Math 302 Module 3 - College of the Redwoodsmsenux2.redwoods.edu/mathjam/302/math302_3.pdfthat we get...

Math 302

Module 3

Department of MathematicsCollege of the Redwoods

June 17, 2011

Contents

3 Equations 13a Solving Systems by Graphing . . . . . . . . . . . . . . . . . . . 2

Approximations . . . . . . . . . . . . . . . . . . . . . . . . . 3Exceptional Cases . . . . . . . . . . . . . . . . . . . . . . . . 5

3b Solving Systems of Two Equations Algebraically . . . . . . . . 8Solving Systems by Substitution . . . . . . . . . . . . . . . . 8Exceptional Cases Revisited . . . . . . . . . . . . . . . . . . 11Solving Systems by Elimination . . . . . . . . . . . . . . . . 13Exceptional Cases . . . . . . . . . . . . . . . . . . . . . . . . 16Solving Systems by Substitution . . . . . . . . . . . . . . . . 18Exceptional Cases Revisited . . . . . . . . . . . . . . . . . . 21

3c Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

ii

Module 3

Equations

1

2 MODULE 3. EQUATIONS

3a Solving Systems by Graphing

In this section we introduce a graphical technique for solving systems of twolinear equations in two unknowns. As we saw in the previous chapter, if a pointsatisfies an equation, then that point lies on the graph of the equation. If weare looking for a point that satisfies two equations, then we are looking for apoint that lies on the graphs of both equations; that is, we are looking for apoint of intersection. What better way to find a point of intersection than todraw the graph and look to see where they intersect? Let’s try an example.

You Try It!

EXAMPLE 1. Solve the following system of equations:

3x+ 2y = 12

y = x+ 1(3.1)

Solution: We are looking for the point (x, y) that satisfies both equations;that is, we are looking for the point that lies on the graph of both equations.Therefore, the logical approach is to plot the graphs of both lines, then identifythe point of intersection.

In 3x+ 2y = 12, let x = 0. Then 2y = 12 or y = 6, making the y-intercept(0, 6). Secondly, let y = 0 in 3x + 2y = 12. Then 3x = 12 or x = 4, makingthe x-intercept (4, 0). These intercepts are plotted in Figure 3.1 and the line3x+ 2y = 12 is drawn through them.

!2 8

!2

8

x

y

(4, 0)

(0, 6)

Figure 3.1: Drawing the graph of3x+ 2y = 12.

!2 8

!2

8

x

y

1

1

(0, 1)

Figure 3.2: Adding the graph ofy = x+ 1.

The second equation y = x + 1 has slope-intercept form y = mx + b. Hence,the slope is m = 1 and the y-intercept is (0, 1). Plot the intercept (0, 1), thengo up 1 and right 1, then draw the line (see Figure 3.2).

3A. SOLVING SYSTEMS BY GRAPHING 3

Finally, we examine the completed graph containing the lines representedby the system equations, then approximate the coordinates of the point ofintersection. It appears that the lines intersect at the point (2, 3), making(x, y) = (2, 3) the solution of System 3.1 (see Figure 3.3).

!2 8

!2

8

x

y

(2, 3)

Figure 3.3: The coordinates of the point of intersection is the solution ofSystem 3.1.

Check: To show that (x, y) = (2, 3) is a solution of System 3.1, we must showthat we get true statements when we substitute 2 for x and 3 for y in bothequations of System 3.1.

Substituting 2 for x and 3 for y in3x+ 2y = 12, we get:

3x+ 2y = 12

3(2) + 2(3) = 12

6 + 6 = 12

12 = 12

Hence, (2, 3) satisfies the equation3x+ 2y = 12.

Substituting 2 for x and 3 for y iny = x+ 1, we get:

y = x+ 1

3 = 2 + 1

3 = 3

Hence, (2, 3) satisfies the equationy = x+ 1.

Because (2, 3) satisfies both equations, this makes (2, 3) a solution of System 3.1.

!

Approximations

Sometimes we have to be satisfied with an approximation of the solution of thesystem of equations.


You Try It!


3x! 5y = !15

2x+ y = !4(3.2)

Solution: Once again, we are looking for the point that satisfies both equationsof the System 3.2. Thus, we need to find the point that lies on the graphs ofboth lines represented by the equations of System 3.2. The approach will be tograph both lines, then approximate the coordinates of the point of intersection.

In 3x ! 5y = !15, let x = 0. Then !5y = !15 or y = 3, making they-intercept (0, 3). Secondly, let y = 0 in 3x ! 5y = !15. Then 3x = !15 orx = !5, so the x-intercept is (!5, 0). These intercepts are plotted in Figure 3.4and the line 3x! 5y = !15 is drawn through them.

With the second equation 2x + y = !4, let x = 0, then y = !4. Hence,the y-intercept is (0,!4). Secondly, let y = 0 in 2x+ y = !4. Then 2x = !4or x = !2. Thus, the x-intercept is (!2, 0). These intercepts are plotted inFigure 3.5 and the line 2x+ y = !4 is drawn through them.

