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Question
Solve the system of equations by using the method of cross multiplication:
`x/6 + y/15 – 4 = 0, x/3 - y/12 – 19/4 = 0`
Solution
The given equations may be written as:
`x/6 + y/15 – 4 = 0` …….(i)
`x/3 - y/12 – 19/4 = 0` …….(ii)
Here` a_1 = 1/6 , b_1 = 1/15 , c_1 = -4, a_2 = 1/3 , b_2 = - 1/12 and c_2 = - 19/4`
By cross multiplication, we have:
`∴ x/([1/15 × (−19/4)− (−1/12) ×(−4)]) = y/([(−4) × 1/3− (1/6) × (−19/4)]) = 1/([1/6 × (−1/12) × 1/3 × 1/15])`
`⇒ x/((−19/60− 1/3)) = y/((−4/3+ 19/34)) = 1/((−1/72 − 1/45))`
`⇒ x/((−39/60)) = y/((−13/24)) = 1/((−13/360)`
`⇒x = [(−39/60) × (−360/13)] = 18, y = [(−13/24) × (−360/13)] = 15`
Hence, x = 18 and y = 15 is the required solution.
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