72 Chapter 1 Linear Equations and Inequalities in One Variable Solving a Literal Equation Example 5 Given solve for y. Solution: Add 2x to both sides. Divide by 3 on both sides. 2x 3y 5, 3y 2x 5 3y 3 y 2x 5 3 Skill Practice Solve for y. 6. 5x 2y 11 Sometimes the variable we want to isolate may appear in more than one term in a literal equation. In such a case, isolate all terms with that variable on one side of the equation. Then apply the distributive property as demonstrated in Example 6. Solving a Literal Equation Example 6 Solve the equation for x. ax 3 cx 7 Solution: Collect the terms containing x on one side of the equation. Collect the remaining terms on the other side. The variable x appears twice in the equation. To isolate x, we want x to appear in only one term. To accomplish this, we apply the distributive property in reverse. Apply the distributive property. The variable x now appears one time in the equation. ax 3 cx 7 ax cx 10 x1a c2 10 1a c2 Divide both sides by . x 10 a c x1a c2 1a c2 10 1a c2 or y 2 3 x 5 3 2x 5 3 2x 3y 5 TIP: Applying the distributive property in reverse is called factoring. Factoring will be studied in detail in Chapter 4. Answer 6. y 11 5x 2 or y 5 2 x 11 2 TIP: When solving a literal equation for a specified variable, there is sometimes more than one way to express your final answer. This flexibility often presents difficulty for students. Students may leave their answer in one form, but the answer given in the text looks different. Yet both forms may be correct. To know if your answer is equivalent to the form given in the text you must try to manipulate it to look like the answer in the book, a process called form fitting. The literal equation from Example 4 can be written in several different forms. The quantity can be split into two fractions. b1 2A b2h h 2A h b2h h 2A h b2 12A b2h2h
miller_intermediate_algebra_4e_ch1_3
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