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Question
Solve each of the following systems of equations by the method of cross-multiplication :
(a + 2b)x + (2a − b)y = 2
(a − 2b)x + (2a + b)y = 3
Solution
The given system of equations may be written as
(a + 2b)x + (2a − b)y - 2 = 0
(a − 2b)x + (2a + b)y - 3 = 0
Here,
`a_1 = a + 2b, b_1 = 2a - b, c_1 = -2`
`a_2 = a - 2b, b_2 = 2a + b, c_2 = -3`
By cross multiplication, we have
`=> x/(-3(2a - b) - (-2)(2a + b)) = (-y)/(3(a + 2b) - (-2)(a - 2b)) = 1/((a + 2b)(2a + b)-(2a - b)(a - 2b))`
`=> x/(-6a + 3b +4a + 2b) = (-y)/(-3a - 6b + 2a - 4b) = 1/(2a^2 + ab + 4ab + 2b^2 -(2a^2 - 4ab - ab + 2b^2))`
`=> x/(-2a + 5b) = (-y)/(-a - 10b) = 1/(2a^2 + ab+ 4ab + 2b^2 - (2a^2 - 4ab - ab + 2b^2))`
`=> x/(-2a + 5b) = (-y)/(-a + 10b) = 1/(10ab)`
`=> x/(-2a + 5b) = y/(a+10b) = 1/(10ab)`
Now
`x/(-2a + 5b) = 1/(10ab)`
`=> y = (a + 10b)/(10ab)`
And
`y/(a + 10b) = 1/(0ab)`
`=> y = (a + 10b)/(10ab)`
Hence `x = (5b - 2a)/(10ab), y = (a + 10b)/(10ab)` is the solution of the given system of equations
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