!10 10

!10

10

x

y

(!5, 0)

(0, 3)

Figure 3.4: The line 3x! 5y = !15has intercepts at (!5, 0) and (0, 3).

!10 10

!10

10

x

y

(!2, 0)

(0,!4)

Figure 3.5: The line 2x + y =!4 has intercepts at (!2, 0) and(0,!4).

The solution of the system is the point of intersection of the two lines. However,unlike Example 1, we’ll have to be content with an approximation of thesecoordinates. It appears that the coordinates of the point of intersection areapproximately (!2.6, 1.4).

Check: Because we only have an approximation of the solution of the system,we cannot expect the solution to check exactly in each equation. However, wedo hope that the solution checks approximately.


!10 10

!10

10

x

y

(!2.6, 1.4)

Figure 3.6: The approximate coordinates of the point of intersection are(!2.6, 1.4).

Substitute (x, y) = (!2.6, 1.4) intothe first equation of System 3.2.

3x! 5y = !15

3(!2.6)! 5(1.4) = !15

!7.8! 7 = !15

!14.8! 15

Note that the solution does notcheck exactly, but it is pretty closeto being a true statement.

Substitute (x, y) = (!2.6, 1.4) intothe second equation of System 3.2.

2x+ y = !4

2(!2.6) + 1.4 = !4

!5.2 + 1.4 = !4

!3.8 = !4

Again, note that the solution doesnot check exactly, but it is prettyclose to being a true statement.

!

Exceptional Cases

Most of the time, given the graphs of two lines, they will intersect in exactlyone point. But there are two exceptions.

You Try It!


2x+ 3y = 6

2x+ 3y = !6(3.3)

Solution: Let’s solve both equations for y.


Solve 2x+ 3y = 6 for y:

2x+ 3y = 6

2x+ 3y ! 2x = 6! 2x

3y = 6! 2x

3y

3=

6! 2x

3

y = !

2

3x+ 2

Solve 2x+ 3y = !6 for y:

2x+ 3y = !6

2x+ 3y ! 2x = !6! 2x

3y = !6! 2x

3y

3=

!6! 2x

3

y = !

2

3x! 2

Both lines have the same slope !2/3, but they have di!erent y-intercepts.Hence, the two lines are parallel distinct lines (see Figure 3.7). They will nothave any points of intersection and System 3.3 has no solution.


!5 5

!5

5

x

y

(0,!2)

(0, 2)

Figure 3.7: The lines 2x+3y = 6 and 2x+3y = !6 are parallel, distinct lines.

!

You Try It!


x! y = 3

!2x+ 2y = !6(3.4)

Solution: Let’s solve both equations for y.

Solve x! y = 3 for y:

x! y = 3

x! y ! x = 3! x

!y = !x+ 3

!1[!y] = !1[!x+ 3]

y = x! 3

Solve !2x+ 2y = !6 for y:

!2x+ 2y = !6

!2x+ 2y + 2x = !6 + 2x

2y = 2x! 6

2y

2=

2x! 6

2y = x! 3

Both lines have slope m = 1, and both have the same y-intercept (0,!3).Hence, the two lines are identical. Hence, System 3.4 has an infinite numberof points of intersection. Any point on either line is a solution of the system.Examples of points of intersection (solutions) are (3, 0), (0,!3), and (4, 1).

!


3b Solving Systems of Two Equations Algebraically

Solving Systems by Substitution

In this section we introduce an algebraic technique for solving systems of twoequations in two unknowns called the substitution method.

Substitution method. The substitution method involves two consecutivesteps:

1. Solve either equation for either variable.

2. Substitute the result from step one into the other equation.

You Try It!


2x! 5y = !8 (3.5)

y = 3x! 1 (3.6)

Solution: Equation (3.30) is already solved for y. Substitute equation (3.30)into equation (3.29).

2x! 5y = !8 Equation (3.29).

2x! 5(3x! 1) = !8 Substitute (3.30) in (3.29).

Now solve for x.

2x! 15x+ 5 = !8 Distribute !5.

!13x+ 5 = !8 Simplify.

!13x = !13 Subtract 5 from both sides.

x = 1 Divide both sides by !13.

Substitute x = 1 in equation (3.30), then solve for y.

y = 3x! 1 Equation (3.30).

y = 3(1)! 1 Substitute 1 for x.

y = 2 Simplify.

Hence, (x, y) = (1, 2) is the solution of the system.

Check: To show that the solution (x, y) = (1, 2) is a solution of the system,we need to show that (x, y) = (1, 2) satisfies both equations (3.29) and (3.30).

3B. SOLVING SYSTEMS OF TWO EQUATIONS ALGEBRAICALLY 9

Substitute (x, y) = (1, 2) in equa-tion (3.29):

2x! 5y = !8

2(1)! 5(2) = !8

2! 10 = !8

!8 = !8

Thus, (1, 2) satisfies equa-tion (3.29).


y = 3x! 1

2 = 3(1)! 1

2 = 3! 1

2 = 2


!

