## For a pair of linear equation, a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0. If a₁/a₂ = b₁/b₂ = c₁/c₂, then the lines are​

Question

For a pair of linear equation, a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0. If a₁/a₂ = b₁/b₂ = c₁/c₂, then the lines are​

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1 month 2021-08-12T06:21:55+00:00 2 Answers 0 views 0

then the lines will be coincident

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:

simultaneous equations in two variables, simultaneous equations

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(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