8.7 systems of equations and...

456 CHAPTER 8 Graphs, Functions, and Systems of Equations and Inequalities

Systems of Equations and ApplicationsLinear Systems in Two Variables The worldwide personal computer mar-ket share for different manufacturers has varied, with first one, then another obtain-ing a larger share. As shown in Figure 36, Hewlett-Packard’s share rose during1995–1998, while Packard Bell, NEC saw its share decline. The graphs intersect atthe point when the two companies had the same market share.

Source: Intelliquest; IDC.

FIGURE 36

We could use a linear equation to model the graph of Hewlett-Packard’s marketshare and another linear equation to model the graph of Packard Bell, NEC’s mar-ket share. Such a set of equations is called a system of equations. The point wherethe graphs in Figure 36 intersect is a solution of each of the individual equations. Itis also the solution of the system of equations.

The definition of a linear equation given earlier can be extended to more vari-ables. Any equation of the form

for real numbers (not all of which are 0), and b, is a linear equationin n variables. If all the equations in a system are linear, the system is a system oflinear equations, or a linear system.

In Figure 37, the two linear equations and are graphedin the same coordinate system. Notice that they intersect at the point . Because

is the only ordered pair that satisfies both equations at the same time, we saythat is the solution set of the system

Since the graph of a linear equation is a straight line, there are three possi-bilities for the number of solutions in the solution set of a system of two linearequations, as shown in Figure 38 on the next page.

2x � y � 4 .

x � y � 5

��3, 2��3, 2�

�3, 2�2x � y � 4x � y � 5

a1, a2, … , an

a1x1 � a2 x2 � � � � � an xn � b

1995 1996 19981997Year

0

3.5

4.0

4.5

5.0

5.5

6.5

7.0

6.0

MARKET SHARE

Mar

ket

Shar

e (i

n pe

rcen

t)

Hewlett-Packard

Packard Bell, NEC

8.7

FIGURE 37

0

(3, 2)x + y = 5

2x – y = 4

x

y

8.7 Systems of Equations and Applications 457

In most cases we cannot rely on graphing to solve systems. There are algebraicmethods to do this, and one such method, called the elimination method, is ex-plained in the following examples. The elimination method involves combining thetwo equations of the system so that one variable is eliminated. This is done using the following fact.

, then

The general method of solving a system by elimination is summarized as follows.

a � c � b � d .If a � b and c � d

A graphing calculator supports ourstatement that is the solutionof the system

2x � y � 4 . x � y � 5

�3, 2�

10

–10

–10 10

x + y = 5

2x – y = 4Graphs of a Linear System (The Three Possibilities)1. The two graphs intersect in a single point. The coordinates of

this point give the only solution of the system. In this case, thesystem is consistent, and the equations are independent. This isthe most common case. See Figure 38(a).

FIGURE 38

2. The graphs are parallel lines. In this case the system isinconsistent; that is, there is no solution common to bothequations of the system, and the solution set is . SeeFigure 38(b).

3. The graphs are the same line. In this case the equations aredependent, since any solution of one equation of the system isalso a solution of the other. The solution set is an infinite set ofordered pairs representing the points on the line. SeeFigure 38(c).

FIGURE 38

x

y

0

(c)

Infinitenumber ofsolutions

0�

x

y

0

(b)

Nosolution

x

y

0

(a)

Onesolution


E X A M P L E 1 Solve the system.

(1)

(2)

Step 1: Both equations are already in standard form.

Step 2: Our goal is to add the two equations so that one of the variables is elimi-nated. Suppose we wish to eliminate the variable x. Since the coefficientsof x are not opposites, we must first transform one or both equations so thatthe coefficients are opposites. Then, when we combine the equations, theterm with x will have a coefficient of 0, and we will be able to solve for y.We begin by multiplying equation (1) by 2 and equation (2) by .

2 times each side of equation (1)

times each side of equation (2)

Step 3: Now add the two equations to eliminate x.

Add.

Step 4: Solve the equation from Step 3 to get .

Step 5: To find x, we substitute 3 for y in either of the original equations. Substi-tuting in equation (2) gives

(2)

Let .

Subtract 9.

Divide by 2. x � 2 .

