Question

Solve the following by cross multiplication method.

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Answer 1

Cross Multiplication Method:

Let us consider two equations

• a₁x + b₁y + c₁ = 0
• a₂x + b₂y + c₂ = 0

By cross multiplication method, we can get the required solution as

x = (b₁c₂ - b₂c₁)/(a₁b₂ - a₂b₁)

y = (c₁a₂ - c₂a₁)/(a₁b₂ - a₂b₁)

(3)

The given equations are

• 15x - 7y - 66 = 0
• 7x + 2y - 15 = 0

Comparing with the previous equations of the formula, we get

a₁ = 15, b₁ = - 7, c₁ = - 66

a₂ = 7, b₂ = 2, c₂ = - 15

Then,

x = (b₁c₂ - b₂c₁)/(a₁b₂ - a₂b₁)

= {(- 7) (- 15) - 2 (- 66)}/{(15 × 2) - 7 (- 7)}

= (105 + 132)/(30 + 49)

= 237/79

= 3

y = (c₁a₂ - c₂a₁)/(a₁b₂ - a₂b₁)

= {7 (- 66) - 15 (- 15)}/{(15 × 2) - 7 (- 7)}

= (- 462 + 225)/(30 + 49)

= - 237/79

= - 3

∴ the required solution is

x = 3 , y = - 3

(4)

The given equations are

• 7x + 4y - 12 = 0
• 3x + 4y + 4 = 0

Comparing with the previous equations of the formula, we get

a₁ = 7, b₁ = 4, c₁ = - 12

a₂ = 3, b₂ = 4, c₂ = 4

Then,

x = (b₁c₂ - b₂c₁)/(a₁b₂ - a₂b₁)

= {(4 × 4) - 4 (- 12)}/{(7 × 4) - (3 × 4)}

= (16 + 48)/(28 - 12)

= 64/16

= 4

y = (c₁a₂ - c₂a₁)/(a₁b₂ - a₂b₁)

= {3 (- 12) - (4 × 7)}/{(7 × 4) - (3 × 4)}

= (- 36 - 28)/(28 - 12)

= - 64/16

= - 4

∴ the required solution is

x = 4 , y = - 4