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Question
Solve the following pairs of equations by reducing them to a pair of linear equations
`1/(3x+y) + 1/(3x-y) = 3/4`
`1/(2(3x-y)) - 1/(2(3x-y)) = (-1)/8`
Solution
`1/(3x+y) + 1/(3x-y) = 3/4`
`1/(2(3x-y)) - 1/(2(3x-y)) = (-1)/8`
Putting `1/(3x+y) = p ` in the given equations, we get
`p + q = 3/4 ... (i)`
`p/2 - q/2 = (-1)/8`
`p - q = (-1)/4 ... (ii)`
Adding (i) and (ii), we get
`2p = 3/4 - 1/4`
2p = 1/2
p = 1/4
Putting the value in equation (ii), we get
1/4 - q = -1/4
`q = 1/4 + 1/4 = 1/2`
`p = 1/(3x+y) = 1/4`
3x + y = 4 ... (iii)
`q = 1/(3x-y) = 1/2`
3x - y = 2 ... (iv)
Adding equations (iii) and (iv), we get
6x = 6
x = 1 ... (v)
Putting the value in equation (iii), we get
3(1) + y = 4
y = 1
Hence, x = 1 and y = 1
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