2x � 4

2x � 9 � 13

y � 3 2x � 3(3) � 13

2x � 3y � 13

y � 3

�19y � �57

�10x � 15y � �65

10x � 4y � 8

�5�10x � 15y � �65

10x � 4y � 8

�5

2x � 3y � 13

5x � 2y � 4

The solution of systems ofequations of graphs morecomplicated than straight lines isthe principle behind the mattang(shown on this stamp), a stickchart used by the people of theMarshall Islands in the Pacific. Amattang is made of roots tiedtogether with coconut fibers, and itshows the wave patterns foundwhen approaching an island.

The solution in Example 1 issupported by a graphing calculator.We solve each equation for y,graph them both, and find thepoint of intersection of the twolines: (2, 3).

10

–10

–10 10

2x + 3y = 13

5x – 2y = 4

Solving Linear Systems by EliminationStep 1: Write in standard form. Write both equations in the form

Step 2: Make the coefficients of one pair of variable terms oppo-sites. Multiply one or both equations by appropriate num-bers so that the sum of the coefficients of either x or y is zero.

Step 3: Add. Add the new equations to eliminate a variable. Thesum should be an equation with just one variable.

Step 4: Solve. Solve the equation from Step 3.

Step 5: Find the other value. Substitute the result of Step 4 intoeither of the given equations and solve for the other variable.

Step 6: Find the solution set. Check the solution in both of thegiven equations. Then write the solution set.

Ax � By � C .

mmi_bounce08.html?2_7



Step 6: The solution appears to be . To check, substitute 2 for x and 3 for y inboth of the original equations.

(1) (2)

? ?

? ?

True True

The solution set is . �

Linear systems can also be solved by the substitution method. This method ismost useful for solving linear systems in which one variable has coefficient 1 or �1.As shown in more advanced algebra courses, the substitution method is the bestchoice for solving many nonlinear systems.

The method of solving a system by substitution is summarized as follows.

��2, 3��

13 � 13 4 � 4

4 � 9 � 13 10 � 6 � 4

2(2) � 3(3) � 13 5(2) � 2(3) � 4

2x � 3y � 13 5x � 2y � 4

�2, 3�

Solving Linear Systems by SubstitutionStep 1: Solve for one variable in terms of the other. Solve one of

the equations for either variable. (If one of the variables hascoefficient 1 or , choose it, since the substitution methodis usually easier this way.)

Step 2: Substitute. Substitute for that variable in the other equa-tion. The result should be an equation with just one variable.

Step 3: Solve. Solve the equation from Step 2.Step 4: Find the other value. Substitute the result from Step 3 into

the equation from Step 1 to find the value of the other variable.Step 5: Find the solution set. Check the solution in both of the

given equations. Then write the solution set.

�1


(1)

(2)

Step 1: To use the substitution method, first solve one of the equations for either xor y. Since the coefficient of y in equation (2) is , it is easiest to solve fory in equation (2).

(2)

Step 2: Substitute for y in equation (1) to get an equation in x.

(1)

Let .

Step 3: Solve for x in the equation just obtained.

Distributive property

Combine terms; subtract 2.

Divide by 11. x � 1

11x � 11

3x � 2 � 8x � 13

y � 1 � 4x 3x � 2�1 � 4x� � 13

3x � 2y � 13

1 � 4x

y � 1 � 4x �y � �1 � 4x

�1

4x � y � �1

3x � 2y � 13

The following example illustrates this method.




Step 4: Now solve for y. Since , .

Step 5: Check to show that ordered pair satisfies both equations. The solutionset is �

The next example illustrates special cases that may result when systems aresolved. (We will use the elimination method, but the same conclusions will followwhen the substitution method is used.)

��1, 5��.�1, 5�

y � 1 � 4�1� � 5y � 1 � 4x

The graphs of the equations inExample 3(a) are parallel. Thereare no solutions.

The graphs of the equations inExample 3(b) coincide. We seeonly one line. There are infinitelymany solutions.

10

–10

–10 10

8x – 2y = –4–4x + y = 2

10

–10

–10 10

3x – 2y = 4

–6x + 4y = 7 E X A M P L E 3 (a) Solve the system

(1)

(2)

The variable x can be eliminated by multiplying both sides of equation (1) by 2and then adding.

2 times equation (1)

(2)

False

Both variables were eliminated here, leaving the false statement , indi-cating that these two equations have no solutions in common. The system is in-consistent, with the empty set as the solution set.

(b) Solve the system

(1)

(2)

Eliminate x by multiplying both sides of equation (1) by 2 and then adding theresult to equation (2).

2 times equation (1)

(2)

True

This true statement, , indicates that a solution of one equation is also a so-lution of the other, so the solution set is an infinite set of ordered pairs. The twoequations are dependent.

