Download Chapter 6 – Systems of Linear Equations

Chapter 6 – Systems of Linear Equations
6.1 – Solving systems of 2 linear equations using a graphing tool
There are often times in technological settings when two linear equations (which would graph as lines)
are part of a system. When this happens, certain applications require knowing when the two lines
intersect, as this point of intersection would provide the (often unique) solution to the system. For
example, consider the following two lines:
y  3x  2
y  12 x  4
When graphed on the TI-83, these two lines look as follows:
If the point of intersection needs to be determined, it can be done on the calculator using the intersect
functionality resulting in the following image:
The intersection point calculated by the TI-83 is (2.4, 5.2). What this means is that while there are an
infinite number of points that satisfy the line y  3x  2 , and an infinite number of points that satisfy
the line y  12 x  4 , there is ONE AND ONLY ONE points that satisfies both of them – specifically the point
(2.4, 5.2) referenced above.
The following example illustrates the use of a TI-83 to solve a system in more detail.
Example 1 – Solve the following system of equations using a TI-83
y  1.5 x  1
y  2 x  3.2
Solution:
1. Enter the equations into the calculator (press the ‘Y=’ button to do this)
2. Graph the equations (press the ‘GRAPH’ button to do this)
3. Use the TI-83 to calculate the point of intersection by doing the following:
Press ‘2nd' and then ‘TRACE’
Press ‘5’ to choose ‘intersect’
Press ‘ENTER’ to choose the first graph
Press ‘ENTER’ to choose the second graph
Move the cursor close to the point of intersection and press ‘ENTER’
4. The calculator will now display the point of intersection:
Hence, these two lines intersect at the point (1.2, –0.8)
6.2 – Solving systems of 2 equations using the Addition-Subtraction method
When a graphing tool is not available, systems of equations can be solved algebraically. Two techniques
for doing this are discussed here:
1. Algebraically solving the system using the Addition-Subtraction method
2. Algebraically solving the system using the Substitution method
The Addition-Subtraction method is discussed in this section. The following algorithm is used when
implementing this method:
1. If necessary, algebraically manipulate both equations so they are in the form ax  by  c
2. If necessary, multiply one or both of the equations so that either the coefficients on the x terms
or the coefficients on the y terms are the same.
3. Add or subtract (as appropriate) one equation from the other in order to remove one of the
variables from the equations
4. Solve the resulting linear equation (which will only have 1 variable) to find the first necessary
value
5. Substitute the solution from step 4 into either of the original equations to find the second
necessary value
6. Check the solution by substituting the values into both of the original equations
This algorithm is demonstrated in the following example which uses the equations graphed previously:
Example 1 – Solve the following system of equations using the Addition-Subtraction method:
y  3x  2
y  12 x  4
Solution:
1. Algebraically manipulate the equations (step 1 from algorithm)
y  3x  2
y  12 x  4
2. Multiply the second equation by 6 so the coefficients on x are the same (step 2 from algorithm)
(note that this step is technically not necessary for this problem since the y variables already
have a common coefficient, but is done to demonstrate the process)
y  3x  2
6 y  3x  24
3. Subtract equation 2 from equation 1 to eliminate the terms with x (step 3 from algorithm)
y  3x  2
6 y  3x  24
---------------------5 y  26
4. Solve 5 y  26 for y (step 4 from the algorithm)
5 y  26
y  5.2
5. Substitute the value 5.2 in for y in either of the original equations (step 5 from algorithm)
y  3x  2
5.2  3 x  2
7.2  3 x
x  2.4
6. Check (step 6 from algorithm)
y  3x  2
5.2  3(2.4)  2
5.2  7.2  2
5.2  5.2
y  12 x  4
5.2  12 (2.4)  4
5.2  1.2  4
5.2  5.2
The solutions check out, and hence the solution is: x  2.4 and y  5.2 which is another way to
notate the point (2.4, 5.2). Note that this is the same solution found using the graphing technique
above.
In the next example, both equations must be multiplied by a value in step 2:
Example 2 – Solve the following system of equations using the Addition-Subtraction method:
2 x  7 y  10
3x  4 y  12
Solution:
1. Algebraically manipulate the equations – this step is unnecessary as the equations are already in
the appropriate form.
