Solve the following systems of equations:
`1/(5x) + 1/(6y) = 12`
`1/(3x) - 3/(7y) = 8, x ~= 0, y != 0`
Solution
Taking 1/x = u and 1/y = v the given equations become
`u/5 + v/6 = 12`
`=> (6u + 5v)/30 = 12`
`=> 6u + 5v = 360` .....(i)
And `u/3 - (3v)/7 = 8`
`=> (7u + 9v)/21 = 8`
=> 7u - 9v = 168 ....(ii)
Let us eliminate ‘v’ from equation (i) and (ii), Multiplying equation (i) by 9 and equation (ii) by 5, we get
54u + 45y = 3240 .......(iii)
35u - 45v = 840 .....(iv)
Adding equation (i) adding equation (ii), we get
54u + 35u = 3240 + 840
=> 89u = 4080
`=> u = 4080/89`
Putting u = 4080/89 in equation (i) we get
`6 xx 4080/89 + 5v = 360`
`=> 24480/89 + 5v = 360`
`=> 5v = 360 - (24480)/89`
`=> 5v = (32040 - 24480)/89`
`=> 5v = 7560/89`
`=> v = 7560/(5 xx 89)`
`=> c = 1512/89`
Hence `x = 1/u = 89/4080` and `y = 1/v = 89/1512`
So, the solution of the given system of equation `x = 89/4080, y = 89/1512`