We will write the solution set of a system of dependent equations as a set ofordered pairs by expressing x in terms of y as follows. Choose either equationand solve for x. Choosing equation (1) gives

The solution set is written as

By selecting values for y and calculating the corresponding values for x, indi-vidual ordered pairs of the solution set can be found. For example, if ,

and the ordered pair is a solution. ��1, �2�x � ��2 � 2��4 � �1y � �2

�� y � 2

4, y .

x �2 � y

�4�

y � 2

4.

�4x � y � 2

0 � 0

0 � 0

8x � 2y � �4

�8x � 2y � 4

8x � 2y � �4 .

�4x � y � 2

0�

0 � 15

0 � 15

�6x � 4y � 7

6x � 4y � 8

�6x � 4y � 7 .

3x � 2y � 4


Linear Systems in Three Variables A solution of an equation in threevariables, such as , is called an ordered triple and is written

. For example, the ordered triples and are both solutionsof , since the numbers in these ordered triples satisfy the equationwhen used as replacements for x, y, and z, respectively. The methods of solving sys-tems of two equations in two variables can be extended to solving systems of equa-tions in three variables such as

Theoretically, a system of this type can be solved by graphing. However, the graphof a linear equation with three variables is a plane and not a line. Since the graph ofeach equation of the system is a plane, which requires three-dimensional graphing,this method is not practical. However, it does illustrate the number of solutions pos-sible for such systems, as Figure 39 shows.

2x � 3y � 2z � 3 .

x � 7y � 3z � �14

4x � 8y � z � 2

2x � 3y � z � 4�10, �3, 7��1, 1, 1��x, y, z�

2x � 3y � z � 4

Graphs of Linear Systems in Three Variables1. The three planes may meet at a single, common point that forms

the solution set of the system. See Figure 39(a).2. The three planes may have the points of a line in common so

that the set of points along that line is the solution set of thesystem. See Figure 39(b).

3. The planes may have no points common to all three so that thereis no solution for the system. See Figure 39(c).

4. The three planes may coincide so that the solution set of thesystem is the set of all points on that plane. See Figure 39(d).

(a) A single solution (b) Points of a line in common

(c) One case of no points in common (d) All points in common

FIGURE 39

I, IIIII,III

II

I

I IIIII

PI IIIII


There are other illustrations of the above cases. For example, two of the planesmight intersect in a line, while the third plane is parallel to one of these planes, againresulting in no points common to all three planes. We give only one example of eachcase in Figure 39.

Since graphing to find the solution set of a system of three equations in threevariables is impractical, these systems are solved with an extension of the elimina-tion method, summarized as follows.

Solving Linear Systems in Three Variables by EliminationStep 1: Eliminate a variable. Use the elimination method to elim-

inate any variable from any two of the given equations. Theresult is an equation in two variables.

Step 2: Eliminate the same variable again. Eliminate the samevariable from any other two equations. The result is an equa-tion in the same two variables as in Step 1.

Step 3: Eliminate a different variable and solve. Use the elimi-nation method to eliminate a second variable from the twoequations in two variables that result from Steps 1 and 2.The result is an equation in one variable that gives the valueof that variable.

Step 4: Find a second value. Substitute the value of the variablefound in Step 3 into either of the equations in two variablesto find the value of the second variable.

Step 5: Find a third value. Use the values of the two variablesfrom Steps 3 and 4 to find the value of the third variable bysubstituting into any of the original equations.

Step 6: Find the solution set. Check the solution in all of the orig-inal equations. Then write the solution set.


(1)

(2)

(3)

Step 1: As before, the elimination method involves eliminating a variable from thesum of two equations. The choice of which variable to eliminate is arbi-trary. Suppose we decide to begin by eliminating z. To do this, multiplyboth sides of equation (1) by 3 and then add the result to equation (2).

Multiply both sides of equation (1) by 3.

(2)

Add. 13x � 31y � �8

x � 7y � 3z � �14

12x � 24y � 3z � 6

2x � 3y � 2z � 3

x � 7y � 3z � �14

4x � 8y � z � 2



Step 2: The new equation has only two variables. To get another equation withoutz, multiply both sides of equation (1) by and add the result to equation(3). It is essential at this point to eliminate the same variable, z.

Multiply both sides of equation (1) by .

(3)

Add.