You Try It!


5x! 2y = 12 (3.7)

4x+ y = 6 (3.8)

Solution: The first step is to solve either equation for either variable. Thismeans that we can solve the first equation for x or y, but it also means that wecould first solve the second equation for x or y. Of these four possible choices,solving equation (3.32) for y seems the easiest way to start.

4x+ y = 6 Equation (3.32).

y = 6! 4x Subtract 4x from both sides.

Next, substitute y = 6! 4x for y in equation (3.31).

5x! 2y = 12 Equation (3.31).

5x! 2(6! 4x) = 12 Substitute y = 6! 4x for y.

5x! 12 + 8x = 12 Distribute !2.

13x! 12 = 12 Simplify.

13x = 24 Add 12 to both sides.

x =24

13Divide both sides by 13.

Finally, to find the y-value, substitute x = 24/13 into the equation y =


6! 4x (you can also substitute x = 24/13 into equation (3.31) or (3.32)).

y = 6! 4x

y = 6! 4

!

24

13

"

Substitute x = 24/13 in y = 6! 4x.

y =78

13!

96

13Multiply, then make equivalent fractions.

y = !

18

13Simplify.

Hence, (x, y) = (24/13,!18/13) is the solution of the system.

!

You Try It!


3x! 2y = 6 (3.9)

4x+ 5y = 20 (3.10)

Solution: Dividing by 2 gives easier fractions to deal with than dividing by3, 4, or 5, so let’s start by solving equation (3.33) for y.

3x! 2y = 6 Equation (3.33).

!2y = 6! 3x Subtract 3x from both sides.

y =6! 3x

!2Divide both sides by !2.

y = !3 +3

2x Divide both terms by !2

using distributive property.

Substitute y = !3 + 3

2x for y in equation (3.34).

4x+ 5y = 20 Equation (3.34).

4x+ 5

!

!3 +3

2x

"

= 20 Substitute !3 +3

2x for y.

4x! 15 +15

2x = 20 Distribute the 5.

8x! 30 + 15x = 40 Clear fractions by multiplyingboth sides by 2.

23x = 70 Simplify. Add 30 to both sides.

x =70



To find y, substitute x = 70/23 for x into equation y = !3 + 3

2x.

y = !3 +3

2x

y = !3 +3

2

!

70

23

"

Substitute 70/23 for x.

y = !

69

23+

105

23Multiply. Make equivalent fractions.

y =36

23Simplify.

Hence, (x, y) = (70/23, 36/23) is the solution of the system.

!

Exceptional Cases Revisited

It is entirely possible that you might apply the substitution method to a systemof equations that either have an infinite number of solutions or no solutions atall. Let’s see what happens should you do that.

You Try It!


2x+ 3y = 6 (3.11)

y = !

2

3x+ 4 (3.12)

Solution: Equation (3.36) is already solved for y, so let’s substitute y =!

2

3x+ 4 for y in equation 3.35.

2x+ 3y = 6 Equation (3.35).

2x+ 3

!

!

2

3x+ 4

"

= 6 Substitute !

2

3x+ 4 for y.

2x! 2x+ 12 = 6 Distribute the 3.

12 = 6 Simplify.

Goodness! What happened to the x? How are we supposed to solve for x inthis situation?

However, the resulting statement, 12 = 6, is false, no matter what we usefor x and y. This should give us a clue that there are no solutions. Perhaps weare dealing with parallel lines?

Let’s solve equation (3.35) to determine the situation.

2x+ 3y = 6 Equation (3.35).

3y = !2x+ 6 Subtract 2x from both sides.

y = !

2

3x+ 2 Divide both sides by 3.


Thus, our system is equivalent to the following two equations.

y = !

2

3x+ 2 (3.13)

y = !

2

3x+ 4 (3.14)

These lines have the same slope !2/3, but di!erent y-intercepts (one has y-intercept (0, 2), the other has y-intercept (0, 4)). Hence, these are two distinctparallel lines and the system has no solution.

!

You Try It!


2x! 6y = !8 (3.15)

x = 3y ! 4 (3.16)

Solution: Equation (3.40) is already solved for x, so let’s substitute x = 3y!4for x in equation (3.39).

2x! 6y = !8 Equation (3.39).

2(3y ! 4)! 6y = !8 Substitute 3y ! 4 for x.

6y ! 8! 6y = !8 Distribute the 2.

!8 = !8 Simplify.

Goodness! What happened to the x? How are we supposed to solve for x inthis situation? However, note that the resulting statement, !8 = !8, is a truestatement this time. Perhaps this is an indication that we are dealing with thesame line?

Let’s put both equations (3.39) and (3.40) into slope-intercept form so thatwe can compare them.

