Question

Solve the following pair of equations by cross multiplication method.

11x + 5y = 7, 6x − 3y = 21

Answer 1

Given equations:

11x + 5y = 7

11x + 5y − 7 = 0 ...(1)

6x − 3y = 21

6x − 3y − 21 = 0 ...(2)

Let’s write equations in standard form:

a₁x + b₁y₁ + c₁ = 0

a₂x + b₂y + c₂ = 0

Here, they are in the form of,

a₁ = 11, b₁ = 5, c₁ = −7

a₂ = 6, b₂ = −3, c₂ = −21

Using the identity:

`x/(b_1c_2 - b_2c_1) = y/(c_1a_2 - c_2a_1) = 1/(a_1b_2 - a_2b_1)`

Now, substituting the values,

⇒ b₁c₂ − b₂c₁

= (5)(−21) − (−3)(−7)

= −105 − 21

∴ b₁c₂ − b₂c₁ = −126

⇒ c₁a₂ − c₂a₁

= (−7)(6) − (−21)(11)

= −42 + 231

∴ c₁a₂ − c₂a₁ = 189

⇒ a₁b₂ − a₂b₁

= (11)(−3) − (6)(5)

= −33 − 30

∴ a₁b₂ − a₂b₁ = −63

So, the value becomes,

`x/-126 = y/189 = 1/-63`

Hence, finding x and y,

`x/-126 = 1/-63`

`x = (-126)(1/-63)`

`x = (-126)/-63`

∴ x = 2

`y/189 = 1/-63`

`y = (189)(1/-63)`

`y = (189)/-63`

∴ y = −3

Thus, solving equations 11x + 5y = 7, 6x − 3y = 21 by cross multiplication method we get, x = 2 and y = −3.