Question

write the condition for solvability for the fair of linear equation which is consistent

Answer 1

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

Answer 2

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