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Section 2.5 Solving Linear Equations: ax + b = cx + d HAWKES LEARNING SYSTEMS Students Matter. Success Counts. Copyright © 2013 by Hawkes Learning Systems/Quant Systems, Inc. All rights reserved. Objectives o Solve equations of the form ax + b = cx + d. o Understand the terms conditional equations, identities, and contradictions. HAWKES LEARNING SYSTEMS Students Matter. Success Counts. Copyright © 2013 by Hawkes Learning Systems/Quant Systems, Inc. All rights reserved. Solve Equations of the Form ax + b = cx + d General Procedure for Solving Linear Equations that Simplify to the Form ax + b = cx + d 1. Simplify each side of the equation by removing any grouping symbols and combining like terms on both sides of the equation. 2. Use the addition principle of equality and add the opposite of a constant term and/or variable term to both sides so that variables are on one side and constants are on the other side. HAWKES LEARNING SYSTEMS Students Matter. Success Counts. Copyright © 2013 by Hawkes Learning Systems/Quant Systems, Inc. All rights reserved. Solve Equations of the Form ax + b = cx + d General Procedure for Solving Linear Equations that Simplify to the Form ax + b = cx + d (cont.) 3. Use the multiplication (or division) principle of equality to multiply both sides by the reciprocal of the coefficient of the variable (or divide both sides by the coefficient itself). The coefficient of the variable will become +1. 4. Check your answer by substituting it for the variable in the original equation. HAWKES LEARNING SYSTEMS Students Matter. Success Counts. Copyright © 2013 by Hawkes Learning Systems/Quant Systems, Inc. All rights reserved. Example 1: Solving Equations with Variables on Both Sides Solve the following equations. a. 5x + 3 = 2 x 18 Solution 5x + 3 = 2x 18 5x + 3 3 = 2x 18 3 5x = 2x 21 5x 2x = 2x 21 2x 3x = 21 HAWKES LEARNING SYSTEMS Students Matter. Success Counts. Write the equation. Add 3 to both sides. Simplify. Add 2 x to both sides. Simplify. Copyright © 2013 by Hawkes Learning Systems/Quant Systems, Inc. All rights reserved. Example 1: Solving Equations with Variables on Both Sides (cont.) 3x 21 = 3 3 x = 7 Check: Divide both sides by 3. Simplify. 5x + 3 = 2x 18 ? 5 7 + 3 = 2 7 18 ? 35 + 3 = 14 18 32 = 32 HAWKES LEARNING SYSTEMS Students Matter. Success Counts. Substitute x = 7. Simplify. True statement Copyright © 2013 by Hawkes Learning Systems/Quant Systems, Inc. All rights reserved. Example 1: Solving Equations with Variables on Both Sides (cont.) b. 4 x + 1 x = 2 x 13 + 5 Solution 4 x + 1 x = 2x 13 + 5 3x + 1 = 2x 8 3x + 1 1 = 2 x 8 1 3x = 2 x 9 3x 2 x = 2 x 9 2 x x = 9 HAWKES LEARNING SYSTEMS Students Matter. Success Counts. Write the equation. Combine like terms. Add 1 to both sides. Simplify. Add 2 x to both sides. Simplify. Copyright © 2013 by Hawkes Learning Systems/Quant Systems, Inc. All rights reserved. Example 1: Solving Equations with Variables on Both Sides (cont.) Check: 4 x + 1 x = 2x 13 + 5 ? 4 9 + 1 9 = 2 9 13 + 5 ? 36 + 1 + 9 = 18 13 + 5 26 = 26 HAWKES LEARNING SYSTEMS Students Matter. Success Counts. Substitute x = 9. Simplify. True statement Copyright © 2013 by Hawkes Learning Systems/Quant Systems, Inc. All rights reserved. Example 2: Solving Linear Equations Involving Decimals 6y + 2.5 = 7y 3.6 Solution 6y + 2.5 = 7y 3.6 6y + 2.5 + 3.6 = 7y 3.6 + 3.6 6y + 6.1 = 7y 6y + 6.1 6y = 7y 6y 6.1 = y HAWKES LEARNING SYSTEMS Students Matter. Success Counts. Write the equation. Add 3.6 to both sides. Simplify. Add 6y to both sides. Simplify. Copyright © 2013 by Hawkes Learning Systems/Quant Systems, Inc. All rights reserved. Example 2: Solving Linear Equations Involving Decimals (cont.) Check: 6y + 2.5 = 7y 3.6 ? 6 6.1 + 2.5 = 7 6.1 3.6 ? 36.6 + 2.5 = 42.7 3.6 39.1 = 39.1 HAWKES LEARNING SYSTEMS Students Matter. Success Counts. Substitute y = 6.1. Simplify. True statement Copyright © 2013 by Hawkes Learning Systems/Quant Systems, Inc. All rights reserved. Example 3: Solving Linear Equations with Fractional Coefficients 1 1 2 7 x+ = x 3 6 5 10 Solution 1 1 2 7 x+ = x 3 6 5 10 7 1 1 2 30 x + = 30 x 3 6 5 10 Write the equation. Multiply both sides by the LCM 30. 1 1 2 7 30 x + 30 = 30 x 30 Apply the distributive property. 