Step 3: Now solve the system of equations from Steps 1 and 2 for x and y. (Thisstep is possible only if the same variable is eliminated in the first two steps.)

Multiply both sides of by 6.

Multiply both sides of by 13.

Add.

Step 4: Substitute 1 for y in either equation from Steps 1 and 2. Choosing gives

Let .

Step 5: Substitute for x and 1 for y in any one of the three given equations tofind z. Choosing equation (1) gives

Let and .

Step 6: It appears that the ordered triple is the only solution of the sys-tem. Check that the solution satisfies all three equations of the system. Weshow the check here only for equation (1).

(1)

?

?

True

Because also satisfies the equations (2) and (3), the solution set is. ��3, 1, 6��

��3, 1, 6�

2 � 2

�12 � 8 � 6 � 2

4��3� � 8�1� � 6 � 2

4x � 8y � z � 2

��3, 1, 6�

z � 6 .

y � 1x � �3 4��3� � 8�1� � z � 2

4x � 8y � z � 2

�3

x � �3 .

�6x � 18

�6x � 19 � �1

y � 1 �6x � 19�1� � �1

�6x � 19y � �1

19y � �1�6x �

y � 1

�61y � �61

�6x � 19y � �1 �78x � 247y � �13

13x � 31y � �8 78x � 186y � �48

�6x � 19y � �1

2x � 3y � 2z � 3

�2 �8x � 16y � 2z � �4

�2

A graphing calculator can beprogrammed to solve a systemsuch as the one in Example 4.Compare the result here to thesolution in the text.


Applications of Linear Systems

Problem Solving

Many problems involve more than one unknown quantity. Although some prob-lems with two unknowns can be solved using just one variable, many times it iseasier to use two variables. To solve a problem with two unknowns, we write twoequations that relate the unknown quantities; the system formed by the pair ofequations then can be solved using the methods of this section.

The following steps, based on the six-step problem-solving method first introducedin Chapter 7, give a strategy for solving problems using more than one variable.

Here is a way to find a person’sage and the date of his or herbirth. Ask the person to do thefollowing.

1. Multiply the number of themonth of birth by 100.

2. Add the date of the month.

3. Multiply the sum by 2.

4. Add 8.

5. Multiply this result by 5.

6. Add 4.

7. Multiply by 10.

8. Add 4.

9. Add the person’s age to this.

Ask the person the numberobtained. You should then subtract444 from the number you aregiven. The final two figures givethe age, the next figure (or figures)will identify the date of the month,and the first figure (or figures) willgive the number of the month.

Solving an Applied Problem by Writing a System of EquationsStep 1: Read the problem carefully until you understand what is

given and what is to be found.

Step 2: Assign variables to represent the unknown values, using diagrams or tables as needed. Write down what each variablerepresents.

Step 3: Write a system of equations that relates the unknowns.

Step 4: Solve the system of equations.

Step 5: State the answer to the problem. Does it seem reasonable?

Step 6: Check the answer in the words of the original problem.

Problems about the perimeter of a geometric figure often involve two unknownsand can be solved using systems of equations.

E X A M P L E 5 Unlike football, where the dimensions of a playing field can-not vary, a rectangular soccer field may have a width between 50 and 100 yd and alength between 50 and 100 yd. Suppose that one particular field has a perimeter of320 yd. Its length measures 40 yd more than its width. What are the dimensions ofthis field? (Source: Microsoft Encarta Encyclopedia 2000.)

Step 1: Read the problem again. We are asked to find the dimensions of the field.

Step 2: Assign variables. Let the length and the width. Figure 40 on thenext page shows a soccer field with the length labeled L and the widthlabeled W.

Step 3: Write a system of equations. Because the perimeter is 320 yd, we find oneequation by using the perimeter formula:

2L � 2W � 320 .

W �L �


Because the length is 40 yd more than the width, we have

The system is, therefore,

(1)

(2)

Step 4: Solve the system of equations. Since equation (2) is solved for L, we canuse the substitution method. We substitute for L in equation (1),and solve for W.

(1)


Combine like terms.

Subtract 80.

Divide by 4.

Let in the equation to find L.

Step 5: State the answer. The length is 100 yd, and the width is 60 yd. Both dimensions are within the ranges given in the problem.

Step 6: Check. The perimeter of this soccer field is

yd,

and the length, 100 yd, is indeed 40 yd more than the width, since

.

The answer is correct.

FIGURE 40 �

Professional sport ticket prices increase annually. Average per-ticket prices inthree of the four major sports (football, basketball, and hockey) now exceed $30.00.

