Chapter 7 - XYZ Custom Plus

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440<strong>Chapter</strong> 7 Solving EquationsThis general method of solving linear equations involves using the two propertiesdeveloped in Sections 7.2 and 7.3. We can add any number to both sides ofan equation or multiply (or divide) both sides by the same nonzero number andalways be sure we have not changed the solution to the equation. The equationsmay change in form, but the solution to the equation stays the same. Lookingback to Example 1, we can see that each equation looks a little different from thepreceding one. What is interesting, and useful, is that each of the equations saysthe same thing about x. They all say that x is −5. The last equation, of course, isthe easiest to read. That is why our goal is to end up with x isolated on one sideof the equation.2. Solve: 6a + 7 = 4a − 3Example 2Solve: 4a + 5 = 2a − 7Solution Neither side can be simplified any further. What we have to do is getthe variable terms (4a and 2a) on the same side of the equation. We can eliminatethe variable term from the right side by adding −2a to both sides:4a + 5 = 2a − 7{4a + (−2a) + 5 = 2a + (−2a) − 7 Add −2a to both sidesStep 2 2a + 5 = −7 Addition2a + 5 + (−5) = −7 + (−5) Add −5 to both sides2a = −12AdditionStep 3{_2a = −12_2 2Divide by 2a = −6Division3. Solve: 5(x − 2) + 3 = −12Example 3Solve: 2(x − 4) + 5 = −11Solutionx − 4:We begin by applying the distributive property to multiply 2 andStep 1Step 2Step 3{{{2(x − 4) + 5 = −112x − 8 + 5 = −11 Distributive property2x − 3 = −11 Addition2x − 3 + 3 = −11 + 3 Add 3 to both sides2x = −8 Addition_2x = −8_2 2Divide by 2x = −4 Division4. Solve: 3(4x − 5) + 6 = 3x + 9Answers2. −5 3. −1 4. 2Example 4Solve: 5(2x − 4) + 3 = 4x − 5Solution We apply the distributive property to multiply 5 and 2x − 4. We thencombine similar terms and solve as usual:Step 1Step 2Step 3{5(2x − 4) + 3 = 4x − 510x − 20 + 3 = 4x − 5Distributive property10x − 17 = 4x − 5Simplify the left side10x + (−4x) − 17 = 4x + (−4x) − 5 Add −4x to both sides{6x − 17 = −5Addition6x − 17 + 17 = −5 + 17 Add 17 to both sides6x = 12Addition{_6x = 12_6 6Divide by 6x = 2Division
7.4 Linear Equations in One Variable441B Equations Involving FractionsWe will now solve some equations that involve fractions. Because integers areusually easier to work with than fractions, we will begin each problem by clearingthe equation we are trying to solve of all fractions. To do this, we will use themultiplication property of equality to multiply each side of the equation by theLCD for all fractions appearing in the equation. Here is an example.Example 5Solve the equation x _2 + x _= 8.6Solution The LCD for the fractions _x 2 and x _is 6. It has the property that both 26and 6 divide it evenly. Therefore, if we multiply both sides of the equation by 6, wewill be left with an equation that does not involve fractions.6 x _2 + x _6 = 6(8) Multiply each side by 66 x _2 + 6 x _6 = 6(8) Apply the distributive property3x + x = 48 Multiplication4x = 48 Combine similar termsx = 12 Divide each side by 45. Solve: x _3 + x _6 = 9Instructor NoteExamples 5–7 are different from theprevious examples in that we usethe multiplication property first toclear the equations of fractions. Isometimes take two days to coverthis section and wait until the secondday to cover these problems.We could check our solution by substituting 12 for x in the original equation. If wedo 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 equationthat involves fractions is the first step. In that first step we multiply both sides ofthe equation by the LCD for all the fractions in the equation. After we have doneso, the equation is clear of fractions because the LCD has the property that all thedenominators divide it evenly.Example 6Solve the equation 2x + _1 2 = 3 _4 .Solution This time the LCD is 4. We begin by multiplying both sides of theequation by 4 to clear the equation of fractions.6. Solve: 3x + 1 _4 = 5 _8 4 2x + 1 _2 = 4 3 _4 Multiply each side by the LCD, 44(2x) + 4 1 _2 = 4 3 _4 Apply the distributive property8x + 2 = 38x = 1MultiplicationAdd −2 to each sideExample 7Solve for x: 3 _x = _1 Divide each side by 88x + 2 = 1 _. (Assume x is not 0.)2Solution This time the LCD is 2x. Following the steps we used in Examples 5and 6, we have2x 3 _x + 2 = 2x 1 _2 Multiply through by the LCD, 2x2x 3 _x + 2x(2) = 2x 1 _2 Distributive property6 + 4x = x Multiplication6 = −3x Add −4x to each side−2 = xDivide each side by −37. Solve: _4 + 3 = 11_x 5 Instructor NoteThe equation shown in Example 7is not a linear equation in one variable.Sometimes I point this out tomy classes. If I were doing this kindof problem in my algebra classes,I would mention the possibility ofobtaining extraneous solutions.However, I never mention this fact inprealgebra or basic math.Answers5. 18 6. _1 7. −58
Page 1 and 2: Solving Equations7IntroductionImage
Page 3 and 4: The Distributive Propertyand Algebr
Page 5 and 6: 7.1 The Distributive Property and A
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Page 9 and 10: 7.1 Problem Set417Problem Set 7.1A
Page 11 and 12: 7.1 Problem Set419Find the value of
Page 13 and 14: 7.1 Problem Set42191. Temperature a
Page 15 and 16: Introduction . . .The Addition Prop
Page 17 and 18: 7.2 The Addition Property of Equali
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Page 23 and 24: The Multiplication Property of Equa
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Page 29 and 30: 7.3 Problem Set437Find the value of
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Page 49 and 50: 7.5 Problem Set457Coin Problems [Ex
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Page 74 and 75: Chapter 7 TestSimplify each express
Page 76: RESEARCH PROJECTAlgebraic Symbolism

Chapter 7 - XYZ Custom Plus

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