Solve the following equations by graphical method- x + y = 4
2x – 3y = 3
For solving these equations by the graphical method, we need to form separate tables for each equation.
We have the equations,
x + y = 4 …(i)
2x – 3y = 3 …(ii)
Take equation (i), we have
x + y = 4
We can write it as,
x = (4 – y) …(iii)
Now, assign values of y and compute values for x.
We can assign values of y = …, -3, -2, -1, 0, 1, 2, 3, 4,…
It is not necessary to put all values. But to form an accurate graph, it is necessary to put at least three values.
For equation (iii):
Say, we put y = 0.
Then, x = 4 – 0
⇒ x = 4
We have, (4, 0).
Now, put y = 1.
Then, x = 4 – 1
⇒ x = 3
We have, (3, 1).
Now, put y = 2.
Then, x = 4 – 2
⇒ x = 2
We have, (2, 2).
We can further find out x by putting values of y = 3, 4, 5,… but here we have just put three values.
Record it in a table,

Now, take equation (ii),
2x – 3y = 3
We can write it as,
2x = 3y + 3
![]()
…(iv)
Assign values for y and compute x.
For equation (iv):
Say, we put y = 0.
Then, ![]()
![]()
![]()
⇒ x = 1.5
We have, (1.5, 0).
Now, put y = 1.
Then, ![]()
![]()
⇒ x = 3
We have, (3, 1).
Now, put y = 2.
Then, ![]()
![]()
⇒ x = 3 × 2
⇒ x = 6
We have, (6, 2).
Record it in a table,
![]()
Represent the two tables on a graph, we get

Notice the intersection point of these two lines, x + y = 4 and 2x – 3y = 3.
These two lines intersect each other at (3, 1).
⇒ (3, 1) is the solution of these equations.
Thus, solution is x = 3 and y = 1.
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