Section 2.1 Solving Linear Equations Involving Two Operations

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Section 2.1 Solving Linear Equations Involving Two OperationsObjectivesIn this section, you will learn to: To successfully complete this section,you need to understand:Define the term equation.Apply the Addition Property of Equality.Apply the Multiplication Property of Equality.Isolate variables.Solve equations involving one operation.Solve equations involving two operations. The reciprocal (R.3)Additive identity (1.1)Multiplicative identity (1.1)Adding and subtracting realnumbers (1.3 and 1.4)Multiplying and dividing realnumbers (1.5)The main operation (1.10) INTRODUCTIONThe equal sign ( ) is used primarily two different ways in mathematics. In Chapter 1, we usedcontinued equal signs when simplifying an expression to indicate that one step is equivalent the nextstep.This is illustrated in Example 1 from Section 1.7, simplifying an expression using the order ofoperations:(4 – 16) 22Subtract first. -12 22Apply the exponent -12 4Divide. -348We also sometimes see continued equal signs when simplifying a fraction, such as 72 . We might needto divide out some smaller common factors before it is completely simplified:4872 Simplify by afactor of 22436 46 Simplify by afactor of 62 3Simplify by afactor of 2An equal sign is also used to indicate the equivalence of two expressions that may or may not containvariables. This equivalence is called an equation. The equations that we see in this text includenumbers and variables, and we are required to find the value of the variable that makes the equationtrue. These equations are often true for only one or two numbers.2.1 Solving Equations Involving Two Operations159 Robert H. Prior, 2014Copying is prohibited.

For example, the equation 5 x 7 is true only when x 2.We can verify this by replacing x with 2 and evaluating each side:5 (2) 77 7 True!The equation x 7 2x 1 is true only when x 6.(6) 7 2(6) 1We can verify this by replacing each x with 6 and evaluating each side:13 12 113 13 True!Some equations, called identities, are true for all numbers.For example, the equation x 3 x 3 is true whenx 2:(2) 3 (2) 35 5 True x 10:x -7:(10) 3 (10) 3(-7) 3 (-7) 313 13 True -4 -4 True Because x 3 x 3 is true for all values of x, the equation is an identity.Equations that have only one or two solutions are called conditional equations. The equationsintroduced at the beginning of this section, 5 x 7 and x 7 2x 1 , are conditional equationsbecause they each have only one solution.The discussion in this section is about solving conditional equations by finding the value of the variablethat makes an equation true.PREPARING TO SOLVE LINEAR EQUATIONSThere is a class of conditional equations called linear equations. In a linear equation, the highest powerof any variable is 1. Here are some examples of linear equations:(a)3x 5 23(b)-9 15 – 2y(c)6w 11 -4w – 9To be well prepared to solve linear equations, we must be familiar with the following ideas fromChapter 1.2.1 Solving Equations Involving Two Operations160 Robert H. Prior, 2014Copying is prohibited.

SectionPropertyExamplea 0 a and 0 a a9 0 91.The additive identity is 0.1.12.The multiplicative identity is 1.1.13.For addition: the sum of anumber and its opposite is 0.1.3a (-a) 04 (-4) 04.For multiplication: the product ofa number and its reciprocal is 1.1.5a bb · a 1344 · 3b·1 band 1 · b b1 · 15 15-2-55 · 2If a number is negative, itsreciprocal is also negative.5.The sign of a term.1.8 1 1The sign in front of a term belongs In the expression3w – 4, the secondto that term.term is -4.6.If a variable is in the numeratorof a fraction, the coefficient ofthe variable is the whole fraction,not just the numerator.1.8axac c x7.The main operation.1.10The main operation of anexpression is the last operation tobe applied, according to the orderof operations.3x34 4 xThe coefficient of x3is 4 .In the expression6y 5, the mainoperation is addition.THE VOCABULARY OF EQUATIONSAn equation is a mathematical sentence in whichone expression another expressionEach expression has one or more terms, and each term is either a constant orcontains a variable. Any term that contains a variable is called a variableterm.Note:If the highest exponent of the variable is 1, such as x1 or x, then the equation isa linear equation. A linear equation typically has only one solution.At right are some examples of linear equations. Noticethat the variable terms can be on the left side, on the rightside, or on both sides:2.1 Solving Equations Involving Two Operations161Linear is pronouncedlih -nee-er.a)n 3 -8b)18 6vc)6x 5 xd)64 y 178 – 2y Robert H. Prior, 2014Copying is prohibited.