Solve equation (3.39) for y:

2x! 6y = !8

!6y = !2x! 8

y =!2x! 8

!6

y =1

3x+

4

3


x = 3y ! 4

x+ 4 = 3y

x+ 4

3= y

y =1

3x+

4

3

Hence, the lines have the same slope and the same y-intercept and they areexactly the same lines. Thus, there are an infinite number of solutions. Indeed,


any point on either line is a solution. Examples of solution points are (0, 4/3),(!4, 0), and (!1, 1).

!

Tip. When you substitute one equation into another and the variable disap-pears, consider:

1. If the resulting statement is false, suspect that the lines are parallel linesand there is no solution.

2. If the resulting statement is true, suspect that you have the same linesand there are an infinite number of solutions.

Solving Systems by Elimination

When both equations of a system are in standard form Ax + By = C, then aprocess called elimination is the best procedure to use to find the solution ofthe system. Elimination is based on two simple ideas, the first of which shouldbe familiar.

1. Multiplying both sides of an equation by a non-zero number does notchange its solutions. Thus, the equation

x+ 3y = 7 (3.17)

will have the same solutions (it’s the same line) as the equation obtainedby multiplying equation (3.17) by 2.

2x+ 6y = 14 (3.18)

2. Adding two true equations produces another true equation. For example,consider what happens when you add 4 = 4 to 5 = 5.

4 = 45 = 59 = 9

Even more importantly, consider what happens when you add two equa-tions that have (2, 1) as a solution. The result is a third equation whosegraph also passes through the solution.


x + y = 3x ! y = 1

2x = 4x = 2

!5 5

!5

5

x

y

x = 2x! y = 1

x+ y = 3

Elimination. Adding a multiple of an equation to a second equation producesand equation that passes through the same solution as the first two equations.

Let’s use these ideas to solve a system of equations.


You Try It!

EXAMPLE 6. Solve the following system of equations.

x+ 2y = !5 (3.19)

2x! y = !5 (3.20)

Solution: Our focus will be on eliminating the variable x. We note that if wemultiply equation (3.19) by !2, then add the result to equation (3.20), the xterms will be eliminated.

!2x ! 4y = 10 Multiply equation (3.19) by !2.2x ! y = !5 Equation (3.20).

! 5y = 5 Add the equations.

Thus, y = !1.To find the corresponding value of x, substitute y = !1 in equation (3.19)

(or equation (3.20)) and solve for x.

x+ 2y = !5 Equation (3.19)

x+ 2(!1) = !5 Substitute y = !1.

x = !3 Solve for x.

Check: To check, we need to show that the point (x, y) = (!3, 1) satisfiesboth equations.

Substitute (x, y) = (!3,!1) intoequation (3.19).

x+ 2y = !5

!3 + 2(!1) = !5

!5 = !5

Substitute (x, y) = (!3,!1) intoequation (3.20).

2x! y = !5

2(!3)! (!1) = !5

!5 = !5

Thus, the point (x, y) = (!3,!1) satisfies both equations and is therefore thesolution of the system.

!

To show that you have the option of which variable you choose to eliminate,let’s try Example 6 a second time, this time eliminating y instead of x.

You Try It!


x+ 2y = !5 (3.21)

2x! y = !5 (3.22)


Solution: This time we focus on eliminating the variable y. We note that ifwe multiply equation (3.22) by 2, then add the result to equation (3.21), the yterms will be eliminated.

x + 2y = !5 Equation (3.21).4x ! 2y = !10 Multiply equation (3.22) by 2.

5x = !15 Add the equations.

Thus, x = !3.To find the corresponding value of y, substitute x = !3 in equation (3.21)

(or equation (3.22)) and solve for y.

x+ 2y = !5 Equation (3.21)

!3 + 2y = !5 Substitute x = !3.

2y = !2 Add 3 to both sides.

y = !1 Divide both sides by 2.

Hence, (x, y) = (!3,!1), just as in Example 6, is the solution of the system.

!

Sometimes the elimination method requires a process similar to that offinding a common denominator.

You Try It!


3x+ 4y = 12 (3.23)

2x! 5y = 10 (3.24)

Solution: Let’s focus on eliminating the x-terms. Note that if we multiplyequation (3.23) by 2, then multiply equation (3.24) by !3, the x-terms willbe eliminated when we add the resulting equations. Note that 6 is the leastcommon multiple of 3 and 2.

6x + 8y = 24 Multiply equation (3.23) by 2.!6x + 15y = !30 Multiply equation (3.24) by !3.

23y = !6 Add the equations.

Hence, y = !6/23.At this point, we could substitute y = !6/23 in either equation, then solve

the result for x. However, working with y = !6/23 is a bit daunting, partic-ularly in the light of elimination being easier. So let’s use elimination again,this time focusing on eliminating y. Similarly, if we multiply equation (3.23)by 5, then multiply equation (3.24) by 4, when we add the results, the y-termswill be eliminated.


15x + 20y = 60 Multiply equation (3.23) by 5.8x ! 20y = 40 Multiply equation (3.24) by 4.

