Question

find the value of x and y,148/x+231/y=527/xy ,231/x+148/y=610/xy ,where x and y is not equal to zero

Answer 1

please mark as brainliest answer

Answer 2

The value of x and y are 2 and 1 respectively where x and y are not equal to zero.

Given:

$\frac { 148 }{ x } +\frac { 231 }{ y } =\frac { 527 }{ xy }$

To find:

The value of x and y

Solution:

$\frac { 148 }{ x } +\frac { 231 }{ y } =\frac { 527 }{ xy } \quad \quad \longrightarrow (1)$

$\frac { 231 }{ x } +\frac { 148 }{ y } =\frac { 610 }{ xy } \quad \quad \longrightarrow (2)$

$(1) \rightarrow 148y + 231x = 527$

$(2) \rightarrow 231y + 148x = 610$

Adding (1) and (2)

$379x + 379y =1137$

$379 (x+y) =1137$

$x + y = 3\quad \quad \longrightarrow (3)$

Subtracting (1) and (2)

$-83 (x-y)= -83$

$x - y =1 \quad \quad \longrightarrow (4)$

Adding (3) and (4)

$2x =4$

$x =2; y=1$