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प्रश्न
Solve the following systems of equations:
`1/(3x + y) + 1/(3x - y) = 3/4`
`1/(2(3x + y)) - 1/(2(3x - y)) = -1/8`
उत्तर
The given equation is
`1/(3x + y) + 1/(3x - y) = 3/4`
`1/(2(3x + y)) - 1/(2(3x - y)) = -1/8`
Let `1/(3x + y) = u and 1/(3x - y) = v` then equation are
`u + v = 3/4` ...(i)
`u/2 - v/2 = 1/8` ....(ii)
Multiply equation (ii) by 2 and add both equations, we get
`u + v = 3/4`
`u - v = -1/4`
`2u = 1/2`
`u = 1/4`
Put the value of u in equation (i) we get
`1 xx 1/4 + v = 3/4`
`v= 1/2`
Then
`1/(3x + y) = 1/4` ....(iii)
3x + y = 4
`1/(3x -y) = 1/2` ...(iv)
3x - y = 2
Add both equation we get
3x + y = 4
3x - y = 2
_________
6x = 6
x = 1
Put the value of x in equation (iii) we get
3 x 1 + y = 4
y = 1
Hence value of x = 1 and y = 1
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