23x = 100 Add the equations.

Thus, x = 100/23, and the system of the system is (x, y) = (100/23,!6/23).

!

Exceptional Cases

In the previous section, we saw that if the substitution method led to a falsestatement, we should suspect that we have parallel lines. The same thing canhappen with the elimination method of this section.

You Try It!


x+ y = 3 (3.25)

2x+ 2y = !6 (3.26)

Solution: Let’s focus on eliminating the x-terms. Note that if we multiplyequation (3.25) by!2, the x-terms will be eliminated when we add the resultingequations.

!2x ! 2y = !6 Multiply equation (3.25) by !2.2x + 2y = !6 Equation (3.26).

0 = !12 Add the equations.

Because of our experience with this solving this exceptional case with substi-tution, the fact that both variables have disappeared should not be completelysurprising. Note that this last statement is false, regardless of the values of xand y. Hence, the system has no solution.

Indeed, if we find the intercepts of each equation and plot them, then wecan easily see that the lines of this system are parallel (see Figure 3.8). Parallellines never intersect, so the system has no solutions.

!

You Try It!


x! 7y = 4 (3.27)

!3x+ 21y = !12 (3.28)


!5 5

!5

5

x

y

(3, 0)

(0, 3)

(!3, 0)

(0,!3)

x+ y = 3

2x+ 2y = !6

Figure 3.8: The lines x+ y = 3 and 2x+ 2y = !6 are parallel.

Solution: We might recognize that the equations 3.27 and (3.28) are identi-cal. But it’s also conceivable that we don’t see that right away and begin theelimination method. Let’s multiply the first equation by 3, then add. This willeliminate the x-terms.

3x ! 21y = 12 Multiply equation (3.27) by 3.!3x + 21y = !12 Equation (3.28).

0 = 0 Add the equations.

Again, all of the variables have disappeared! However, this time the last state-ment is true, regardless of the values of x and y.

Notice that if we multiply equation (3.27) by !3, then we have two identicalequations.

!3x + 21y = !12 Multiply equation (3.27) by 3.!3x + 21y = !12 Equation (3.28).

The equations are identical! Hence, there are an infinite number of points ofintersection. Indeed, any point on either line is a solution. Example points ofsolution are (4, 0), (0,!4/7), and (18, 2).

!


Solving Systems by Substitution

In this section we introduce an algebraic technique for solving systems of twoequations in two unknowns called the substitution method.

Substitution method. The substitution method involves two consecutivesteps:

1. Solve either equation for either variable.

2. Substitute the result from step one into the other equation.

You Try It!


2x! 5y = !8 (3.29)

y = 3x! 1 (3.30)

Solution: Equation (3.30) is already solved for y. Substitute equation (3.30)into equation (3.29).

2x! 5y = !8 Equation (3.29).

2x! 5(3x! 1) = !8 Substitute (3.30) in (3.29).

Now solve for x.

2x! 15x+ 5 = !8 Distribute !5.

!13x+ 5 = !8 Simplify.

!13x = !13 Subtract 5 from both sides.

x = 1 Divide both sides by !13.

Substitute x = 1 in equation (3.30), then solve for y.

y = 3x! 1 Equation (3.30).

y = 3(1)! 1 Substitute 1 for x.

y = 2 Simplify.

Hence, (x, y) = (1, 2) is the solution of the system.

Check: To show that the solution (x, y) = (1, 2) is a solution of the system,we need to show that (x, y) = (1, 2) satisfies both equations (3.29) and (3.30).



2x! 5y = !8

2(1)! 5(2) = !8

2! 10 = !8

!8 = !8



y = 3x! 1

2 = 3(1)! 1

2 = 3! 1

2 = 2


!

You Try It!


5x! 2y = 12 (3.31)

4x+ y = 6 (3.32)

Solution: The first step is to solve either equation for either variable. Thismeans that we can solve the first equation for x or y, but it also means that wecould first solve the second equation for x or y. Of these four possible choices,solving equation (3.32) for y seems the easiest way to start.

4x+ y = 6 Equation (3.32).

y = 6! 4x Subtract 4x from both sides.

Next, substitute y = 6! 4x for y in equation (3.31).

5x! 2y = 12 Equation (3.31).

5x! 2(6! 4x) = 12 Substitute y = 6! 4x for y.

5x! 12 + 8x = 12 Distribute !2.

13x! 12 = 12 Simplify.

13x = 24 Add 12 to both sides.

x =24


Finally, to find the y-value, substitute x = 24/13 into the equation y = 6! 4x


(you can also substitute x = 24/13 into equation (3.31) or (3.32)).

y = 6! 4x

y = 6! 4

!

24

13

"

Substitute x = 24/13 in y = 6! 4x.

y =78

13!

96

13Multiply, then make equivalent fractions.

y = !

18

13Simplify.

Hence, (x, y) = (24/13,!18/13) is the solution of the system.

!

You Try It!


