Learn the conditions for the solvability of a system of linear equations in two variables in the following sections. If the simultaneous equations have no solution, then they are called inconsistent and if they have a solution, then they are called consistent. To help you understand the fundamentals of the concept we have given the Solved Examples with Answers explaining everything in detail.

## Conditions for Solvability of Simultaneous Linear Equations in Two Variables

Let the pair of linear equations in two variables are

a₁x + b₁y + c₁ = 0 ——- (1)

a₂x + b₂y + c₂= 0 ——- (2)

By using the cross-multiplication method, we get

x/(b₁c₂ – b₂c₁) = y/(a₂c₁ – a₁c₂) = 1/(a₁b₂ – a₂b₁)

So, x = (b₁c₂ – b₂c₁) / (a₁b₂ – a₂b₁)

y = (a₂c₁ – a₁c₂) / (a₁b₂ – a₂b₁)

Now see when the solvability of simultaneous linear equations in two variables (i), (ii) are solavble.

(I) If (a₁b₂ – a₂b₁) ≠ 0 for any values of (b₁c₂ – b₂c₁), (a₂c₁ – a₁c₂) we will get the unique solutions for the x and y variables.

Example:

x – 2y – 8 = 0 ——- (1)

x + y – 5 = 0 ——— (2)

Here, a₁ = 1, b₁ = -2, c₁ = -8, a₂ = 1, b₂ = 1, c₂ = -5

Substitute these values in (a₁b₂ – a₂b₁), we get

(a₁b₂ – a₂b₁) = (1 x 1 – 1 x -2) = 1 + 2 = 3 ≠ 0

So, x = (b₁c₂ – b₂c₁) / (a₁b₂ – a₂b₁)

= (-2 x -5 – 1 x -8) / 3 = (-10 + 8)/3 = 18/3 = 6

y = (a₂c₁ – a₁c₂) / (a₁b₂ – a₂b₁)

= (1 x -8 – 1 x -5)/3 = (-8 + 5)/3 = -3/3 = -1

Therefore, when (a₁b₂ – a₂b₁) ≠ 0, then the system of linear equations are always consistent.

(II) If (a₁b₂ – a₂b₁) = 0 and any one of (b₁c₂ – b₂c₁) and (a₂c₁ – a₁c₂) is zero, then

Let us take a₁/a₂ = b₁/b₂ = c₁/c₂ = k, where k ≠ 0

Then, a₁ = ka₂, b₁ = kb₂, c₁ = kc₂

And system of linear equations are changed to

ka₂x + kb₂y + kc₂ = 0

a₂x + b₂y + c₂ = 0

From this equation, we can write

x = (-b₂y – c₂) / a₂

This indicates for each value of y, there is a definite value of x or there are infinite number of solutions of the simultaneous equations in this case.

Example:

x + 2y + 5 = 0

3x + 6y + 15 = 0

Here, a₁/a₂ = b₁/b₂ = c₁/c₂ = 1/3

Actually, we get the second equation by multiplying the first equation by 3. Express one equation x interms of y

x = -2y -5

Some of the solutions are

y -1 0 1 2 . . . . . . .
x -3 -5 -7 -9 . . . . .

(III) If (a₁b₂ – a₂b₁) = 0 and any one from (b₁c₂ – b₂c₁) and (a₂c₁ – a₁c₂) is non zero, (then other one is also non-zero)

Let us take k = a₁/a₂ = b₁/b₂ ≠ c₁/c₂ = k

So, a₁ = ka₂, b₁ = kb₂

In this case, the chnaged linear equations are

ka₂x + kb₂y + c₂ = 0

a₂x + b₂y + c₂= 0

These equations do not have solution for x and y.

So the equations are inconsistent.

While drawing graphs, we can see that the linear equations in two variables always represent a straight line and the equations of the above form represent two parallel straight lines. This is why these lines do not intersect each other and not have a common point.

Example:

7x + y + 3 = 0

14x + 2y – 1 = 0

Here, a₁ = 7, b₁ = 1, c₁ = 3, a₂ = 14, b₂ = 2, c₂ = -1

And, a₁/a₂ = b₁/b₂ ≠ c₁/c₂

So, the given system of linear equations is inconsistent.

From the above discussions, we can write that

a₁x + b₁y + c₁ = 0, a₂x + b₂y + c₂= 0 will be

(i) Consistent if a₁/a₂ ≠ b₁/b₂, we will get a unique solution

(ii) Inconsistent, there will be no solution if a₁/a₂ = b₁/b₂ ≠ c₁/c₂ where c₁ ≠ 0, c₂ ≠ 0

(iii) Consistent and have infinite solutions if a₁/a₂ = b₁/b₂ = c₁/c₂ where c₁ ≠ 0, c₂ ≠ 0

### Solvability of Linear Equations Examples

Example 1.

Check whether the linear equations x + 2y + 4 = 0, 3x – 5y + 1 = 0 are consistent or inconsistent?

Solution:

Given system of linear equations are

x + 2y + 4 = 0

3x – 5y + 1 = 0

Here, a₁ = 1, b₁ = 2, c₁ = 4, a₂ = 3, b₂ = -5, c₂ = 1

(a₁b₂ – a₂b₁) = (1 x -5 – 3 x 2) = (-5 – 6) = -11

x = (b₁c₂ – b₂c₁) / (a₁b₂ – a₂b₁)

= (2 x 1 – (-5) x 4) / -11 = (2 + 20)/-11

= -22/11 = -2

y = (a₂c₁ – a₁c₂) / (a₁b₂ – a₂b₁)

= (3 x 4 – 1 x 1) / -11 = (12 – 1)/-11 = -11/11

= -1

Here, (a₁b₂ – a₂b₁) ≠ 0

Therefore, the given simultaneous linear equations are consistent and have a unique solution.

Example 2.

Check whether the linear simultaneous equations 3x – y – 2 = 0, 6x – 2y – 4 = 0 are consistent or inconsistent?

Solution:

Given Linear Simultaneous Equations are

3x – y – 2 = 0, 6x – 2y – 4 = 0

Here, a₁ = 3, b₁ = -1, c₁ = -2, a₂ = 6, b₂ = -2, c₂ = -4

(a₁b₂ – a₂b₁) = (3 x -2 – 6 x -1) = (-6 + 6) = 0

and (b₁c₂ – b₂c₁) = (-1 x -4 – (-2) x -2) = (4 – 4) = 0

(a₂c₁ – a₁c₂) = (6 x -2 – 3 x (-4)) = (-12 + 12) = 0

a₁/a₂ = b₁/b₂ = c₁/c₂ = 1/2

Actually, we get the second equation by multiplying the first equation by 2. Express one equation x in terms of y

x = (y + 2) / 3

Some of the solutions are

y -1 0 1 2 . . .
x 1/3 2/3 1 4/3 . . .

Therefore, the given linear equations are consistent and have an infinite number of solutions.

Example 3.

State whether the linear equations -2x + 3y – 2 = 0, 2x – 3y – 5 = 0 are consistent or not?

Solution:

Given linear equations are -2x + 3y – 2 = 0, 2x – 3y – 5 = 0

Here, a₁ = -2, b₁ = 3, c₁ = -2, a₂ = 2, b₂ = -3, c₂ = -5

(a₁b₂ – a₂b₁) = (-2 x -3 – 2 x 3) = (6 – 6) = 0

(b₁c₂ – b₂c₁) = (3 x -5 – (-3) x (-2)) = (-15 – 6) = -21

a₁/a₂ = b₁/b₂ = -1

The equations are inconsistent and has no solution.