Solve the following system of equations: `4/x+3y=8,\ \ \ 6/x-4y=-5`

Options

A. `x = 1` , `y = 2`
B. `x = 2` , `y = 1`
C. `x = 2` , `y = 2`
D. `x = 1` , `y = 1`

Answer

C

Explanation

Putting ` (1)/(x) = u`, the given equations become :
` 4u + 3y = 8 " " `… (i)
` 6u - 4y = - 5 " " ` … (ii)
Multiplying (i) by 4 and (ii) by 3 and adding the results, we get
` 16 u + 18 u = 32 - 15 `
` rArr 34 u = 17" " `... (iii)
` rArr u = (17)/(34) = (1)/(2)`
Puttinhg `u = (1)/(2) ` in (i), we get
` ( 4 xx (1)/(2)) + 3y = 8 rArr 2 + 3y = 8 rArr 3y = 8 - 2 rArr 3y = 6 rArr y = 2`.
Now , `u = (1)/(2) rArr (1)/(x) = (1)/(2) rArr x = 2`.
Hence, ` x = 2 and y = 2 `.