3x! 2y = 6 (3.33)

4x+ 5y = 20 (3.34)

Solution: Dividing by 2 gives easier fractions to deal with than dividing by3, 4, or 5, so let’s start by solving equation (3.33) for y.

3x! 2y = 6 Equation (3.33).

!2y = 6! 3x Subtract 3x from both sides.

y =6! 3x

!2Divide both sides by !2.

y = !3 +3

2x Divide both terms by !2

using distributive property.

Substitute y = !3 + 3

2x for y in equation (3.34).

4x+ 5y = 20 Equation (3.34).

4x+ 5

!

!3 +3

2x

"

= 20 Substitute !3 +3

2x for y.

4x! 15 +15

2x = 20 Distribute the 5.

8x! 30 + 15x = 40 Clear fractions by multiplyingboth sides by 2.

23x = 70 Simplify. Add 30 to both sides.

x =70



To find y, substitute x = 70/23 for x into equation y = !3 + 3

2x.

y = !3 +3

2x

y = !3 +3

2

!

70

23

"

Substitute 70/23 for x.

y = !

69

23+

105

23Multiply. Make equivalent fractions.

y =36

23Simplify.

Hence, (x, y) = (70/23, 36/23) is the solution of the system.

!

Exceptional Cases Revisited

It is entirely possible that you might apply the substitution method to a systemof equations that either have an infinite number of solutions or no solutions atall. Let’s see what happens should you do that.

You Try It!


2x+ 3y = 6 (3.35)

y = !

2

3x+ 4 (3.36)

Solution: Equation (3.36) is already solved for y, so let’s substitute y =!

2

3x+ 4 for y in equation 3.35.

2x+ 3y = 6 Equation (3.35).

2x+ 3

!

!

2

3x+ 4

"

= 6 Substitute !

2

3x+ 4 for y.

2x! 2x+ 12 = 6 Distribute the 3.

12 = 6 Simplify.

Goodness! What happened to the x? How are we supposed to solve for x inthis situation?

However, the resulting statement, 12 = 6, is false, no matter what we usefor x and y. This should give us a clue that there are no solutions. Perhaps weare dealing with parallel lines?

Let’s solve equation (3.35) to determine the situation.

2x+ 3y = 6 Equation (3.35).

3y = !2x+ 6 Subtract 2x from both sides.

y = !

2

3x+ 2 Divide both sides by 3.


Thus, our system is equivalent to the following two equations.

y = !

2

3x+ 2 (3.37)

y = !

2

3x+ 4 (3.38)

These lines have the same slope !2/3, but di!erent y-intercepts (one has y-intercept (0, 2), the other has y-intercept (0, 4)). Hence, these are two distinctparallel lines and the system has no solution.

!

You Try It!


2x! 6y = !8 (3.39)

x = 3y ! 4 (3.40)

Solution: Equation (3.40) is already solved for x, so let’s substitute x = 3y!4for x in equation (3.39).

2x! 6y = !8 Equation (3.39).

2(3y ! 4)! 6y = !8 Substitute 3y ! 4 for x.

6y ! 8! 6y = !8 Distribute the 2.

!8 = !8 Simplify.

Goodness! What happened to the x? How are we supposed to solve for x inthis situation? However, note that the resulting statement, !8 = !8, is a truestatement this time. Perhaps this is an indication that we are dealing with thesame line?

Let’s put both equations (3.39) and (3.40) into slope-intercept form so thatwe can compare them.


2x! 6y = !8

!6y = !2x! 8

y =!2x! 8

!6

y =1

3x+

4

3


x = 3y ! 4

x+ 4 = 3y

x+ 4

3= y

y =1

3x+

4

3

Hence, the lines have the same slope and the same y-intercept and they areexactly the same lines. Thus, there are an infinite number of solutions. Indeed,


any point on either line is a solution. Examples of solution points are (0, 4/3),(!4, 0), and (!1, 1).

Tip. When you substitute one equation into another and the variable disap-pears, consider:

1. If the resulting statement is false, suspect that the lines are parallel linesand there is no solution.

2. If the resulting statement is true, suspect that you have the same linesand there are an infinite number of solutions.

3C. APPLICATIONS 25

3c Applications

In this section we create and solve applications that lead to systems of linearequations. As we create and solve our models, we’ll follow the Requirementsfor Word Problem Solutions. However, instead of setting up a single equation,we set up a system of equations for each application.

You Try It!

EXAMPLE 1. In geometry, two angles that sum to 90! are called comple-mentary angles. If the second of two complementary angles is 30 degrees largerthan twice the first angle, find the degree measure of both angles.

Solution: In the solution, we address each step of the Requirements for WordProblem Solutions.

1. Set up a Variable Dictionary. Our variable dictionary will take the formof a diagram, naming the two complementary angles ! and ".

!

"

2. Set up a Systems of Equations. The “second angle is 30 degrees largerthan twice the first angle” becomes

" = 30 + 2! (3.41)

Secondly, the angles are complementary, meaning that the sum of theangles is 90!.

