Solve the following equations by graphical method- 3x + 2y = 0; 2x + y= -1

For solving these equations by graphical method, we need to form separate tables for each equation.

We have the equations,

3x + 2y = 0 …(i)

2x + y = -1 …(ii)

Take equation (i), we have

3x + 2y = 0

We can write it as,

2y = -3x

…(iii)

Now, assign values of x and compute values of y.

We can assign values of x = …, -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 atleast three values.

For equation (iii):

Say, we put x = -4.

Then,

⇒ y = 3 × 2

⇒ y = 6

We have, (-4, 6).

Now, put x = -3.

Then,

⇒ y = 4.8

We have, (-3, 4.8).

Now, put x = -2.

Then,

⇒ y = 3

We have, (-2, 3).

Now, put x = -1.

Then,

⇒ y = 1.5

We have, (-1, 1.5).

We put x = 0.

Then,

⇒ y = 0

We have, (0, 0).

Now, put x = 1.

Then,

⇒ y = -1.5

We have, (1, -1.5).

Now, put x = 2.

Then,

⇒ y = -3

We have, (2, -3).

We can further find out y by putting values of x = 3, 4, 5,… but here we have just put seven values.

Record it in a table,

Now, take equation (ii),

2x + y = -1

We can write it as,

y = -2x – 1 …(iv)

Assign values for x and compute y.

For equation (iv):

Say, we put x = -2.

Then, y = -2(-2) – 1

⇒ y = 4 – 1

⇒ y = 3

We have, (-2, 3).

Now, put x = -1.

Then, y = -2(-1) – 1

⇒ y = 2 – 1

⇒ y = 1

We have, (-1, 1).

Now, we put x = 0.

Then, y = -2(0) – 1

⇒ y = 0 – 1

⇒ y = -1

We have, (0, -1).

Now, put x = 1.

Then, y = -2(1) – 1

⇒ y = -2 – 1

⇒ y = -3

We have, (1, -3).

Now, put x = 2.

Then, y = -2(2) – 1

⇒ y = -4 – 1

⇒ y = -5

We have, (2, -5).

Record it in a table.

Represent the two tables on a graph, we get

Notice the intersection point of these two lines, 3x + 2y = 0 and 2x + y = -1.

These two lines intersect each other at (-2, 3) in the 2^nd quadrant.

⇒ (-2, 3) is the solution of these equations.

Thus, solution is x = -2 and y = 3.