Each of these equations is a conditional equation and is true for only one value. To solve an equation isto find the one number that makes the equation true; this number is called the solution.Throughout this section, and the next two, we look at avariety of strategies for solving linear equations. If amistake is made during the solving process, then we canstill get an answer, but it won’t be the correct answer. Inother words, it won’t be the solution.After we solve an equation and arrive at an answer, wecan verify if the answer is, indeed, the solution byplacing the answer into the equation, as demonstratedin Example 1.Example 1:Note:In the process of determining if areplacement value is a solution, we use the?symbol until we can verify whether ornot the last statement is true.For each equation, a student solved the equation and arrived at the answer shown.Verify whether the answer is a solution or not.a)12 8 – 4x; x 5b)3w – 2 2w – 5; w -3Procedure:For each equation, replace the variable with the given answer and evaluate eachside. If the two sides are equal, then the answer makes the equation true, and it isthe solution. If they are not equal, then the answer is not the solution.Answer:a)12 8 – 4xReplace x with 5.?12 8 – 4(5)?12 8 – 2012 -12 Falseb)3w – 2 2w – 5No, 5 is not the solution.Replace each w with -3.?3(-3) – 2 2(-3) – 5?-9 – 2 -6 – 5-11 -11 True 2.1 Solving Equations Involving Two Operations162Yes, -3 is the solution. Robert H. Prior, 2014Copying is prohibited.

You Try It 1a)For each equation, a student solved the equation and arrived at the answer shown.Verify whether or not the answer is a solution. Use Example 1 as a guide.15 – 5m 30; m 3b)18 5p – 2; p 4c)6v 3 4v – 1; v -2BALANCING EQUATIONSThe strategy for solving a linear equation is to isolate the variable. To isolate the variable means to getthe variable alone on one side of the equation and a single number on the other, such as x -5 or 3 y.For example, to isolate x in the equation x 3 8, we must remove the constant, 3, fromthe left side. This will leave x alone (isolated) on that side.In the solving process, as we work our way toward isolating the variable, we must often write severalequations, each one a bit simpler than the previous one. This is called reducing the equation. Eachequation in this series of reduced equations is equivalent to the others, and these equivalent equationsall have the same solution.To create simpler, reduced, equations we make changes to the expressions in the equation. To maintainthe solution, though, we must keep the equations balanced by performing the same operation—addition,subtraction, multiplication, or division—on each side of the equation.This idea of keeping an equation balanced can be visualized using a scale.Here we see the equation x 3 8. The leftside, x 3, and the right side, 8, are equal to eachother, so they are in balance.To isolate the variable, we might want to removethe 3 from the left side by subtracting it away.However, if we subtract 3 from only one side,then that side becomes lighter and the scale (andthe equation) is out of balance.2.1 Solving Equations Involving Two Operations163x 38x 3–38 Robert H. Prior, 2014Copying is prohibited.

If we subtract 3 from each side of the scale at thesame time, then the scale remains in balance andwe maintain a balanced equation.x 3–38–3In the example above, subtracting 3 from each side creates a constant term of 0 on the left side, therebyisolating the variable:x 3 8Subtract 3 from each side.x 3–3 8–3Simplify each sidex 0 5Simplify the left side.x 5The solution is 5.Although the solution itself is 5, it is common to state it as x 5.In general, to isolate the variable or the variable term, we create 0 by adding the opposite of any term wewant to remove, or clear. The property that allows us do this is the Addition Property of Equality:The Addition Property of EqualityWe may add any number or term, c, to each side of an equation,and the equality remains:Ifthena b,a c b cWhen clearing the constant, it is important to keep inmind the notion that the sign in front of the termbelongs to that term.This helps us to identify the sign of the constant term,and we can then add its opposite to each side of theequation.2.1 Solving Equations Involving Two Operations164Note:As we see in Example 2b), it is commonto write the final equation with thevariable on the left and the constant onthe right.This will be most beneficial in Section2.7, Solving Inequalities, so we should getin the habit now of placing the variableon the left side in the answer. Robert H. Prior, 2014Copying is prohibited.