!+ " = 90 (3.42)

Thus, we have a system of two equations in two unknowns ! and ".

3. Solve the System. As equation (3.41) is already solved for ", let use thesubstitution method and substitute 30 + 2! for " in equation (3.42).

!+ " = 90 Equation (3.42).

!+ (30 + 2!) = 90 Substitute 30 + 2! for ".

3!+ 30 = 90 Combine like terms.

3! = 60 Subtract 30 from both sides.

! = 20 Divide both sides by 3.


4. Answer the Question. The first angle is ! = 20 degrees. The secondangle is:

" = 30 + 2! Equation (3.41).

" = 30 + 2(20) Substitute 20 for !.

" = 70 Simplify.

5. Look Back. Certainly 70! is 30 degrees larger than twice 20!. Also, notethat 20! + 70! = 90!, so the angles are complementary. We have thecorrect solution.

!

You Try It!

EXAMPLE 2. The perimeter of a rectangle is 280 feet. The length of therectangle is 10 feet less than twice the width. Find the width and length of therectangle.


1. Set up a Variable Dictionary. Our variable dictionary will take the formof a diagram, naming the width and length W and L, respectively.

L

W

L

W

2. Set up a System of Equations. The perimeter is found by summing thefour sides of the rectangle.

P = L+W + L+W

P = 2L+ 2W

We’re told the perimeter is 280 feet, so we can substitute 280 for P inthe last equation.

280 = 2L+ 2W

We can simplify this equation by dividing both sides by 2, giving thefollowing result:

L+W = 140 (3.43)

3C. APPLICATIONS 27

Secondly, we’re told that the “length is 10 feet less than twice the width.”This translates to:

L = 2W ! 10 (3.44)

3. Solve the System. As equation (3.44) is already solved for L, let use thesubstitution method and substitute 2W ! 10 for L in equation (3.43).

W + L = 140 Equation (3.43).

W + (2W ! 10) = 140 Substitute 2W ! 10 for L.

3W ! 10 = 140 Combine like terms.

3W = 150 Add 10 to both sides.

W = 50 Divide both sides by 3.

4. Answer the Question. The width is W = 50 feet. The length is:

L = 2W ! 10 Equation (3.44).

L = 2(50)! 10 Substitute 50 for W .

L = 90 Simplify.

Thus, the length is L = 90 feet.

5. Look Back. Perhaps a picture, labeled with our answers might bestdemonstrate that we have the correct solution. Remember, we foundthat the width was 50 feet and the length was 90 feet.

90

50

90

50

Note that the perimeter is P = 90 + 50 + 90 + 50 = 280 feet. Secondly,note that the length 90 feet is 10 feet less than twice the width. So wehave the correct solution.

!

You Try It!

EXAMPLE 3. Pascal has $3.25 in change in his pocket, all in dimes andquarters. He has 22 coins in all. How many dimes does he have?



1. Set up a Variable Dictionary. Let D represent the number of dimes andlet Q represent the number of quarters.

2. Set up a System of Equations. Using a table to summarize information isa good strategy. In the first column, we list the type of coin. The secondcolumn gives the number of each type of coin, and the third columncontains the value (in cents) of the number of coins in Pascal’s pocket.

Number of Coins Value (in cents)

Dimes D 10D

Quarters Q 25Q

Totals 22 325

Note that D times, valued at 10 cents apiece, is worth 10D cents. Sim-ilarly, Q quarters, valued at 25 cents apiece, is worth 25Q cents. Notealso how we’ve change $3.25 to 325 cents.

The second column of the table gives us our first equation.

D +Q = 22 (3.45)

The third column of the table gives us our second equation.

10D + 25Q = 325 (3.46)

3. Solve the System. Because equations (3.45) and (3.46) are both in stan-dard form Ax + By = C, we’ll use the elimination method to find asolution. Because the question asks us to find the number of dimes inPascal’s pocket, we’ll focus on eliminating the Q-terms and keeping theD-terms.

!25D ! 25Q = !550 Multiply equation (3.45) by !25.10D + 25Q = 325 Equation (3.46).

!15D = !225 Add the equations.

Dividing both sides of the last equation by !15 give us D = 15.

4. Answer the Question. The previous solution tells us that Pascal has 15dimes in his pocket.

5. Look Back. Again, summarizing results in a table might help us see if wehave the correct solution. First, because we’re told that Pascal has 22coins in all, and we found that he had 15 dimes, this means that he musthave 7 quarters.

3C. APPLICATIONS 29

Number of Coins Value (in cents)

Dimes 15 150

Quarters 7 175

Totals 22 325

Fifteen dimes are worth 150 cents, and 7 quarters are worth 175 cents.That’s a total of 22 coins and 325 cents, or $3.25. Thus we have thecorrect solution.

!

You Try It!

