Q3 of 92 Page 71

Let us solve the following simultaneous equations by the method of comparison and check whether the solutions satisfy the equations.


expressing the equations in terms of one variable.

………[1]


………[2]



⇒


⇒ 2xy + 3 = 5y


⇒ 2xy = 5y – 3


………[3]


by eq. 2



⇒ 5xy – 2 = 3y


⇒ 5xy = 3y + 2


………[4]


on comparing eq. [3] and eq. [4]



⇒ 5y(5y – 3) = 2y(3y + 2)


⇒ 25y2 – 15y = 6y2 + 4y


cancelling one power of y from both sides


⇒ 25y – 15 = 6y + 4


⇒ 19y = 19


⇒ y = 1


by eq. [3]




⇒ x = 1


∴ solution of this pair is x = 1, y = 1


verifying by eq. [1]



L.H.S



= 5 = R.H.S


by eq. [2]



L.H.S



= 3 = R.H.S


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