Example 2:Solve each equation by clearing the constant. Also, verify the solution.28x – 6 4b) -9 4 yc) w – 5 5a)Procedure:Apply the Addition Property of Equality. Recognize the constant that must becleared, and add its opposite to isolate the variable. Verify the solution by using itas a replacement value for the variable in the original equation.Answer:a)x – 6 4x – 6 6 4 6Clear the constant -6 byadding 6 to each side.Verify the solution, 10:x–6 4?10 – 6 Simplify each side; -6 6 0.44 4b)x 0 10x 0 is just x.x 10Verify x 10-9 4 y-9 (-4) 4 (-4) y Clear the constant 4 byadding -4 to each side.Verify the solution, -13:-9 4 y?-9 Simplify each side; 4 (-4) 0.4 (-13)-9 -9-13 0 y0 y is just y.-13 yWrite the equation with thevariable on the left side.y -13c)28w – 5 5 Verify y -13 2Clear the constant - 5 by2adding 5 to each side.10Verify the solution, 2 or 5 :28w – 5 52282w – 5 5 5 510 2 ? 5 –5Simplify each side;22- 5 5 0.10w 0 5w 22.1 Solving Equations Involving Two Operations85858 5 Simplify the right side.10To verify the solution, use w 5165 Robert H. Prior, 2014Copying is prohibited.

Look back at Example 2 and notice these three things:1. Each step in the solving process is directly below the preceding step.2. The equal signs are lined up, one below the other.3. We isolated the variable by creating 0, the additive identity, on the side of the variable term.You Try It 2Solve each equation by clearing the constant. Also, verify the solution. Use Example2 as a guide.a)x 7 20b)-19 b – 12c)w – 9 -9d)837 m 3ANOTHER ADDITION TECHNIQUEAnother technique for clearing a constant is to add the constant’s opposite to each side below theconstant term. For instance, in Example 2a), in the equation x – 6 4, we cleared the constant, -6, byadding it’s opposite, 6, to each side: x – 6 6 4 6.The other technique, though, allows us to write the 6 directly below the constants on each side:x – 6 4 6 6Notice the equal sign between 6’sthat we are adding to each side.x 0 10Notice that the equal signs are allaligned, one below the other.x – 6 4 6 6x 0 10The equal sign between the 6’s is required to show that we are adding 6 to each side. This isespecially important when the equations have more terms, as in Section 2.2.Example 3 demonstrates this technique, “adding the opposite” below the term we are intending to clear.2.1 Solving Equations Involving Two Operations166 Robert H. Prior, 2014Copying is prohibited.

Example 3:Solve each of these equations by clearing the constant.a)w – 8 -3Procedure:b)We must recognize the constant that needs to be cleared and add its opposite toeach side.a)w – 8 8w 0wb) -3 8 5 5To clear the -8 we’ll add 8 to each side.-5 m 7To clear the 7 we’ll add -7 to each side.-7 Notice that we write -7 -7Notice that we write 8 8We still want to get the identity for addition, 0.Check the answer: 5 – 8 -3 is true.-7-12 m 0-12 mm -12You Try It 3:-5 m 7We still want to get the identity for addition, 0.Check the answer: -12 7 -5 is true.Solve each equation by clearing the constant. Use Example 3 as a guide.a)p 4 15b)-3 w 7c)11 y – 9d)v – 8 -10SOLVING EQUATIONS INVOLVING MULTIPLICATIONSome linear equations involve multiplication but no addition or subtraction, such as32 y 18and5x -35.In each of these equations, the coefficient of the variable is not 1. To isolate the variable, we mustcreate a coefficient of 1. We do so by multiplying each side of the equation by the reciprocal of thecoefficient.2.1 Solving Equations Involving Two Operations167 Robert H. Prior, 2014Copying is prohibited.

3For example, in the equation 2 y 18, we can create a coefficient of 1 by multiplying each side by233 (the reciprocal of the coefficient, 2 ) :32 y 1823Multiply each side by 3 , which is the reciprocal of 2 .Verify:32 y 182323 · 2 y 3 · 181 · y 12y 122Simplify. 332182612· 2 1; 3 · 1 1 · 1 1 12.1·y y3 12 ? 2 · 1183 6 ? 1 ·11812To verify the answer, use y 1 .18 18 The property that allows us to multiply each side of the equation by the same number is theMultiplication Property of Equality.The Multiplication Property of EqualityEach side of an equation may be multiplied by any non-zero number, c,and the equality remains:Ifthena b,c · a c · b, c 0If the coefficient is an integer, as in 5x -35, then we have two options:1.We can multiply each side of the equation1by the reciprocal of 5, which is 5 .2.We can divide each side by thecoefficient, 5:1Multiply each side by 5 .5x -35Divide each side by 5.1 -35Simplify. 5 · 1 -75x-355 5Simplify each fraction.1 · x -71x -7x has a coefficient of 1.x -7x -75x -35151 -355 · 1 x 5 · 12.1 Solving Equations Involving Two Operations168 Robert H. Prior, 2014Copying is prohibited.

