Question

if (1,1) is the solution of the equation px+qy+(a-b)=0 and px+qy+(b-c)=0 then find a+c

Answer 1

Given :  (1,1) is the solution of the equation px+qy+(a-b)=0 and px+qy+(b-c)=0  

To Find : a+c

Solution:

px+qy+(a-b)=0  

(1,1) is the solution

=> p + q  + (a - b)  = 0  

px+qy+(b-c)=0  

(1,1) is the solution

=> p + q + ( b - c)  = 0

Equating both

p + q  + (a - b)   =   p + q + ( b - c)

=> a - b  = b  - c

=> a + c  = 2b

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Answer 2

Given :- if (1,1) is the solution of the equation px+qy+(a-b)=0 and px+qy+(b-c)=0 then find a+c ?

Answer :-

putting x = 1 and y = 1 in first given equation, we get,

→ px+ qy + (a - b) = 0

→ p * 1 + q * 1 + (a - b) = 0

→ p + q + a - b = 0

→ a - b = -(p + q) ------ Eqn.(1)

and, putting x = 1 and y = 1 in second given equation, we get,

→ px+ qy + (b - c) = 0

→ p * 1 + q * 1 + (b - c) = 0

→ p + q + b - c = 0

→ b - c = -(p + q) ------ Eqn.(2)

comparing both equations we get,

→ a - b = b - c

→ a + c = b + b

→ a + c = 2b (Ans.)

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