Question

ax+ bx+c=0
eliminate x by comparision method​

Answer 1

Answer:

1) ax+by+c=0

2) bx+ay+b=0

Multiplying 1) and 2) by b and a respectively.

abx+b²y+ac=0

abx+a²y+ab=0

- - -

Subtracting

(b²-a²)y+a(c-b)=0

(b²-a²)y=-a(c-b)

y=-a(c-b)/b²-a²←

Substitiuting in 1)

ax+b[-a(c-b)/b²-a²]=0

ax+[-ab(c-b)/b²-a²]=0

ax=-[-ab(c-b)/b²-a²]=0

ax=ab(c-b)/b²-a²

x=b(c-b)/b²-a²

Answer 2

Answer:

ax + by + c = 0, (a, b ≠ 0)

Two such equations can be written as:

a₁x + b₁y + c₁ = 0 ----------- (i)

a₂x + b₂y + c₂ = 0 ----------- (ii)

Let us solve the two equations by the method of elimination, multiplying both sides of equation (i) by a₂ and both sides of equation (ii) by a₁, we get:

a₁a₂x + b₁a₂y + c₁a₂ = 0

a₁ a₂x + a₁b₂y + a₁c₂ = 0

Subtracting, b₁a₂y - a₁b₂y + c₁a₂ - c₂a₁ = 0

or, y(b₁ a₂ - b₂a₁) = c₂a₁ - c₁a₂Therefore, y = (c₂a₁ - c₁a₂)/(b₁a₂ - b₂a₁) = (c₁a₂ - c₂a₁)/(a₁b₂ - a₂b₁) where (a₁b₂ - a₂b₁) ≠ 0

Therefore, y/(c₁a₂ - c₂a₁) = 1/(a₁b₂ - a₂b₁), ------------- (iii) Again, multiplying both sides of (i) and (ii) by b₂ and b₁ respectively, we get;

a₁b₂x + b₁b₂y + b₂c₁ = 0

a₂b₁x + b₁b₂y + b₁c₂ = 0

Subtracting, a₁b₂x - a₂b₁x + b₂c₁ - b₁c₂ = 0

or, x(a₁b₂ - a₂b₁) = (b₁c₂ - b₂c₁)

or, x = (b₁c₂ - b₂c₁)/(a₁b₂ - a₂b₁)

Therefore, x/(b₁c₂ - b₂c₁) = 1/(a₁b₂ - a₂b₁) where (a₁b₂ - a₂b₁) ≠ 0 -------------- (iv)

From equations (iii) and (iv), we get:

x/(b₁c₂ - b₂c₁) = y/(c₁a₂) - c₂a₁ = 1/(a₁b₂ - a₂b₁) where (a₁b₂ - a₂b₁) ≠ 0