Can anyone teach me crossmultiplication for class 10
Answers
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)
Note:
If the value of x or y is zero, that is, (b₁c₂ - b₂c₁) = 0 or (c₁a₂ - c₂a₁) = 0, it is not proper to express in the formula for cross- multiplication, because the denominator of a fraction can never be 0.
From the two simultaneous equations, it appears that the formation of relation (A) by cross-multiplication is the most important concept.
At first, express the co-efficient of the two equations as in the following form:
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) Note: If the value of x or y is zero, that is, (b₁c₂ - b₂c₁) = 0 or (c₁a₂ - c₂a₁) = 0, it is not proper to express in the formula for cross- multiplication, because the denominator of a fraction can never be 0. From the two simultaneous equations, it appears that the formation of relation (A) by cross-multiplication is the most important concept. At first, express the co-efficient of the two equations as in the following form: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)
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)