2. Multiply the first equation by 3 and the second equation by 2 so the coefficients on x are the
same
6 x  21y  30
6 x  8 y  24
3. Subtract equation 2 from equation 1 to eliminate the terms with x
6 x  21y  30
6 x  8 y  24
---------------------13 y  6
4. Solve 5 y  26 for y (step 4 from the algorithm)
13 y  6
y  136
5. Substitute the value
2 x  7 y  10
2 x  7( 136 )  10
42
2 x  13
 10
88
2 x  13
x
44
13
6
13
in for y in either of the original equations
6. Check (step 6 from algorithm)
2 x  7 y  10
3 x  4 y  12
44
2( 13
)  7( 136 )  10
44
3( 13
)  4( 136 )  12
88
13
42
 13
 10
130
13
 10
132
13
24
 13
 12
156
13
 12
12  12
10  10
The solutions check out, and hence the solution is: x  1344 and y  136 which is another way to notate
44
6
the point ( 13 , 13 ).
6.3 – Solving systems of 2 equations using the Substitution method
As discussed above, the two techniques for solving a system of equations are:
1. Algebraically solving the system using the Addition-Subtraction method
2. Algebraically solving the system using the Substitution method
In this section, the Substitution method is discussed. The following algorithm is used when
implementing this method:
1. If necessary, algebraically manipulate one of the equations so that one of the variables is
isolated
2. Substitute the corresponding expression for the isolated variable into the second equation
3. Solve the resulting linear equation (which will only have 1 variable) to find the first necessary
value
4. Substitute the solution from step 4 into either of the original equations to find the second
necessary value
5. Check the solution by substituting the values into both equations
This algorithm is demonstrated in the following example which uses the equations solved in the sections
above:
Example 1 – Solve the following system of equations using the Substitution method:
y  3x  2
y  12 x  4
Solution:
1. Algebraically manipulate one of the equations to isolate a variable (step 1 from algorithm) – this
step is unnecessary since the equations already have the y variable isolated
2. Substitute the corresponding value in for y (step 2 from algorithm)
y  3x  2
1
2
x  4  3x  2
3. Solve
1
2
x  4  3 x  2 for x (step 3 from the algorithm)
1
2
x  4  3x  2
1
2
x  3 x  2  4
 52 x  6
5 x  12
x  2.4
4. Substitute the value 2.4 in for x in either of the original equations (step 4 from algorithm)
y  3x  2
y  3(2.4)  2
y  7.2  2
y  5.2
5. Check (step 5 from algorithm)
y  3x  2
5.2  3(2.4)  2
5.2  7.2  2
5.2  5.2
y  12 x  4
5.2  12 (2.4)  4
5.2  1.2  4
5.2  5.2
The solutions check out, and hence the solution is: x  2.4 and y  5.2 which is another way to
notate the point (2.4, 5.2). Note that this is the same solution found using the graphing technique
and the addition-subtraction technique above.
In the next example, step 1 is required as neither equation is isolated for a variable. In this case, the
addition-subtraction method is typically the preferred method but any method can be used.
Example 2 – Solve the following system of equations using the Substitution method:
2 x  5 y  27
11x  3 y  41
Solution:
1. Algebraically manipulate one of the equation to isolate a variable
2 x  5 y  27
2 x  5 y  27
x
y  272
5
2
2. Substitute the corresponding value in for x into the other equation (step 2 from algorithm)
11x  3 y  41
11( 52 y  272 )  3 y  41
3. Solve 11( 52 y  272 )  3 y  41 for y
11( 52 y  272 )  3 y  41
55
2
6
82
y  297
2  2 y  2
49
2
y
379
2
49 y  379
y
379
49
4. Substitute the value
2 x  5 y  27
1323
2 x  5( 379
49 )   49
1323
2 x  1895
49   49
2x 
x
572
49
286
49
379
49
in for y in either of the original equations
5. Check (step 5 from algorithm)
2 x  5 y  27
11x  3 y  41
379
2( 286
49 )  5( 49 )  27
379
11( 286
49 )  3( 49 )  41
3146
49
 1137
49  41
 1323
49  27
2009
49
 41
27  27
41  41
572
49
 1895
49  27
The solutions check out, and hence the solution is: x 
379 .
notate the point ( 286
49 , 49 )
286
49
and y 
379
49
which is another way to
6.4 – Applications associated with systems of equations
There are many technological situations where a system of equations is needed to find solutions. The
following provide examples of this situation.
