Download Systems of Equations: Introduction

Systems of Equations
Solving a system of linear equations means to determine all the points that are common to all the
lines in that system of equations. In Grade 10 students learned how to solve a 2x2 system of
linear equations by several methods. These methods were:
a) comparison
b) graphing
c) substitution
In Grade 11, students will expand these methods to include addition and subtraction
(elimination) and the inverse matrix. Before introducing these new methods, a review of the
previously learned methods is necessary.
Examples:
1. Solve the following system of linear equations by the comparison method.
 x  4 y  1 


 4 x  5 y  15
Solution:
Solve each equation in terms of the same variable. Usually students
will solve the equations in terms of the variable y since they have done
this when determining the equation of a line and when graphing.
Express each equation in the form y  mx  b
x  4 y  1
x  x  4 y  x 1
 4y   x 1
 4 y  1x  1


1
1 1
4 y  x 1
4
1
1
y  x
4
4
4
1
1
y  x
4
4
 4 x  5 y  15
 4 x  4 x  5 y  4 x  15
5 y  4 x  15
5
4
15
y  x
5
5
5
4
y  x3
5
Since both equations are equal to y, they are equal to each other.
Set both mx + b parts equal to each other and solve the linear equation.
1
1 4
x  x3
4
4 5
20 1 x  20 1  20 4 x  203
4
4
5
5x  5  16x  60
5x  16x  5  16x  16x  60
 11x  5  60
 11x  5  5  60  5
 11x  55
 11
55
x
 11
 11
x  5
Substitute this value into one of the equations to determine the value of y.
4
x3
5
4
y   5  3
5
y  4  3
y
The common point or the solution is (-5,-1)
y  1
2. Solve the following system of linear equations by graphing.
2


2 x  5 y  5 
y  x 1 
5


 Put each equation in the form y  mx  b 
4 x  2 y  22
 y  2 x  11
Graph both lines on the same axis:
The point of intersection of the two lines is the common point or the solution.
The solution is (5, 1).
3. Solve the following system of linear equations by using the substitution method.
2 x  3 y  27 


4 x  y  5 
Solution:
Solve one of the equations in terms of one of the variables. The
process of substitution will involve fewer and simpler computations if
one of the variables has a coefficient of 1 or -1. Since the coefficient
of y is -1, solve for that variable.
4x  y  5
4 x  4 x  y  4 x  5
 y  4 x  5
1
4
5
y
x
1
1
1
y  4x  5
Substitute this expression for y into the other equation wherever the variable y appears. Solve
the linear equation.
2 x  3 y  27
2 x  34 x  5  27
2 x  12 x  15  27
14 x  15  27
14 x  15  15  27  15
14 x  42
14
42
x
14
14
x3
Substitute this value into the equation to determine the value of y.
y  4x  5
y  43  5
y  12  5
y7
The common point or the solution is (3, 7)
Exercises:
1. Solve the following system of linear equations by using the method of comparison.
 x  3 y  15  0


7 y  3x  21

2. Solve the following system of linear equations by using the method of substitution.
 x  3 y  11


3x  2 y  30
3. Solve the following system of linear equations by using the method of graphing.
2 x  y  7  0


x  2 y  1  0 
Solving a system of linear equations by adding and subtracting (elimination).
To solve a system of linear equations by elimination requires one of the variables, in both
equations, to have opposite numerical coefficients. This can be achieved by multiplying one or
both of the equations by an appropriate value. When this is done, the equations are then
combined. This will eliminate the variable with the opposite coefficients and will leave one
variable in the answer equation. This equation must be solved. The solution will then be
substituted into one of the given equations of the system to determine the value of the variable
that was eliminated. To facilitate this procedure, the system should be aligned as
a1 x  b1 y  c1 


a2 x  b2 y  c2 
Example1: Solve the following system of linear equations by using the method of
elimination.
3x  5 y  12 


2 x  10 y  4  The system is properly aligned.
Choose one variable and perform the necessary multiplication to produce opposite coefficients
for that variable in both equations.
Solution:
23x  5 y  12  6x – 10y = 24 This equation was multiplied by 2 in
2 x  10 y  4
2x + 10y = 4
order to change the coefficient of y
to a negative 10. This makes the
8x
= 28 coefficients of y opposites. When the
8
8
equations were combined (added),
the y’s were eliminated.
x = 3.5
Substitute this value into one of the given equations of the system to determine the value of y.
2(3.5) + 10y = 4
7 + 10y = 4
7 – 7 + 10y = 4 – 7
10y = -3
10
10
y = – 0.3
The common point or the solution is (3.5, – 0.3)
Example 2:
Solve the following system of linear equations by using the method of
elimination.
2 x  9 y  1


4 x  y  15 

 22 x  9 y  1

4 x  y  15
 4 x  18 y  2
4 x  y  15
 17 y  17
 17
17
y
 17
 17
y  1
Substitute this value into one of the given equations of the system to determine the value of x.
4 x  y  15
4 x   1  15
4 x  1  15
4 x  1  1  15  1
The common point or the solution is (4,-1)
4 x  16
4
16
x
4
4
x4
Example 3: Solve the following system of linear equations by using the method of
elimination.
3x  2 y  9  0
4x  3y  5
3x  2 y  9  0

  3x  2 y  9  9  0  9 4 x  3 y  3 y  3 y  5
4 x  3 y  5 
3x  2 y  9
4x  3y  5
3 x  2 y  9
4x  3y  5
The system is properly aligned.
33 x  2 y  9 
24 x  3 y  5

