Solve the following system of equations: `5/(x-1)+1/(y-2)=2,\ \ \ \ 6/(x-1)-3/(y-2)=1`

Options

A. `x = 2` and `y = 3`
B. `x = 4` and `y = 5`
C. `x = 6` and `y = 7`
D. `x = 8` and `y = 9`

Answer

B

Explanation

Putting `(1)/(x - 1)= y and (1)/(y - 2) = v `, the given equations become
` 5u + v = 2 " " `… (i)
` 6u - 3v = 1 " " `… (iii)
Multiplying (i) by 3 and adding the result with (ii) we get ` 15 u + 6u = 6 + 1 `
` rArr 21 u = 7`
` rArr u = ( 7)/( 21) = (1)/(3)`
Putting `u = (1)/(3)` in (i) , we get
` ( 5 xx (1)/(3)) + v = 2 rArr (5)/(3) + v = 2 rArr v = 2 - (5)/(3) rArr v = (1)/(3). `
Now, ` u = (1)/(3) rArr (1)/(x- 1 ) = (1)/(3) rArr x - 1 = 3 rArr x = 4 ` [ by cross multiprication ]
Similarly, `v = (1)/(3) rArr (1)/( y - 2) = (1)/(3) rArr y - 2 = 3 rArr y = 5 ` [ by cross multiplication ]
Hence, ` x = 4 and y = 5`.