Inequality by Clearing Fractions 3 4 – m – < – 1 8 Original inequality (– m – ) < (– ) 3 4 1 8 8 8 Multiply each side by LCD of fractions. 8(– m) – 8( ) < 8(– ) 3 4 1 8 Distributive property –6m – 1 < –2 Simplify. –6m – 1 < –2 + 1 + 1 Add 1 to each side. –6m < –1 Simplify. –6m < –1 Divide each side by –6. Reverse inequality symbol. –6 –6 > 1 6 m > Simplify.

Lesson Quiz Solve the equation or inequality. 1. k = 6 2. 8.6 = 5.8y – 3 y = 2 3. b < –14 4. 5. Challenge Solve the equation x = –63