Solving Simultaneous Equations Graphically | Graphical Method Examples with Solution

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.