3 6 5 10 HAWKES LEARNING SYSTEMS Students Matter. Success Counts. Copyright © 2013 by Hawkes Learning Systems/Quant Systems, Inc. All rights reserved. Example 3: Solving Linear Equations with Fractional Coefficients (cont.) 10 x + 5 = 12x 21 10 x + 5 5 = 12 x 21 5 10 x = 12x 26 10 x 12 x = 12 x 26 12 x 2x = 26 2 x 26 = 2 2 x = 13 HAWKES LEARNING SYSTEMS Students Matter. Success Counts. Simplify. Add 5 to both sides. Simplify. Add 12x to both sides. Simplify. Divide both sides by 2. Simplify. Copyright © 2013 by Hawkes Learning Systems/Quant Systems, Inc. All rights reserved. Example 4: Solving Equations with Parentheses Solve the following equations. a. 2 y 7 = 4 y + 1 26 Solution 2 y 7 = 4 y + 1 26 Write the equation. 2y 14 = 4 y + 4 26 Use the distributive property. 2y 14 = 4 y 22 Combine like terms. 2y 14 + 22 = 4 y 22 + 22 HAWKES LEARNING SYSTEMS Students Matter. Success Counts. Add 22 to both sides. Here we will put the variable on the right side to get a positive coefficient of y. Copyright © 2013 by Hawkes Learning Systems/Quant Systems, Inc. All rights reserved. Example 4: Solving Equations with Parentheses (cont.) 2y + 8 = 4 y 2y + 8 2y = 4 y 2y 8 = 2y 8 2y = 2 2 4=y HAWKES LEARNING SYSTEMS Students Matter. Success Counts. Simplify. Add 2y to both sides. Simplify. Divide both sides by 2. Simplify. Copyright © 2013 by Hawkes Learning Systems/Quant Systems, Inc. All rights reserved. Example 4: Solving Equations with Parentheses (cont.) b. 2 5x + 13 2 = 6 3x 2 41 Solution 2 5x + 13 2 = 6 3x 2 41 10 x 26 2 = 18 x + 12 41 10x 28 = 18x 29 10x 28 + 18x = 18 x 29 + 18x HAWKES LEARNING SYSTEMS Students Matter. Success Counts. Write the equation. Use the distributive property. Be careful with the signs. Combine like terms. Add 18x to both sides. Copyright © 2013 by Hawkes Learning Systems/Quant Systems, Inc. All rights reserved. Example 4: Solving Equations with Parentheses (cont.) 8 x 28 = 29 8x 28 + 28 = 29 + 28 8 x = 1 8 x 1 = 8 8 1 x= 8 HAWKES LEARNING SYSTEMS Students Matter. Success Counts. Simplify. Add 28 to both sides. Simplify. Divide both sides by 8. Simplify. Copyright © 2013 by Hawkes Learning Systems/Quant Systems, Inc. All rights reserved. Example 5: Solutions of Equations Determine whether each of the following equations is a conditional equation, an identity, or a contradiction. a. 3x + 16 = 11 Solution 3x + 16 = 11 Write the equation. 3x + 16 16 = 11 16 Add 16 to both sides. 3x = 27 Simplify. 3x 27 = Divide both sides by 3. 3 3 x = 9 Simplify. The equation has one solution and it is a conditional equation. HAWKES LEARNING SYSTEMS Students Matter. Success Counts. Copyright © 2013 by Hawkes Learning Systems/Quant Systems, Inc. All rights reserved. Example 5: Solutions of Equations (cont.) b. 3 x 25 + 3x = 6 x + 10 Solution 3 x 25 + 3x = 6 x + 10 3x 75 + 3x = 6x + 60 6x 75 = 6 x + 60 Write the equation. Use the distributive property. Combine like terms. 6 x 75 6 x = 6 x + 60 6 x Add 6x to both sides. 75 = 60 Simplify. The last equation is never true. Therefore, the original equation is a contradiction and has no solution. HAWKES LEARNING SYSTEMS Students Matter. Success Counts. Copyright © 2013 by Hawkes Learning Systems/Quant Systems, Inc. All rights reserved. Example 5: Solutions of Equations (cont.) c. 2 x 7 + x = 14 x Solution 2 x 7 + x = 14 x 2x + 14 + x = 14 x Write the equation. Use the distributive property. 14 x = 14 x Combine like terms. 14 x + x = 14 x + x Add x to both sides. 14 = 14 Simplify. The last equation is always true. Therefore, the original equation is an identity and has an infinite number of solutions. Every real number is a solution. HAWKES LEARNING SYSTEMS Students Matter. Success Counts. Copyright © 2013 by Hawkes Learning Systems/Quant Systems, Inc. All rights reserved. Practice Problems Solve the following linear equations. 1. x + 14 6x = 2x 7 2. 6.4 x + 2.1 = 3.1x 1.2 2x 1 1 3 1 1 1 3. =x+ 4. n+ = n 3 2 6 14 4 7 4 5. 5 y 3 = 14 4 y + 2 Determine whether each of the following equations is a conditional equation, an identity, or a contradiction. 6. 7 x 3 + 42 = 7x + 21 HAWKES LEARNING SYSTEMS Students Matter. Success Counts. 7. 2x + 14 = 2 x + 1 + 10 Copyright © 2013 by Hawkes Learning Systems/Quant Systems, Inc. All rights reserved. Practice Problem Answers 1. x = 3 2. x = 1 3. x = 2 4. n = 7 2 5. y = 3 6. Identity 7. Contradiction HAWKES LEARNING SYSTEMS Students Matter. Success Counts. Copyright © 2013 by Hawkes Learning Systems/Quant Systems, Inc. All rights reserved.
