Systems of Equations-Solve by Graphing, Comparison, and Substitution
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
y
y
y
 4x  5
 43  5
 12  5
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 