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प्रश्न
Solve each of the following systems of equations by the method of cross-multiplication :
6(ax + by) = 3a + 2b
6(bx - ay) = 3b - 2a
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उत्तर
The given system of equation is
6(ax + by) = 3a + 2b .....(i)
6(bx - ay) = 3b - 2a .....(ii)
From equation (i), we get
6ax + 6by - (3a + 2b) = 0 .....(iii)
From equation (ii), we get
6bx - 6ay - (3b - 2a) = 0 .....(iv)
Here
`a_1 = 6a, b_1 = 6b,c_1 = -(3a + 2b)`
`a_2 = 6b, b_2 = -6a, c_2 = -(3b - 2a)`
By cross multiplication, we have
`x/(-6b(3b - 2a) - 6a (3a + 2b)) = (-y)/(-6a(3b - 2a) + 6b(3a + 2b)) = 1/(-36a^2 - 36b^2)`
`=> x/(-18b^2 + 12ab - 18a^2 - 12ab) = (-y)/(-18an + 12a^2 + 18ab + 12b^2) = 1/(-36a^2 - 36b^2)`
`=> x/(-18a^2 - 18b^2) = (-y)/(12a^2 + 12b^2) = 1/(-36(a^2 + b^2))`
`=> x/(-18(a^2 + b^2)) = (-y)/(12(a^2 + b^2)) = (-1)/(36(a^2 + b^2))`
Now
`x/(-18(a^2 + b^2)) = (-1)/(36(a^2 + b^2))`
`=> x = (18(a^2 + b))/(36(a^2 + b^2))`
= 1/2
And
`(-y)/(12(a^2+b^2)) = (-1)/(36(a^2 + b^2))`
`=> y = (12(a^2 + b^2))/(36(a^2 + b^2))`
y = 1/3
Hence x = 1/2, y = 1/3 is the solution of the given system of equations.
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