Question

activity aim : verify the condition of of solvability for a pair of linear equation graphilly​

Answer 1

Answer:

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)

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.

(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

Step-by-step explanation:

hope this helps u,have a nice day :)