Solvability of Linear Simultaneous Equations | How to Determine if a Linear Equation is Solvable or Not?

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.

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