EXAMPLE 4. Rosa inherits $10,000 and decides to invest the money in twoaccounts, part in a certificate of deposit that pays 4% interest per year, andthe rest in a mutual fund that pays 5% per year. At the end of the first year,Rosa’s investments earn a total of $420 in interest. Find the amount investedin each account.


1. Set up a Variable Dictionary. Let C represent the amount invested in thecertificate of deposit and M represent the amount invested in the mutualfund.

2. Set up a System of Equations. We’ll again use a table to summarizeinformation.

Rate Amount invested Interest

Certificate of Deposit 4% C 0.04C

Mutual Fund 5% M 0.05M

Totals 10,000 420

At 4%, the interest earned on a C dollars investment is found by taking4% of C (i.e., 0.04C). Similarly, the interest earned on the mutual fundis 0.05M .

The third column of the table gives us our first equation. The totalinvestment is $10,000.

C +M = 10000 (3.47)

The fourth column of the table gives us our second equation. The totalinterest earned is the sum of the interest earned in each account.

0.04C + 0.05M = 420 (3.48)


Let’s clear the decimals from equation (3.48) by multiplying both sidesof the equation by 100.

4C + 5M = 42000 (3.49)

3. Solve the System. Because equations (3.47) and (3.49) are both in stan-dard form Ax + By = C, we’ll use the elimination method to find asolution. We’ll focus on eliminating the C-terms.

!4C ! 4M = !40000 Multiply equation (3.47) by !4.4C + 5M = 42000 Equation (3.49).

M = 2000 Add the equations.

Thus, the amount invested in the mutual fund in M = $2, 000.

4. Answer the Question. The question asks us to find the amount investedin each account. So, substitute M = 2000 in equation (3.47) and solvefor C.

C +M = 10000 Equation (3.47).

C + 2000 = 10000 Substitute 2000 for M .

C = 8000 Subtract 2000 from both sides.

Thus C = $8, 000 was invested in the certificate of deposit.

5. Look Back. First, note that the investments in the certificate of depositand the mutual fund, $8,000 and $2,000 respectively, total $10,000. Let’scalculate the interest on each investment: 4% of $8,000 is $320 and 5%of $2,000 is $100.

Rate Amount invested Interest

Certificate of Deposit 4% 8000 320

Mutual Fund 5% 2000 100

Totals 10,000 420

Note that the total interest is $420, as required in the problem statement.Thus, our solution is correct.

!

You Try It!

EXAMPLE 5. Peanuts retail at $0.50 per pound and cashews cost $1.25 perpound. If you were a shop owner, how many pounds of peanuts and cashews

3C. APPLICATIONS 31

should you mix to make 50 pounds of a peanut-cashew mixture costing $0.95per pound?


1. Set up a Variable Dictionary. Let P be the number of pounds of peanutsused and let C be the number of pounds of cashews used.

2. Set up a System of Equations. We’ll again use a table to summarizeinformation.

Cost per pound Amount (pounds) Cost

Peanuts $0.50 P 0.50P

Cashews $1.25 C 1.25C

Totals $0.95 50 0.95(50)=47.50

At $0.50 per pound, P pounds of peanuts cost 0.50P . At $1.25 per pound,C pounds of cashews cost 1.25C. Finally, at $0.95 per pound, 50 poundsof a mixture of peanuts and cashews will cost 0.95(50), or $47.50.

The third column of the table gives us our first equation. The totalnumber of pounds of mixture is given by the following equation:

P + C = 50 (3.50)

The fourth column of the table gives us our second equation. The totalcost is the sum of the costs for purchasing the peanuts and cashews.

0.50P + 1.25C = 47.50 (3.51)

Let’s clear the decimals from equation (3.51) by multiplying both sidesof the equation by 100.

50P + 125C = 4750 (3.52)

3. Solve the System. Because equations (3.50) and (3.52) are both in stan-dard form Ax + By = C, we’ll use the elimination method to find asolution. We’ll focus on eliminating the P -terms.

!50P ! 50C = !2500 Multiply equation (3.50) by !50.50P + 125C = 4750 Equation (3.52).

75C = 2250 Add the equations.

Divide both sides by 75 to get C = 30 pounds of cashews are in the mix.


4. Answer the Question. The question asks for both amounts, peanuts andcashews. Substitute C = 30 in equation (3.50) to determine C.

P + C = 50 Equation (3.50).

P + 30 = 50 Substitute 30 for P .

P = 20 Subtract 30 from both sides.

Thus, there are P = 20 pounds of peanuts in the mix.

5. Look Back. First, note that the amount of peanuts and cashews in themix is 20 and 30 pounds respectively, so the total mixture weighs 50pounds as required. Let’s calculate the costs: for the peanuts, 0.50(20),or $10, for the cashews, 1.25(30) = 37.50.

Cost per pound Amount (pounds) Cost

Peanuts $0.50 20 $10.00

Cashews $1.25 30 $37.50

Totals $0.95 50 47.50

Note that the total cost is $47.50, as required in the problem statement.Thus, our solution is correct.

!

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