Chapter 9 - XYZ Custom Plus

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540 <strong>Chapter</strong> 9 Solving Equations and Inequalities This general method of solving linear equations involves using the two properties developed in Sections 8.2 and 8.3. We can add any number to both sides of an equation or multiply (or divide) both sides by the same nonzero number and always be sure we have not changed the solution to the equation. The equations may change in form, but the solution to the equation stays the same. Looking back to Example 1, we can see that each equation looks a little different from the preceding one. What is interesting, and useful, is that each of the equations says the same thing about x. They all say that x is 25. The last equation, of course, is the easiest to read. That is why our goal is to end up with x isolated on one side of the equation. 2. Solve: 6a 1 7 5 4a 2 3 Example 2 Solve: 4a 1 5 5 2a 2 7 Solution Neither side can be simplified any further. What we have to do is get the variable terms (4a and 2a) on the same side of the equation. We can eliminate the variable term from the right side by adding 22a to both sides: 4a 1 5 5 2a 2 7 54a 1 (−2a) 1 5 5 2a 1 (−2a) 2 7 Add 22a to both sides Step 2 2a 1 5 5 27 Addition 2a 1 5 1 (−5) 5 27 1 (−5) Add 25 to both sides 2a 5 212 Addition Step 3 5 } 2 a } 5 } 2 12 } Divide by 2 2 2 a 5 26 Division 3. Solve: 5(x 2 2) 1 3 5 212 Example 3 Solve: 2(x 2 4) 1 5 5 211 Solution x 2 4: We begin by applying the distributive property to multiply 2 and Step 1 Step 2 Step 3 5 5 5 2(x 2 4) 1 5 5 211 2x 2 8 1 5 5 211 2x 2 3 5 211 2x 2 3 1 3 5 211 1 3 2x 5 28 Distributive property Addition Add 3 to both sides Addition } 2 x } 5 } 2 8 } Divide by 2 2 2 x 5 24 Division 4. Solve: 3(4x 2 5) 1 6 5 3x 1 9 Answers 2. 25 3. 21 4. 2 Example 4 Solve: 5(2x 2 4) 1 3 5 4x 2 5 Solution We apply the distributive property to multiply 5 and 2x 2 4. We then combine similar terms and solve as usual: Step 1 Step 2 Step 3 5 5(2x 2 4) 1 3 5 4x 2 5 10x 2 20 1 3 5 4x 2 5 10x 2 17 5 4x 2 5 10x 1 (−4x) 2 17 5 4x 1 (−4x) 2 5 5 6x 2 17 5 25 6x 2 17 1 17 5 25 1 17 6x 5 12 5 Distributive property Simplify the left side Add 24x to both sides Addition Add 17 to both sides Addition } 6 x } 5 } 1 2 } Divide by 6 6 6 x 5 2 Division
9.4 Linear Equations in One Variable 541 B Equations Involving Fractions We will now solve some equations that involve fractions. Because integers are usually easier to work with than fractions, we will begin each problem by clearing the equation we are trying to solve of all fractions. To do this, we will use the multiplication property of equality to multiply each side of the equation by the LCD for all fractions appearing in the equation. Here is an example. Example 5 Solve the equation } 2 x } 1 }6 x } 5 8. 5. Solve: } 3 x } 1 }6 x } 5 9 x x Solution The LCD for the fractions } } 2 and }6 } is 6. It has the property that both 2 and 6 divide it evenly. Therefore, if we multiply both sides of the equation by 6, we will be left with an equation that does not involve fractions. 6 1 } 2 x } 1 }6 x }2 5 6(8) Multiply each side by 6 6 1 } 2 x }2 1 6 1 } 6 x }2 5 6(8) Apply the distributive property 3x 1 x 5 48 Multiplication 4x 5 48 Combine similar terms x 5 12 Divide each side by 4 We could check our solution by substituting 12 for x in the original equation. If we do so, the result is a true statement. The solution is 12. As you can see from Example 5, the most important step in solving an equation that involves fractions is the first step. In that first step we multiply both sides of the equation by the LCD for all the fractions in the equation. After we have done so, the equation is clear of fractions because the LCD has the property that all the denominators divide it evenly. Example 6 Solve the equation 2x 1 } 1 2 } 5 }3 4 }. 6. Solve: 3x 1 } 1 4 } 5 }5 8 } Solution This time the LCD is 4. We begin by multiplying both sides of the equation by 4 to clear the equation of fractions. 4 1 2x 1 } 1 2 } 2 5 4 1 }3 4 } 2 4(2x) 1 4 1 } 1 2 } 2 5 4 1 }3 4 } 2 8x 1 2 5 3 8x 5 1 Multiply each side by the LCD, 4 Apply the distributive property Multiplication Add 22 to each side x 5 } 1 } Divide each side by 8 8 7. Solve: } 4 x } 1 3 5 }1 1 } 5 Example 7 Solve for x: } 3 x } 1 2 5 }1 }. (Assume x is not 0.) 2 Solution This time the LCD is 2x. Following the steps we used in Examples 5 and 6, we have 2x 1 } 3 x } 1 2 2 5 2x 1 }1 2 } 2 Multiply through by the LCD, 2x 2x 1 } 3 x } 2 1 2x(2) 5 2x 1 }1 2 } 2 Distributive property 6 1 4x 5 x Multiplication 6 5 23x Add 24x to each side 22 5 x Divide each side by 23 Answers 5. 18 6. } 1 } 7. 25 8
Page 1 and 2: Solving Equations and Inequalities
Page 3 and 4: The Distributive Property and Algeb
Page 5 and 6: 9.1 The Distributive Property and A
Page 7 and 8: 9.1 The Distributive Property and A
Page 9 and 10: 9.1 Problem Set 517 Problem Set 9.1
Page 11 and 12: 9.1 Problem Set 519 Find the value
Page 13 and 14: 9.1 Problem Set 521 91. Temperature
Page 15 and 16: Introduction . . . The Addition Pro
Page 17 and 18: 9.2 The Addition Property of Equali
Page 21 and 22: 9.2 Problem Set 529 Applying the Co
Page 23 and 24: The Multiplication Property of Equa
Page 25 and 26: 9.3 The Multiplication Property of
Page 29 and 30: 9.3 Problem Set 537 Find the value
Page 31: Introduction . . . Linear Equations
Page 37 and 38: 9.4 Problem Set 545 C Applying the
Page 39 and 40: Introduction . . . Applications As
Page 41 and 42: 9.5 Applications 549 Step 2 Assign
Page 43 and 44: 9.5 Applications 551 Example 4 The
Page 45 and 46: 9.5 Applications 553 Step 2 Assign
Page 49 and 50: 9.5 Problem Set 557 Coin Problems 3
Page 51 and 52: Introduction . . . Evaluating Formu
Page 53 and 54: 9.6 Evaluating Formulas 561 The rat
Page 57 and 58: 9.6 Problem Set 565 B Maximum Heart
Page 59 and 60: 9.6 Problem Set 567 Applying the Co
Page 61 and 62: Chapter 9 Summary Combining Similar
Page 63 and 64: Chapter 9 Review Simplify the expre
Page 65 and 66: Chapter 9 Cumulative Review 573 28.
Page 67 and 68: Chapter 9 Projects Solving Equation

Chapter 9 - XYZ Custom Plus

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