Example 1
A chemist needs 50 mL of an 8% acid solution. She only has a 5% and a 13% solution on hand. How
much of each needs to be combined to obtain the necessarily 8% solution?
Solution:
1. The first step is to determine the necessary relationships in order to obtain the relevant system
of equations. Variables are used to symbolize the needed quantities:
A = the mL of the 5% solution needed
B = the mL of the 13% solution needed
Considering the problem, it can be determined that:
A  B  50
(since the amount of the two solutions must add to 50mL)
0.05 A  0.13B  0.08(50) (since the amount of acid will be constant)
2. Now that the system has been found, it is solved using any technique above. Here the AdditionSubtraction technique is used.
A  B  50
0.05 A  0.13B  0.08(50)
A  B  50
A  2.6 B  80
(this equation was multiplied by 20)
1.6 B  30
(the bottom equation was subtracted from the top equation)
B  18.75
A  18.75  50
(the value of B is substitued in to one of the equations)
A  31.25
Since A represented the amount of the 5% solution, and B represented the amount of the 13%
solution, 31.25 mL of the 5% solution must be mixed with 18.75 mL of the 13% solution.
Example 2
If a concrete mix must have 5 times as much gravel as cement, and enough product must be mixed
to fill a volume of 15 m3, how much gravel and how much cement is needed to make this mix?
Solution:
1. Variables are used to symbolize the needed quantities:
C = m3 of cement needed
G = m3 of gravel needed
Considering the problem, it can be determined that:
C  G  15
G  5C
(since the amount of the two quantities must add to 15m3 )
(since there is 5 times more gravel)
2. Now that the system has been found, it is solved using any technique above. Here the
Substitution method is used since the second equation is already solved for G.
C  G  15
C  5C  15
6C  15
C  2.5
2.5  G  15
G  12.5
(5C is substituted in place of G)
(The value of C determined above is substituted into one of the equations)
Since C represented the amount of cement and G represented the amount of gravel, 2.5 m3 of
cement is needed, and 12.5 m3 of gravel is needed.
6.5 – Systems of 3 linear equations
In certain instances there are three unknown quantities which need to be determined. In this case, three
equations are needed. While the process is more complicated than that for systems of two equations,
the actual steps use similar techniques. The algorithm is as follows:
1. If necessary, algebraically manipulate all equations so they are in the form ax  by  cz  d .
2.
3.
4.
5.
6.
Eliminate one of the variables from any pair of the three equations.
Eliminate the same variables from any other pair of the three equations.
Solve the resulting system of two linear equations using any of the three earlier techniques.
Substitute the two values obtained in step 4 to obtain the last unknown.
Check the solution by substituting the values into all three equations.
Example 1 – Solve the following system of equations:
x  y  z  10
2 x  2 y  5z  4
3x  2 y  5 z  20
Solution:
1. All equations are already in the form ax  by  cz  d (step 1 from algorithm)
2. Eliminate y from the first two equations (step 2 from algorithm)
x  y  z  10
2x  2 y  5z  4
2 x  2 y  2 z  20
2x  2 y  5z  4
4 x  7 z  24
3. Eliminate y from the first and third equations (step 3 from algorithm)
x  y  z  10
3 x  2 y  5 z  20
2 x  2 y  2 z  20
3 x  2 y  5 z  20
5 x  7 z  40
4. Solve the system of two equations formed by combining the equations that resulted from the
previous two steps (step 4 from algorithm)
4 x  7 z  24
5 x  7 z  40
 x  16
x  16
4(16)  7 z  24
64  7 z  24
7 z  40
z   407
5. Substitute the values to find y (step 5 from algorithm)
x  y  z  10
16  y  407  10
112
7
72
7
 y  407  707
 y  707
y   72
6. Check (step 6 from algorithm)
2 x  2 y  5z  4
x  y  z  10
16  
2
7
40
7
 10
16  6  10
10  10
2(16)  2( 72 )  5( 407 )  4
32  74  200
7 4
28
7
4
44
3 x  2 y  5 z  20
3(16)  2( 72 )  5( 407 )  20
48  74  200
7  20
336
7
 74  200
7  20
20  20
The solutions check out, and hence the solution is: x  16, y   72 , and z   407 .