We get the same result using either technique. This example shows us that we can divide each side bythe same number.The Division Property of EqualityWe may divide each side of an equation by any non-zero number, c, andequality remains:Ifa b,abc c , c 0thenThis is especially helpful when the coefficient is an integer. In Example 4, there are a variety ofcoefficients, so read each step carefully to see what the best strategy is for each equation.Example 4:a)Solve each equation by clearing the coefficient. Also, verify the solution.-4x 32Procedure:b)-515 8 yc)5w2 -20d)18 12pApply the Multiplication Property of Equality. Recognize the coefficient thatmust be cleared, and either multiply each side by its reciprocal or divide by thecoefficient directly. Verify the solution by using it as a replacement value for thevariable in the original equation.Answer:a)-4x 32The coefficient is an integer, -4,so we divide each side by -4.-4x32-4 -4Simplify each fraction.Verify the solution, -8:-4x 32?-4(-8) 3232 32x -82.1 Solving Equations Involving Two OperationsVerify x -8 169 Robert H. Prior, 2014Copying is prohibited.

b)-515 8 y-8 15-8 -55 · 1 5 · 8 y-24 1yy -24c)5w2 -20-5The coefficient is a fraction, 8 , so we-8multiply each side by its reciprocal, 5 .Verify the solution, -24:15 -58 ySimplify each side.?15 -58 · (-24)-8 15-8 35 · 1 1 · 1 -24.15 -5 -248 · 1Write the equation with thevariable on the left.15 -5 -31 · 1Verify y -24 5First write the left side as 2 w.15 15Verify the solution, -8:5w252 w -20252 -205 · 2 w 5 · 15The coefficient is a fraction, 2, so we2multiply each side by its reciprocal, 5.Simplify each side. -205(-8) ? 2-20-40 ? 2-20 -20 -202 -202 -45 · 1 1 · 1 -8.1w -8d) w -8Verify w -818 12pThe coefficient is an integer, 12,so we divide each side by 12.3Verify the solution, 2 :18 12p1812p12 1232 p3p 22.1 Solving Equations Involving Two OperationsSimplify each fraction.?18 312 · 218312 simplifies by a factor of 6 to 2 .?18 12 31 ·2Write the equation with thevariable on the left.?18 6 31 ·13Verify p 2 17018 18 Robert H. Prior, 2014Copying is prohibited.

Look back at Example 4 and notice three things:1.Each step in the solving process is directly below the preceding step.2.3.The equal signs are lined up, one below the other.We created a coefficient of 1 (the multiplicative identity) to isolate the variable.When the coefficient is -1, we can clear it by either dividing or multiplying each side by -1.Example 5:Solve -x 7 byProcedure:The coefficient is -1.Answer:a)(a) dividing by -1-x 7 Divide each side by -1.-1x7-1 -1x -7You Try It 4-x 7 Multiply each side by -1.-1 · (-1x) -1 · 7 We get the same solution. x -7Solve each of these equations by clearing the coefficient. Verify the solution. UseExamples 4 and 5 as guides.a)56 v -30b)d)-w -14e)Remember:b)(b) multiplying by -1.-20 -5x-21p 6c)-3y12 4b)8 -xWhenever we use one of the properties of equality to clear a constant or coefficient,we must keep the equation balanced by modifying each side in the same way.This is sometimes stated as, “Whatever you do to one side, you must do to theother.”2.1 Solving Equations Involving Two Operations171 Robert H. Prior, 2014Copying is prohibited.

SOLVING LINEAR EQUATIONS IN STANDARD FORMIn the linear equations we have solved so far in this section, we had to clear either a constant or acoefficient to isolate the variable. We now turn our attention to equations that require us to clear both aconstant and a coefficient, such as 3x 5 -7. Equations of this type are considered to be in standardform:The standard form of a linear equation isax b c or c ax b,where x is a variable and a, b, and c are constants, a 0.In the standard form of a linear equation, a cannot be 0, but either b or c (or both) may be 0.Think About It 1In the standard form of a linear equation, why is it important that a 0?Because an equation in standard form involves two operations, the question becomes, “To isolate thevariable, which number should we clear first, the constant or the coefficient?”For example, in the equation 3x 5 -7, should we first clear the constant, 5, by adding -5 to eachside, or should we first clear the coefficient, 3, by dividing each side by 3?There are two reasons we must clear the constant first:1.Because the goal is to isolate the variable, we must prepare this equation—and others like it—byfirst isolating the variable term. This means that we must clear the constant first and reduce theequation.2.In any equation involving more than one operation, we must first clear the main operation. Forexample, because 3x 5 is a sum (the main operation is addition), we must clear the sum byadding the opposite of the constant.In either case, once the constant has been cleared, we have an equation that can be solved by clearing thecoefficient.2.1 Solving Equations Involving Two Operations172 Robert H. Prior, 2014Copying is prohibited.