9 x  6 y  27
8 x  6 y  10
17 x  17
17 x  17

17
17
x  1
Substitute this value into one of the given equations of the system to determine the value of y.
3x  2 y  9  0
3 1  2 y  9  0
 3  2y  9  0
2y  6  0
2y  6  6  0  6
The common point or the solution is (-1,-3)
2 y  6
2
6
y
2
2
y  3
Example 4: Solve the following system of linear equations by using the method of
elimination.
2
3

20 3 x  20 2 y  202
 4 x  5 y  2 
4
5


 
1
3
113
1
3
113
 x y 

14 x  14 y  14
 7
7
2
7
2
7 
 215 x  8 y  40

152 x  21y  226
 30 x  16 y  80
30 x  315 y  3390
331y  3310
331
3310
y
331
331
y  10
15 x  8 y  40
2 x  21y  226
Substitute this value into one of the given equations of the system to determine the value of x.
3
2
x  10   2
4
5
3
x4 2
4
3
x44  24
4
3
x6
4
4 3 x  46
4
3 x  24
The common point or the solution is (8,10)
3
24
x
3
3
x8
Exercises: Solve the following system of linear equations by using the method of
elimination.
16 x  y  181  0
1. 

19 x  y  214 
3
26 

x  y  
4. 
5
5
4 y  61  7 x 
 x  69  6 y 
2. 

3x  4 y  45
 x  7 y  38 
3. 

14 y   x  46
Solutions:
 x  3 y  15  0
1. 

7 y  3x  21

Comparison Method
 x  3 y  15  0
 x  x  3 y  15  0  x
3 y  15  15  x  15
7 y  3x  21
7 y  3x  3x  3x  21
7 y  3x  21
7
3
21
y  x
7
7
7
3
y  x3
7
3 y  x  15
3
1
15
y  x
3
3
3
1
y  x5
3
1
3
x5 x3
3
7
1 
3 
21 x   215  21 x   213
3 
7 
7 x  105  9x  63
7 x  105  105  9x  63  105
7 x  9x  42
7 x  9x  9x  9x  42
 2x  42
2
 42
x
2
2
x  21
1
x5
3
1
y  21  5
3
y  75
y  12
y
The solution is (21, 12)
 x  3 y  11
2. 

3x  2 y  30
x  3 y  11
x  3 y  3 y  3 y  11
x  3 y  11
Substitution Method
x  3 y  11
x  3 9   11
x  27  11
x  16
3x  2 y  30
3 3 y  11  2 y  30
 9 y  33  2 y  30
 7 y  33  30
 7 y  33  33  30  33
 7 y  63
7
63
y
7
7
y  9
The solution is (16,-9).
2 x  y  7  0
3. 

x  2 y  1  0 
2x  y  7  0
2x  2x  y  7  0  2x
y  7  2 x
y  7  7  2 x  7
y  2 x  7
x  2y 1  0
x  x  2y 1  0  x
 2 y 1  x
 2 y 1  1  x  1
 2 y  x  1
2
1
1
y
x
2
2
2
1
1
y  x
2
2
Graphing Method
Following is the graph of y  2 x  7 and y 
1
1
x .
2
2
The solution is the point of intersection (3, 1).
Solutions: Elimination Method
16 x  y  181  0
1. 

19 x  y  214 
16 x  y  181  0
16 x  y  181  181  0  181
16 x  y  181
16 x  y  181
19 x  y  214
 16 x  y  181
19 x  y  214
3x  33
3
33
x
3
3
x  11
The solution is (11, -5).

 1(16 x  y  181)
19 x  y  214
16 x  y  181
16(11)  y  181
176  y  181
176  176  y  181  176
1
5
y
1
1
y  5
 x  69  6 y 
2. 

3x  4 y  45
x  69  6 y
3 x  4 y  45
x  6 y  69  6 y  6 y
3 x  4 y  45  4 y
x  6 y  69
3 x  4 y  45
x  6 y  69
3 x  4 y  45

 3( x  6 y  69)
3 x  4 y  45

 3x  18 y  207
3x  4 y  45
14 y  252
14
 252
y
14
14
y  18

x  69  6 y
x  69  6(18)
x  69  108
x  39
The solution is (-39, -18)
 x  7 y  38 
3. 

14 y   x  46
x  7 y  38
14 y   x  46
x  7 y  7 y  7 y  38
14 y  x   x  x  46
x  7 y  38
x  14 y  46
x  7 y  38
x  14 y  46

2( x  7 y  38)
x  14 y  46

2 x  14 y  76
x  14 y  46
3x  30
3
30
x
3
3
x  10

x  7 y  38
10  7 y  38
10  38  7 y  38  38
 28  7 y
 28 7
 y
7
7
4 y
The solution is (10, -4).
3
26 

x  y  
4. 
5
5
4 y  61  7 x 
3
26
y
5
5
3 
 26 
5 x   5 y   5 
5 
 5 
5 x  3 y  26
x
5 x  3 y  26
7 x  4 y  61

4 y  61  7 x
4 y  7 x  61  7 x  7 x
7 x  4 y  61
4(5 x  3 y  26)
3(7 x  4 y  61)

20 x  12 y  104
21x  12 y  183
41x  287
41
287
x
41
41
x7
The solution is (7, 3)
4 y  61  7 x
4 y  61  7(7)
4 y  61  49
4 y  12
4
12
y
4
4
y3