L

W

100 � 40 � 60

2�100� � 2�60� � 320

L � 60 � 40 � 100

L � W � 40W � 60

W � 60

4W � 240

4W � 80 � 320

2W � 80 � 2W � 320

L � W � 40 2�W � 40� � 2W � 320

2L � 2W � 320

W � 40

L � W � 40 .

2L � 2W � 320

L � W � 40 .


E X A M P L E 6 During recent National Hockey League and National BasketballAssociation seasons, two hockey tickets and one basketball ticket purchased at theiraverage prices would have cost $110.40. One hockey ticket and two basketball tick-ets would have cost $106.32. What were the average ticket prices for the two sports?(Source: Team Marketing Report, Chicago.)

Step 1: Read the problem again. There are two unknowns.

Step 2: Assign variables. Let h represent the average price for a hockey ticket andb represent the average price for a basketball ticket.

Step 3: Write a system of equations. Because two hockey tickets and one basket-ball ticket cost a total of $110.40, one equation for the system is

By similar reasoning, the second equation is

Therefore, the system is

(1)

(2)

Step 4: Solve the system of equations. To eliminate h, multiply equation (2) by and add.

(1)

Multiply each side of (2) by .

Add.

Divide by .

To find the value of h, let in equation (2).

(2)

Let .

Multiply.

Subtract 68.16.

Step 5: State the answer. The average price for one basketball ticket was $34.08.For one hockey ticket, the average price was $38.16.

Step 6: Check that these values satisfy the conditions stated in the problem. �

We solved mixture problems earlier using one variable. For many mixture prob-lems, we can use more than one variable and a system of equations.

h � 38.16

h � 68.16 � 106.32

b � 34.08 h � 2�34.08� � 106.32

h � 2b � 106.32

b � 34.08

�3 b � 34.08

�3b � �102.24

�2 �2h � 4b � �212.64

2h � b � 110.40

�2

h � 2b � 106.32 .

2h � b � 110.40

h � 2b � 106.32 .

2h � b � 110.40 .

E X A M P L E 7 How many ounces each of 5% hydrochloric acid and 20% hydrochloric acid must be combined to get 10 oz of solution that is 12.5% hy-drochloric acid?

Step 1: Read the problem. Two solutions of different strengths are being mixed together to get a specific amount of a solution with an “in-between”strength.


Step 2: Assign variables. Let x represent the number of ounces of 5% solution andy represent the number of ounces of 20% solution. Use a table to summa-rize the information from the problem.

Figure 41 also illustrates what is happening in the problem.

FIGURE 41

Step 3: Write a system of equations. When x ounces of 5% solution and y ouncesof 20% solution are combined, the total number of ounces is 10, so

(1)

The ounces of pure acid in the 5% solution (.05x) plus the ounces of pureacid in the 20% solution (.20y) should equal the total ounces of pure acidin the mixture, which is (.125)10, or 1.25. That is,

(2)

Notice that these equations can be quickly determined by reading down inthe table or using the labels in Figure 41.

Step 4: Solve the system of equations (1) and (2). Eliminate x by first multiplyingequation (2) by 100 to clear it of decimals and then multiplying equation(1) by .

Multiply each side of (2) by 100.

Multiply each side of (1) by .

Add.

Because and , x is also 5.

Step 5: State the answer. The desired mixture will require 5 ounces of the 5% solution and 5 ounces of the 20% solution.

Step 6: Check that these values satisfy both equations of the system. �

x � y � 10y � 5

y � 5

15y � 75

�5 �5x � 5y � �50

5x � 20y � 125

�5

.05x � .20y � 1.25 .

x � y � 10 .

Ounces ofsolution

Ounces ofpure acid

x

.05x

+y

.20y

=10

.125(10)

Problems that can be solved bywriting a system of equations havebeen of interest historically. Thefollowing problem first appearedin a Hindu work that dates back toabout A.D. 850.

The mixed price of 9 citrons [alemonlike fruit shown in thephoto] and 7 fragrant woodapples is 107; again, the mixedprice of 7 citrons and 9 fragrantwood apples is 101. O youarithmetician, tell me quickly the price of a citron and the price of a wood apple here,having distinctly separated those prices well.

Use a system to solve thisproblem. The answer can be foundat the end of the exercises for thissection on page 476.

