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प्रश्न
Solve each of the following systems of equations by the method of cross-multiplication
5ax + 6by = 28
3ax + 4by = 18
उत्तर
The given system of equation is
5ax + 6by = 28
`=> 5ax + 6by - 28 = 0` .....(i)
and 3ax + 4by - 18 = 0
=> 3ax + 4by - 18 = 0 ....(ii)
Here
`a_1 = 5a , b_1 = 6b, c_1 = -28`
`a_2 = 3a, b_2 = 4b, c_2 = -18`
By cross multiplication we have
`= x/(6b xx (-18) - (-28) xx 4b) = (-4)/(5a xx (-18) - (-28) xx 3a) = 1/(5a xx 4b - 6b xx 3a)`
`=> x/(-108b + 112b) = (-y)/(-90a + 80a) = 1/(20ab - 18ab)`
`=> x/(4b) = (-y)/(-6a) = 1/(2ab)`
Now
`x/(4b) = (-y)/(-6a) = 1/(2ab)`
Now
`x/(4b) = 1/(2ab)`
`=> x = (5b - 2a)/10ab`
And
`(-y)/(-6a) = 1/(2ab)`
`=> y = (6a)/(2ab) = 3/b`
Hence `x = 2/a, y = 3/b` is the soluytion of the given system of equation.
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