can someone explain me cross multiplication method?
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General form of a linear equation in two unknown quantities:
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 - c₁a₂
Therefore, 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
This relation informs us how the solution of the simultaneous equations, co-efficient x, y and the constant terms in the equations are inter-related, we can take this relation as a formula and use it to solve any two simultaneous equations. Avoiding the general steps of elimination, we can solve the two simultaneous equations directly.
So, the formula for cross-multiplication and its use in solving two simultaneous equations can be presented as:
If (a₁b₂ - a₂b₁) ≠ 0 from the two simultaneous linear equations
a₁x + b₁y + c₁ = 0 ----------- (i)
a₂x + b₂y + c₂ = 0 ----------- (ii)
we get, by the cross-multiplication method:
x/(b₁c₂ - b₂c₁) = y/(c₁a₂ - c₂a₁) = 1/(a₁b₂ - a₂b₁) ---------- (A)
That means, x = (b₁c₂ - b₂c₁)/(a₁b₂ - a₂b₁)
y = (c₁a₂ - c₂a₁)/(a₁b₂ - a₂b₁)
.....
hope it helps u and if yes then mark it as the brainliest....
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 - c₁a₂
Therefore, 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
This relation informs us how the solution of the simultaneous equations, co-efficient x, y and the constant terms in the equations are inter-related, we can take this relation as a formula and use it to solve any two simultaneous equations. Avoiding the general steps of elimination, we can solve the two simultaneous equations directly.
So, the formula for cross-multiplication and its use in solving two simultaneous equations can be presented as:
If (a₁b₂ - a₂b₁) ≠ 0 from the two simultaneous linear equations
a₁x + b₁y + c₁ = 0 ----------- (i)
a₂x + b₂y + c₂ = 0 ----------- (ii)
we get, by the cross-multiplication method:
x/(b₁c₂ - b₂c₁) = y/(c₁a₂ - c₂a₁) = 1/(a₁b₂ - a₂b₁) ---------- (A)
That means, x = (b₁c₂ - b₂c₁)/(a₁b₂ - a₂b₁)
y = (c₁a₂ - c₂a₁)/(a₁b₂ - a₂b₁)
.....
hope it helps u and if yes then mark it as the brainliest....
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