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

● **Simultaneous Linear Equations**

**Solvability of Linear Simultaneous Equations**

**Word Problems on Simultaneous Linear Equations**

**Word Problems on Simultaneous Linear Equations**

**Practice Test on Word Problems Involving Simultaneous Linear Equations**

● **Simultaneous Linear Equations - Worksheets**

**Worksheet on Simultaneous Linear Equations**

**Worksheet on Problems on Simultaneous Linear Equations**

**8th Grade Math Practice****From Solvability of Linear Simultaneous Equations to HOME PAGE**

**Didn't find what you were looking for? Or want to know more information
about Math Only Math.
Use this Google Search to find what you need.**

## New! Comments

Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.