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प्रश्न
Solve each of the following systems of equations by the method of cross-multiplication :
`b/a x + a/b y - (a^2 + b^2) = 0`
x + y - 2ab = 0
उत्तर
The given system of the equation may be written as
`b/a x + a/b y - (a^2 + b^2) = 0``
x + y - 2ab = 0
Here
`a_1 = b/a, b_1 =- a/b,c_1 = -(a^2 + b^2)`
`a_2 = 1, b_2 = 1, c_2 = -2ab`
By cross multiplication, we have
`x/(-2ab xx a/b + a^2 + b^2) = (-y)/(-2ab xx a/b + a^2 + b^2) = 1/(b/a - a/b)`
`=> x/(-2a^2 + a^2 + b^2) = (-y)/(-2b^2 + a^2 + b^2) = 1/((b^2 - a^2)/(ab))`
`=> x/(b^2 - a^2) = (-y)/(-b^2 + a^2) = 1/((b^2 - a^2)/(ab))`
Now
`x/(b^2 - a^2) = 1/((b^2 - a^2)/(ab))`
`=> x = b^2 - a^2 aa (ab)/(b^2 - a^2)`
=> x = ab
And
`(-y)/(-b^2 + a^2) = 1/((b^2 - a^2)/(ab))`
`=> -y = b^2 + a^2 xx (ab)/(b^2 - a^2)`
`=> -y =-(b^2 - a^2) xx (ab)/(b^2 - a^2)``
=> -y = -ab
=> y = ab
Hence, x = ab,y = ab is the solution of the given system of equations.
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