Chapter 6 Homework problems
Section 6.1 Homework
Use the Virtual TI-83 (or any graphing calculator) to find the point of intersection of the following lines.
1.
y  4x  5
y  3x  9
2. Note: For this problem you will need to manipulate the equation to isolate y, and you will need
to change the Window to:
Xmin = 70
Xmax = 80
Ymin = –30
Ymax = –40
y  2 x  168
y  2 x  128
3. Note: For this problem you will need to manipulate the equation to isolate y, and you will need
to change the Window back to the default values:
Xmin = –10
Xmax = 10
Ymin = –10
Ymax = 10
4 y  27.8  2 x
24 y  153.3  3x
4. Enter the following two equations as written below, and graph them to try to determine the
point of intersection.
y  3x  5
y  3x  2
Can you explain what happened in this case?
Section 6.2 Homework
Use the Addition-Subtraction method to solve the following system of equations (note that the first 3
problems are the same as the preceding section and will have the same answers).
1.
y  4x  5
y  3x  9
2.
y  2 x  168
y  2 x  128
3.
4 y  27.8  2 x
24 y  153.3  3x
4.
3a  11b  172
7a  4b  57
Section 6.3 Homework
Use the Substitution method to solve the following system of equations (note that the first 3 problems
are the same as the preceding sections and will have the same answers).
1.
y  4x  5
y  3x  9
2.
y  2 x  168
y  2 x  128
3.
4 y  27.8  2 x
24 y  153.3  3x
4.
4m  30n  12
7m  3n  17
Section 6.4 Homework – use any of the three methods when solving the following problems.
1. E10 is a common gasohol blend, as typically no modifications need to be made to a vehicle’s
engine in order to use this mixture. E10 means the mixture would have 10% ethanol mixed with
90% gasoline.
If 30 liters of E10 are needed, but only E12 and E5 are on hand (these are common mixtures in
other countries), how much of the E12 must be mixed with the E5 in order to obtain the 30 liters
of E10?
Note: E12 would contain 12% ethanol and E5 would contain 5% ethanol.
2. A rectangular yard has been fenced using a total of 344m of fencing. Find the dimensions of the
yard if the length is 14m more than the width.
3. A plane travels 1400 miles with the wind in 5 hours. It makes the return trip in 6 hours.
Assuming the wind speed is constant, find the speed of the wind and the speed of the plane.
4. A circuit has the following relationships between two currents I1 and I 2 (currents are measured
in mA):
17.6 I1  8I 2  11.2
18I1  4 I 2  22
Find I1 and I 2 .
Section 6.5 Homework
Solve the following systems of equations.
2 x  4 y  6 z  14
1. 4 x  2 y  2 z  8
9 x  6 y  6 z  30
3 x  2 y  20
2.  x  2 y  6 z  24
3 x  4 y  4 z  28
3x  2 y  4 z  14
3. x  10 y  5 z  4
5x  4 y  5z  2
Answers to Homework Problems
Chapter 6 Answers
Section 6.1 Homework
1.
x  2 and y  3
2.
x  74 and y  20
3.
x  1.5 and y  6.2
4. The two lines are parallel and hence there is no point of intersection
Section 6.2 Homework
1.
x  2 and y  3
2.
x  74 and y  20
3.
x  1.5 and y  6.2
4.
a  0.937 and b  15.892
Section 6.3 Homework
1.
x  2 and y  3
2.
x  74 and y  20
3.
x  1.5 and y  6.2
4.
m  2.135 and n  0.685
Section 6.4 Homework
1. 21.4 L of E12 and 8.6 L of E10 are needed
2. Length = 93 m and width = 79 m
3. Plane speed = 256.7 mph and wind speed = 23.3 mph
4.
I1  3mA and I 2  8mA
Section 6.5 Homework
1.
x  2, y  1, and z  1
2.
x
3.
139
x   452
23 , y   23 , and z 
28
17
, y
128
17
, and z 
30
17
350
23