. Using the Principles Together Applying Both Principles a Combining Like Terms a Clearing Fractions and Decimals a Contradictions and Identities EXAMPLE Solve: An important strategy for solving new problems is to find a way to make a new problem look like a problem that we already know how to solve. This is precisely the approach taken in this section. You will find that the last steps of the examples in this section are nearly identical to the steps used for solving the equations of Section.. What is new in this section appears in the early steps of each example. Applying Both Principles The addition and multiplication principles, along with the laws discussed in Chapter, are our tools for solving equations. In this section, we will find that the sequence and manner in which these tools are used is especially important. + x = 7. Were we to evaluate + x, the rules for the order of operations direct us to first multiply by and then add. Because of this, we can isolate x and then x by reversing these operations: We first subtract from both sides and then divide both sides by. Our goal is an equivalent equation of the form x = a. Isolate the x-term. Isolate x. + x = 7 + x - = 7 - + - + x = x = x = x = Using the addition principle: subtracting from both sides (adding -) Using a commutative law. Try to perform this step mentally. Using the multiplication principle: dividing both sides by (multiplying by ) Check: + x = 7 + # 7 + 7 =? 7 We use the rules for order of operations: Find the product, #, and then add. The solution is. 7
8 CHAPTER EQUATIONS, INEQUALITIES, AND PROBLEM SOLVING EXAMPLE Solve: x - 7 =. In x - 7, we multiply first and then subtract. To reverse these steps, we first add 7 and then either divide by or multiply by. x - 7 = Adding 7 to both sides x - 7 + 7 = + 7 STUDY SKILLS Use the Answer Section Carefully When using the answers listed at the back of this book, try not to work backward from the answer. If you frequently require two or more attempts to answer an exercise correctly, you probably need to work more carefully and/or reread the section preceding the exercise set. Remember that on quizzes and tests you have only one attempt per problem and no answer section to consult. Check: The solution is. x = 8 # x = # 8 # x = # # x = x - 7 = # - 7 8-7 =? M Multiplying both sides by 7 EXAMPLE Solve: - t =. We have - t = - t - = - + -t + - = - r + - + -t = - -t = - --t = -- t =. Subtracting from both sides Try to do these steps mentally. Try to go directly to this step. Multiplying both sides by - (Dividing by - would also work.) Check: - t = -? = The solution is. 9 As our skills improve, certain steps can be streamlined.
. USING THE PRINCIPLES TOGETHER 87 EXAMPLE Solve:. - 7.y = -8.8. We have. - 7.y = -8.8. - 7.y -. = -8.8 -. -7.y = -.8-7.y -7. = -.8-7. y =.. Subtracting. from both sides Dividing both sides by -7. Check:. - 7.y = -8.8. - 7.. - 8.8. -.8-8.8? = -8.8 The solution is.. Combining Like Terms If like terms appear on the same side of an equation, we combine them and then solve. Should like terms appear on both sides of an equation, we can use the addition principle to rewrite all like terms on one side. EXAMPLE Solve. a) x + x = - b) -x + = -8x + c) x + - 7x = - x + 7 d) - x + = x - - a) x + x = - 7x = - 7x 7 = - 7 x = - Combining like terms Dividing both sides by 7 The check is left to the student. The solution is -. b) To solve -x + = -8x +, we must first write only variable terms on one side and only constant terms on the other. This can be done by subtracting from both sides, to get all constant terms on the right, and adding 8x to both sides, to get all variable terms on the left. Isolate variable terms on one side and constant terms on the other side. -x + = -8x + -x + 8x + = -8x + 8x + Adding 8x to both sides 7 x + = 7 x + - = - Subtracting from both sides 7 x = Combining like terms 7x 7 = 7 Dividing both sides by 7 x = 7 The check is left to the student. The solution is 7.
