Section 2 3 Solving Linear Equations Involving Fractions

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Section 2. 3 Solving Linear Equations Involving Fractions and Decimals; Classifying Equations

Section 2. 3 Solving Linear Equations Involving Fractions and Decimals; Classifying Equations

Section 2. 3 Objectives 1 Use the Least Common Denominator to Solve a Linear

Section 2. 3 Objectives 1 Use the Least Common Denominator to Solve a Linear Equation Containing Fractions 2 Solve a Linear Equation Containing Decimals 3 Classify a Linear Equation as an Identity, Conditional, or a Contradiction 4 Use Linear Equations to Solve Problems Copyright © 2014, 2010, 2007 Pearson Education, Inc. 2

Solving Equations Containing Fractions Example: Solve the equation Fractions can be removed by multiplying

Solving Equations Containing Fractions Example: Solve the equation Fractions can be removed by multiplying both sides of the equation by the LCD of all the fractions. The LCD is 10. Remove all parentheses. Divide out common factors. Simplify. Copyright © 2014, 2010, 2007 Pearson Education, Inc. Continued. 3

Solving Equations Containing Fractions Example continued: Add – 4 to both sides of the

Solving Equations Containing Fractions Example continued: Add – 4 to both sides of the equation. Divide both sides by 10 Check your answer in the original equation. Copyright © 2014, 2010, 2007 Pearson Education, Inc. 4

Solving Equations Containing Decimals Example: Solve the equation Decimals can be eliminated by multiplying

Solving Equations Containing Decimals Example: Solve the equation Decimals can be eliminated by multiplying both sides of the equation by the LCD of all the decimals. Distribute before eliminating the decimals. Combine like terms. Subtract 3. 6 x from both sides. Subtract 5. 4 from both sides. Continued. Copyright © 2014, 2010, 2007 Pearson Education, Inc. 5

Solving Equations Containing Decimals Example continued: Multiply both sides of the equation by the

Solving Equations Containing Decimals Example continued: Multiply both sides of the equation by the LCD 10 to eliminate the decimal. Simplify. Divide both sides by – 38. Check your answer in the original equation. Copyright © 2014, 2010, 2007 Pearson Education, Inc. 6

Conditional Equations & Contradictions A conditional equation is an equation that is true for

Conditional Equations & Contradictions A conditional equation is an equation that is true for some values of the variable and false for other values of the variable. A contradiction is an equation that is false for every replacement value of the variable. Example: Solve the equation 4 x – 8 = 4 x – 1 – 8 = – 1 Subtract 4 x from both sides. This is a false statement so the equation is a contradiction. The solution set is the empty set, written { } or . Copyright © 2014, 2010, 2007 Pearson Education, Inc. 7

Identities An identity is an equation that is satisfied for all values of the

Identities An identity is an equation that is satisfied for all values of the variable for which both sides of the equation are defined. Example: Solve the equation 5 x + 3 = 2 x + 3(x + 1). 5 x + 3 = 2 x + 3 Distribute to remove parentheses. 5 x + 3 = 5 x + 3 Combine like terms. 3=3 Subtract 5 x from both sides. This is a true statement for all real numbers x. The solution set is all real numbers. Copyright © 2014, 2010, 2007 Pearson Education, Inc. 8

Solving Problems You have $5000 to invest, and your financial advisor advises you to

Solving Problems You have $5000 to invest, and your financial advisor advises you to put part of the money in a Certificate of Deposit (CD) that earns 2% simple interest compounded annually, and the rest in bonds that earn 4% simple interest compounded annually. To determine the amount x you should invest in the CD to earn $170 interest at the end of one year, solve the equation 0. 02 x + 0. 04(5000 – x) = 170 0. 02 x + 200 – 0. 04 x = 170 – 0. 02 x + 200 – 200 = 170 – 200 – 0. 02 x = – 30 Distributive property. Combine like terms. Isolate x. Copyright © 2014, 2010, 2007 Pearson Education, Inc. 9

Solving Problems – 0. 02 x = – 3000 x = 1500 Multiply by

Solving Problems – 0. 02 x = – 3000 x = 1500 Multiply by 100 Divide by – 2 Because x = 1500, you must invest $1500 in Certificate of Deposit to earn $170 in interest at the end of the year. Copyright © 2014, 2010, 2007 Pearson Education, Inc. 1 0