You will no longer feel drawing 2 or more linear equations on a graph difficult anymore with our article. Get a detailed procedure to represent simultaneous equations graphically easily. Furthermore, have a look at the example questions provided below to get clarity on the topic and solve related problems.
To solve a pair of simultaneous equations graphically, we first draw two equations on the graph. Those two equations form straight lines intersecting each other at a common point. This common point gives the solution of the pair of simultaneous equations.
How to Solve Simultaneous Linear Equations Graphically?
Let us take two first-order linear equations
p₁x + q₁y + r₁ = 0
p₂x + q₂y + r₂ = 0
Draw these two lines on a coordinate graph. These lines always form a straight line on the graph. Suppose L₁ represent the graph of p₁x + q₁y + r₁ = 0 and L₂ represent the graph of p₂x + q₂y + r₂ = 0. By solving the simultaneous equations graphically we will get three possible solutions. They are as follows.
1. When the two lines L₁, L₂ interest at a single point
- Here, two lines meet at a point called p (x, y).
- So, the obtained point x coordinate, y coordinate is the unique solution of the given linear equations.
- This system is called independent.
2. When two lines L₁, L₂ are coincident
- Here, two equations represent the single line.
- Therefore, the equations have infinitely many solutions.
- This system is called dependent.
3. When two lines L₁, L₂ are parallel to each other
- The two equations have no common solutions.
- This system is called inconsistent.
Examples of Simultaneous Equations Graphically
1. Solve this system of equations by graphing: y = x + 1 and x + y = 5
Solution:
Given linear equations are y = x + 1 and x + y = 5
On a graph paper to solve simultaneous equations graphically, draw a horizontal line X’OX and a vertical line YOY’ as the x-axis and y-axis respectively.
The first equation y = x + 1 is in the slope intercept form y = mx + c [m = 1, c = 1]
Now apply the trial and error method to get the 3 pairs of points (x, y) which satisfy the equation y = x + 1.
If the value of x = -1 then y = -1 + 1 = 0
If the value of x = 0 then y = 0 + 1 = 1
If the value of x = 1 then y = 1 + 1 = 2
If the value of x = 2 then y = 2 + 1 = 3
Arrange these values of the linear equation y = x + 1 in the table
| x | -1 | 0 | 1 | 2 |
|---|---|---|---|---|
| y | 0 | 1 | 2 | 3 |
Plot the points of the equation y = x + 1; A (-1, 0), B (0, 1), C (1, 2), D (2, 3) on the graph.
Join the points A, B, C, and D to get the line equation AD.
So, the line AD represents the equation y = x + 1.
Convert the second equation x + y = 5 into the slope-intercept form.
y = 5 – x [ here, m =-1, c = 5]
Apply the trial and error method to get the points that satisfy the given equation.
If the value of x = -1 then y = 5 – (-1) = 5 + 1 = 6
If the value of x = 0 then y = 5 – 0 = 5
If the value of x = 1 then y = 5 – 1 = 4
If the value of x = 2 then y = 5 – 2 = 3
Arrange the values of the linear equation in the table.
| x | -1 | 0 | 1 | 2 |
|---|---|---|---|---|
| y | 6 | 5 | 4 | 3 |
Plot the points P (-1, 6), Q (0, 5), R (1, 4), S (2, 3) on the graph
Join the points to get a straight line of x + y = 5
We get two straight lines intersecting each other at (2, 3).
Therefore, x = 2 and y = 3 is the solution of the given system of equation.
2. Solve graphically the system if linear equation y = 2x = 4 and y + 2x = 1.
Solution:
Given two linear equations are y + 2x = 4 and y + 2x = 1
Convert the first equation y + 2x = 4 into the slope intercept form.
y = 4 – 2x [ here m = -2, c = 4]
Apply the trial and error method to get the points on the line.
If the value of x = -1 then y = 4 – 2 (-1) = 4 + 2 = 6
If the value of x = 0 then y = 4 – 2 (0) = 4 – 0 = 4
If the value of x = 1 then y = 4 – 2 (1) = 4 – 2 = 2
If the value of x = 2 then y = 4 – 2 (2) = 4 – 4 = 0
Arrange these values of the linear equation y = 4 – 2x in the table.
| x | -1 | 0 | 1 | 2 |
|---|---|---|---|---|
| y | 6 | 4 | 2 | 0 |
Now plot the points A (-1, 6), B (0, 4), C (1, 2), D (2, 0) on the graph.
Join the points to get the linear equation y + 2x = 4.
Convert the second equation y + 2x = 1 into the slope intercept form.
y = 1 – 2x [m = -2, c = 1]
Apply the trial and error method to get the points on the equation.
When the value of x = -1 then y = 1 – 2(-1) = 1 + 2 = 3
When the value of x = 0 then y = 1 – 2(0) = 1 – 0 = 1
When the value of x = 1 then y = 1 – 2(1) = 1 – 2 = -1
When the value of x = 2 then y = 1 – 2(2) = 1 – 4 = -3
Arrange these values of the equation in the table.
| x | -1 | 0 | 1 | 2 |
|---|---|---|---|---|
| y | 3 | 1 | -1 | -3 |
Plot the points P (-1, 3), Q (0, 1), R (1, -1), S (2, -3) on the graph.
Join the points to form a linear equation y + 2x = 1
We get from the graph that two straight lines are parallel to each other.
Therefore, the given system of equations has no solutions.
3. Solve this system of equations by graphing: x – 2y = 2 and 2y + 2 = x.
Solution:
Given linear equations are x – 2y = 2 and 2y + 2 = x.
Convert the first equation into y = mx + c form
y = x / 2 – 1 [ here m = 1/2, c = -1]
Apply the trial and error method to find 3 pairs of values of (x, y) which satisfy the given equation x – 2y = 2.
If the value of x = -1 then y = -1/2 – 1 = -3/2
If the value of x = 0 then y = -0/2 – 1 = -1
If the value of x = 1 then y = 1/2 – 1 = -1/2
Arrange these values in the table
| x | -1 | 0 | 1 |
|---|---|---|---|
| y | -3/2 | -1 | -1/2 |
Now plot the points of the equation A (-1, -3/2) B (0, -1), C (1, -1/2)
Join the points A, B, C to get the graph line x – 2y = 2
Convert the second equation 2y + 2 = x into the slope-intercept form.
y = ½ (x – 2) [ m = 1/2, c = -1]
Apply the trial and error method to find 3 pairs of values of (x, y) which satisfy the given equation 2y + 2 = x.
If the value of x = -1 then y = ½ (-1 – 2) = -3/2
If the value of x = 0 then y = ½ (0 – 2) = -1
If the value of x = 1 then y = ½ (1 – 2) = -1/2
Arrange these values in the table
| x | -1 | 0 | 1 |
|---|---|---|---|
| y | -3/2 | -1 | -1/2 |
Now plot the points of the equation 2y + 2 = x; A (-1, -3/2) B (0, -1), C (1, -1/2) on the graph paper.
Join the points of A, B, and C; to get the graph line AC.
Thus line AC is the graph of 2y + 2 = x.
We find that the two straight lines coincide.
Therefore, the given system of equations has an infinite number of solutions.