# Solvability of Linear Simultaneous Equations

To understand the condition for solvability of linear simultaneous equations in two variables, if linear simultaneous equations in two variables have no solution, they are called inconsistent whereas if they have solution, they are called consistent.

In the method of cross-multiplication, for the simultaneous equations,

a₁x + b₁y + c₁ = 0 --------- (i)

a₂x + b₂y + c₂ = 0 --------- (ii)

we get: x/(b₁ c₂ - b₂ c₁) = y/(a₂ c₁ - a₁ c₂) = 1/(a₁ b₂ - a₂ b₁)

that is, x = (b₁ c₂ - b₂ c₁)/(a₁ b₂ - a₂ b₁) , y = (a₂ c₁ - a₁ c₂)/(a₁ b₂ - a₂ b₁) --------- (iii)

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

(1) If (a₁ b₂ - a₂ b₁) ≠ 0 for any values of (b₁ c₂ - b₂ c₁) and (a₂ c₁ - a₁ c₂), we get unique solutions for x and y from equation (iii)

For examples:

7x + y + 3 = 0 ------------ (i)

2x + 5y – 11 = 0 ------------ (ii)

Here, a₁ = 7, a₂ = 2, b₁ = 1, b₂ = 5, c₁ = 3, c₂ = -11

and (a₁ b₂ - a₂ b₁) = 33 ≠ 0 from equation (iii)

we get, x = -26/33 , y = 83/33

Therefore, (a₁ b₂ - a₂ b₁) ≠ 0, then the simultaneous equations (i), (ii) are always consistent.

(2) If (a₁ b₂ - a₂ b₁) = 0 and one of (b₁ c₂ - b₂ c₁) and (a₂ c₁ - a₁ c₂) is zero (in that case, the other one is also zero), we get,

a₁/a₂ = b₁/b₂ = c₁/c₂ = k (Let) where k ≠ 0

that is, a₁ = ka₂, b₁ = kb₂ and c₁ = kc₂ and changed forms of the simultaneous equations are

ka₂x + kb₂y + kc₂ = 0

a₂x + b₂y + c₂ = 0

But they are two different forms of the same equation; expressing x in terms of y, we get

x = - b₂y + c₂/a₂

Which indicates that for each definite value of y, there is a definite value of x, in other words, there are infinite number of solutions of the simultaneous equations in this case?

For examples:

7x +   y + 3 = 0

14x + 2y + 6 = 0

Here, a₁/a₂ = b₁/b₂ = c₁/c₂ = 1/2

Actually, we get the second equation when the first equation is multiplied by 2. In fact, there is only one equation and expressing x in term of y, we get:

x = -(y + 3)/7

Some of the solutions in particular:

(3) If (a₁ b₂ - a₂ b₁) = 0 and one of (b₁ c₂ - b₂ c₁) and (a₂ c₁ - a₁ c₂) is non-zero (then the other one is also non-zero) we get,

(let) k = a₁/a₂ = b₁/b₂ ≠ c₁/c₂

That is, a₁ = ka₂ and b₁ = kb₂

In this case, the changed forms of simultaneous equations (i) and (ii) are

ka₂x + kb₂y + c₁ = 0 ………. (v)

a₂x + b₂y + c₂ = 0 ………. (vi)

and equation (iii) do not give any value of x and y. So the equations are inconsistent.

At the time of drawing graphs, we will notice that a linear equation in two variables always represents a straight line and the two equations of the forms (v) and (vi) represent two parallel straight lines. For that reason, they do not have any common point.

For examples:

7x + y + 3 = 0

14x + 2y - 1 = 0

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

and a₁/a₂ = b₁/b₂ ≠ c₁/c₂

So, the given simultaneous equations are inconsistent.

From the above discussion, we can arrive at the following conclusions that the solvability of linear simultaneous equations in two variables

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

(1) Consistent if a₁/a₂ ≠ b₁/b₂: in this case, we will get unique solution

(2) Inconsistent, that is, there will be no solution if

a₁/a₂ = b₁/b₂ ≠ c₁/c₂ where c₁ ≠ 0, c₂ ≠ 0

(3) Consistent having infinite solution if

a₁/a₂ = b₁/b₂ = c₁/c₂ where c₁ ≠ 0, c₂ ≠ 0

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Cross-Multiplication Method

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Worksheet on Problems on Simultaneous Linear Equations