Percent Ounces of Ounces of(as a decimal) Solution Pure Acid

5% � .05 x .05x20% � .20 y .20y

12.5% � .125 10 (.125)10


Problems that use the distance formula were first introduced in Chap-ter 7. In many cases, these problems can be solved with systems of two linear equa-tions. Keep in mind that setting up a table and drawing a sketch will help you solvesuch problems.

d � rt

François Viète, a mathematicianof sixteenth-century France, didmuch for the symbolism ofmathematics. Before his time,different symbols were often usedfor different powers of a quantity.Viète used the same letter with adescription of the power and thecoefficient. According to HowardEves in An Introduction to theHistory of Mathematics, Viètewould have written

as

B 5 in A quad C plano 2 in

cub aequatur D solido.A � A

�

5BA2 � 2CA � A3 � D

E X A M P L E 8 Two executives in cities 400 mi apart drive to a business meet-ing at a location on the line between their cities. They meet after 4 hr. Find the speedof each car if one car travels 20 mph faster than the other.

Step 1: Read the problem carefully.

Step 2: Assign variables.

Let the speed of the faster car,

the speed of the slower car

We use the formula . Since each car travels for 4 hr, the time, t, foreach car is 4. This information is shown in the table. The distance is foundby using the formula and the expressions already entered in thetable.

Find d from .

Draw a sketch showing what is happening in the problem. See Figure 42.

FIGURE 42

Step 3: Write two equations. As shown in the figure, since the total distance trav-eled by both cars is 400 mi, one equation is

Because the faster car goes 20 mph faster than the slower car, the secondequation is

Step 4: Solve. This system of equations,

(1)

(2)

can be solved by substitution. Replace x with in equation (1) andsolve for y.

20 � y

x � 20 � y ,

4x � 4y � 400

x � 20 � y .

4x � 4y � 400 .

Cars meet after 4 hr.

400 mi4x 4y

d � rt

d � rt

d � rt

y �

x �

r t d

Faster car x 4 4xSlower car y 4 4y


Let .


Combine like terms.

Subtract 80.

Divide by 8.

Since , and ,

Step 5: State the answer. The speeds of the two cars are 40 mph and 60 mph.

Step 6: Check the answer. Since each car travels for 4 hr, the total distance traveledis

mi,

as required. �

The problems in Examples 5–8 also could be solved using only one variable.Many students find that the solution is simpler with two variables.

4�60� � 4�40� � 240 � 160 � 400

x � 20 � 40 � 60 .

y � 40x � 20 � y

y � 40

8y � 320

80 � 8y � 400

80 � 4y � 4y � 400

x � 20 � y 4�20 � y� � 4y � 400

E X A M P L E 9 A company produces three color television sets, models X, Y,and Z. Each model X set requires 2 hr of electronics work, 2 hr of assembly time,and 1 hr of finishing time. Each model Y requires 1, 3, and 1 hr of electronics, as-sembly, and finishing time, respectively. Each model Z requires 3, 2, and 2 hr of thesame work, respectively. There are 100 hr available for electronics, 100 hr availablefor assembly, and 65 hr available for finishing per week. How many of each modelshould be produced each week if all available time must be used?

Step 1: Read the problem again. There are three unknowns.

Step 2: Assign variables.

Let

and

Organize the information in a table.

Step 3: Write a system of three equations. The x model X sets require 2x hr ofelectronics, the y model Y sets require 1y (or y) hr of electronics, and the z model Z sets require 3z hr of electronics. Since 100 hr are available forelectronics,

(1)2x � y � 3z � 100 .

z � the number of model Z produced per week.

y � the number of model Y produced per week,

x � the number of model X produced per week,

Each Each EachModel X Model Y Model Z Totals

Hours of electronics work 2 1 3 100Hours of assembly time 2 3 2 100Hours of finishing time 1 1 2 65


Similarly, from the fact that 100 hr are available for assembly,

(2)

and the fact that 65 hr are available for finishing leads to the equation

(3)

Again, notice the advantage of setting up a table. By reading across, we can eas-ily determine the coefficients and constants in the equations of the system.

Step 4: Solve the system

to find , , and .

Step 5: State the answer. The company should produce 15 model X, 10 model Y,and 20 model Z sets per week.

Step 6: Check that these values satisfy the conditions of the problem. �

z � 20y � 10x � 15

x � y � 2z � 65

2x � 3y � 2z � 100

2x � y � 3z � 100

x � y � 2z � 65 .

2x � 3y � 2z � 100 ,

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