Example 5:Solve each equation, and verify the solution.a)Procedure:3x 5 -7b)-15 6 – 7yFirst isolate the variable term by clearing the constant.Answer:a)3x 5 -7The constant term is 5, so add -5 to each side.-5 -53x 0 -12Simplify the left side.3x -123x-123 3x -4Clear the coefficient by dividing each side by 3.3xSimplify each side. 3 simplifies to 1x, or just x.Verify x -4 ?3(-4) 5 -7?-12 5 -7-7 -7b)-15 6 – 7y-6 -6-21 0 – 7yThe constant term is 6, so add -6 to each side.-21 -7yClear the coefficient by dividing each side by -7.-21-7y-7 -7-7ySimplify each side. -7 simplifies to 1y, or just y.0 – 7y is -7y.3 yy 3Verify y 3 ?-15 6 – 7(3)?-15 6 – 21-15 -152.1 Solving Equations Involving Two Operations 173 Robert H. Prior, 2014Copying is prohibited.

You Try It 5Solve each equation by first isolating the variable term. Verify the solution. UseExample 5 as a guide.a)3x – 5 19b)43 y 15 3c)4 3m – 6c)-4v 12 -16e)-1 9 2wf)-2 18 – 5pYou Try It AnswersYou Try It 1:a)c)0 30; No, 3 is not the solution.-9 -9; Yes, -2 is the solution.b)18 18; Yes, 4 is the solution.You Try It 2:a)x 13b)b -7c)w 0d)1m 3You Try It 3:a)p 11b)w -10c)y 20d)v -2You Try It 4:a)b)x 4c)y -16d)w 14e)v -36-2p 7f)x -8a)e)x 8w -5b)f)y -9p 4c)10m 3d)v 7You Try It 5:2.1 Solving Equations Involving Two Operations174 Robert H. Prior, 2014Copying is prohibited.

Section 2.1 ExercisesThink Again.1.To clear a constant, we add its opposite to each side. To clear a coefficient, why don’t we divideby its opposite?2.What extra step(s) would be required to solve this equation: -1 3y – 5y 9?Focus Exercises.For each, replace the variable with 16 and decide whether 16 is the solution.3.29 m 554.114 k 985.5 · d 806.210 x · 1510.-4 6 – 15y22For each, replace the variable with 3 and decide whether 3 is the solution.7.-5 6k – 98.7 3m 119.-12x 7 1Solve each equation. Verify the solution.11.x – 12 612.-7 w – 413.y – 9 -814.2 p 815.m 6 -416.0 x 617.p – 5 018.4 v 419.m – 6 -620.57x 12 1221.115m 6 622.3-5v 16 16Solve each equation. Verify the solution.23.6n 4224.7y -4225.-3k -3626.18 -9p27.-x -928.-r 829.4w 1830.-9y 2131.23 v 1032.-29 y 633.2x8 734.-3k125 - 252.1 Solving Equations Involving Two Operations175 Robert H. Prior, 2014Copying is prohibited.

Solve each equation. Verify the solution.35.3x 5 1736.4x 9 937.4 – 5w -3638.45 4x 939.-7v – 7 1440.-y 12 1041.-8 -p 1242.-8 5x – 843.2y 6 044.3y – 5 045.3v 1 046.0 -20 – 4x47.0 15 – 10x58.2 – 2k -749.4m – 9 -3950.10 -5p 1851.16 12k 152.13 x – 2 353.23 x 9 754.2y6 – 5 10Think Outside the Box.Solve each equation by first combining like terms, as necessary.55.8w – 5w 1 -856.-2y – 3 – 4y -2757.13 2x – 7x 2858.-15 -7q 4q 21Each of these contains a variable term on each side of the equation. To solve, first manipulate eachequation to be in standard form by clearing one of the variable terms (add its opposite to each side ofthe equation). Then finish solving the equation.59.7w 19 4w – 860.-2y 1 4y 1661.13 – 5x -8 – 2x62.-14 – 3p 7p 262.1 Solving Equations Involving Two Operations176 Robert H. Prior, 2014Copying is prohibited.

There is a class of conditional equations called linear equations. In a linear equation, the highest power of any variable is 1. Here are some examples of linear equations: (a) 3x 5 23 (b) -9 15 – 2y (c) 6w 11 -4w – 9 To be well prepared to solve linear equations,

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