88 CHAPTER EQUATIONS, INEQUALITIES, AND PROBLEM SOLVING TECHNOLOGY CONNECTION Most graphing calculators have a TABLE feature that lists the value of a variable expression for different choices of x. For example, to evaluate x + - 7x for x =,,, Á, we first use E to enter x + - 7x as y. We then use F j to specify TblStart =, Tbl =, and select AUTO twice. By pressing Fn, we can generate a table in which the value of x + - 7x is listed for values of x starting at and increasing by ones. X X Y. Create the above table on your graphing calculator. Scroll up and down to extend the table.. Enter - x + 7 as y. Your table should now have three columns.. For what x-value is y the same as y? Compare this with the solution of Example (c). Is this a reliable way to solve equations? Why or why not? c) x + - 7x = - x + 7 Check: -x + = 7 - x -x + + x = 7 - x + x + x = 7 x = x = The student can confirm that checks and is the solution. d) - x + = x - - - x - = x - - Using the distributive law. This is now similar to part (c) above. -x - = x - 7 Combining like terms on each side -x - + 7 = x Adding 7 and x to both sides. This isolates r - + 7 = x + x the x-terms on one side and the constant terms on the other. - = 8x - 8 x = x + - 7x = - x + 7 # + - 7 # - # + 7 + - 8? - + 7 = = 8x 8 Clearing Fractions and Decimals Combining like terms within each side Adding x to both sides. This is identical to Example. Subtracting from both sides Dividing both sides by Dividing both sides by 8 - = x This is equivalent to x = -. The student can confirm that - checks and is the solution. Equations are generally easier to solve when they do not contain fractions or decimals. The multiplication principle can be used to clear fractions or decimals, as shown here. 9. X Y. X X X Y Y 7 9 7. ; not reliable because, depending on the choice of Tbl, it is easy to scroll past a solution without realizing it. Clearing Fractions x + = A x + B = # x + = Clearing Decimals.x + 7 =..x + 7 = #. x + 7 = In each case, the resulting equation is equivalent to the original equation, but easier to solve. The easiest way to clear an equation of fractions is to multiply both sides of the equation by the smallest, or least, common denominator of the fractions in the equation.
. USING THE PRINCIPLES TOGETHER 89 EXAMPLE Solve: (a) x - = x; (b) x + = 8. a) We multiply both sides by, the least common denominator of and. a x - b = # x # x - # = # x Multiplying both sides by CAUTION! Be sure the distributive law is used to multiply all the terms by. x - = x. Note that the fractions are cleared: # and # =, # =, =. - = 8x Subtracting x from both sides - Dividing both sides by 8 8 = 8x 8-8 = x The student can confirm that - 8 checks and is the solution. b) To solve x + = 8, we can multiply both sides by (or divide by ) to undo the multiplication by on the left side. # x + = # 8 x + = x = 8 x = Multiplying both sides by # ; = and = Subtracting from both sides Dividing both sides by The student can confirm that checks and is the solution. # 8 9 STUDENT NOTES EXAMPLE 7 Solve: Compare the steps of Examples and 7. Note that although the two approaches differ, they yield the same solution. Whenever you can use two approaches to solve a problem, try to do so, both as a check and as a valuable learning experience. To clear an equation of decimals, we count the greatest number of decimal places in any one number. If the greatest number of decimal places is, we multiply both sides by ; if it is, we multiply by ; and so on. This procedure is the same as multiplying by the least common denominator after converting the decimals to fractions.. - 7.y = -8.8. The greatest number of decimal places in any one number is two. Multiplying by will clear all decimals.. - 7.y = -8.8. - 7.y = - 8.8-7y = -88-7y = -88 - -7y = -8 y = -8-7 Multiplying both sides by Using the distributive law Subtracting from both sides Combining like terms Dividing both sides by -7 y =. In Example, the same solution was found without clearing decimals. Finding the same answer in two ways is a good check. The solution is.. 7
9 CHAPTER EQUATIONS, INEQUALITIES, AND PROBLEM SOLVING An Equation-Solving Procedure. Use the multiplication principle to clear any fractions or decimals. (This is optional, but can ease computations. See Examples and 7.). If necessary, use the distributive law to remove parentheses. Then combine like terms on each side. (See Example.). Use the addition principle, as needed, to isolate all variable terms on one side. Then combine like terms. (See Examples 7.). Multiply or divide to solve for the variable, using the multiplication principle. (See Examples 7.). Check all possible solutions in the original equation. (See Examples.) Contradictions and Identities All of the equations we have examined so far had a solution. Equations that are true for some values (solutions), but not for others, are called conditional equations. Equations that have no solution, such as x + = x +, are called contradictions. If, when solving an equation, we obtain an equation that is false for any value of x, the equation has no solution. EXAMPLE 8 Solve: x - = x - +. x - = x - + x - = x - + Using the distributive law x - = x - Combining like terms -x + x - = -x + x - Using the addition principle - = - Since the original equation is equivalent to - = -, which is false regardless of the choice of x, the original equation has no solution. There is no solution of x - = (x - ) +. The equation is a contradiction. It is never true. Some equations, like x + = x +, are true for all replacements. Such an equation is called an identity. EXAMPLE 9 Solve: x + 7 = 7x + - x. x + 7 = 7x + - x x + 7 = 7x + 7 - x Using the distributive law x + 7 = x + 7 Combining like terms The equation x + 7 = x + 7 is true regardless of the replacement for x, so all real numbers are solutions. Note that x + 7 = x + 7 is equivalent to x = x, 7 = 7, or =. All real numbers are solutions and the equation is an identity.
. USING THE PRINCIPLES TOGETHER 9. EXERCISE SET For Extra Help i Concept Reinforcement In each of Exercises, match the equation with an equivalent equation from the column on the right that could be the next step in finding a solution.. (c) x - = 7 a) x - =. (e) x + x = b) x + x =. (a) x - = c) x = 7 +. (f) 7x = 9 d) 8x + x = +. (b) x = - x e) 9x =. (d) 8x - = - x f ) x = 9 7 Solve and check. Label any contradictions or identities. 7. x + 9 = 8 8. x - = 8 9. z + = 7 7. z + = 7. 7t - 8 = 7... x - 9 =.. 7. 8. 9... -z - 8 = - 9. -7x - = -9... -.n =..8. - 7x =. x - = x - 9 = -7 = 9t + 8-9 = + t - t = 9 - t = 7 - x = 99 7 8z + = - x + = - -. +.n =.9 7. t - = 8 8. t - = 9 - -7 - - -8 - -.8 Aha! 9.. + 7 x = - + x = - - -8.. a - 8a - 7-8 = - = - -. x = x + x All real numbers; identity. -z + 8z = 9. x - = x -. x - x = x + x 7. All real numbers; identity - y = - y - 8. x - = 7 + x 9. 7a - =.. = x +. 9 = x -. + m - = 8. + m - 8 = 7. x + = x - No solution; contradiction. x - 7 = x - + x No solution; contradiction 7. 8. 9. x + = x +. y + = y +. - x = x - 7x +.. 7 + x - = x + - x. + x - 7 = x - - x. y - + y + = y + - y. y - + y = 7y + 8 - y 7 7. + 7x = 7x + No solution; contradiction 8. t + + t = + t All real numbers; identity 9. ( - t) = r + 8 = r + b - = 7b + - - x = x - x + 9 - x - = 7 - - -
9 CHAPTER EQUATIONS, INEQUALITIES, AND PROBLEM SOLVING. d + = 7d - 7 9. Tyla solves - t = (Example ) by adding t - to both sides. Is this approach preferable to. 7x - = x - -8 the one used in Example? Why or why not?. t + + 8 = t - + - 9. Why must the rules for the order of operations be. t + - = t - t + understood before solving the equations in this section?. x - x + = x - + 8. 7 - x - = 7x - + x All real numbers; identity. x - 7 = x - + x No solution; contradiction 7. 9 - x + = x + + x 8. - c + = c + + c Clear fractions or decimals, solve, and check. 9. + t = 7. - - + x = - - 7. + t = t - 7. + m = m - - 7. x + = + x - 7. - y = 9 - y + - 7..x +. =. - 8.x - 7..9 -.z =. -.z.8 77. 78..7t + 8 -.t =.t -. + 8 79. - x - x = x + 7 8. y + 8 y = + y 7 8. x - = 7 8. x - = 7 9 8. t - = 8 8. x + = - - 8. A x - B = - 8. - A 7 8 - xb - 8 = 8 87..7x + =. - x - 88..7 +.t =.9t -.9.9x - 8 = - x +. - 89. a + a - = a + - a + 9..8 - b - =. + - b -7. - Aha! Skill Review To prepare for Section., review evaluating algebraic expressions (Section.8). Evaluate. [.8] 9. - a, for a = 9., # t, for t = 9. 7x - x, for x = - 9. t8 - t, for t = - Synthesis -7 - -8 97. What procedure would you use to solve an equation like.x + 7 = -.8 + x? Could your procedure be streamlined? If so, how? 98. Dave is determined to solve x + = - by first using the multiplication principle to eliminate the. How should he proceed and why? Solve. Label any contradictions or identities. 99. 8.x -.. -.7x = -.x + 9. 99 or.97..8 + 9.x -.8 =.9x +.8 - x.. -x - + = - x - x No solution; contradiction. = t - - - -7t -. xx + - x + x - = 9 - x - x. xx - = xx + - x + x -. 9 - x = - x - - x. 7-8, -x = x 7. - x + = x - 9 7 x + 8. + - = + x 9. 9 - -x - = x + All real numbers; identity. -9t + = - 9t - 8, + No solution; contradiction -
. FORMULAS 9 Step-by-Step Solutions Focus: Solving linear equations Time: minutes Group size: In general, there is more than one correct sequence of steps for solving an equation. This makes it important that you write your steps clearly and logically so that others can follow your approach. ACTIVITY. Each group member should select a different one of the following equations and, on a fresh sheet of paper, perform the first step of the solution. - x - = 7x + - x - 7x - x - = x + x - 7 + 9 x - 7 + x - + x = - 9-x - 9. Pass the papers around so that the second and third steps of each solution are performed by the other two group members. Before writing, make sure that the previous step is correct. If a mistake is discovered, return the problem to the person who made the mistake for repairs. Continue passing the problems around until all equations have been solved.. Each group should reach a consensus on what the three solutions are and then compare their answers to those of other groups.