Question

solve for X and Y
$\frac{1}{x + 1} + \frac{1}{y + 1} = 10 \\ and \\ \frac{1}{x + 1} - \frac{1}{y + 1} = 4$

Answer 1

2/x+1 = 14( add both)

x+1)/2 = 1/14

x+1 = 1/7

x= -6/7

1/x+1 + 1/y+1 = 10

7 + 1/y+1= 10

1/y+1= 3

y+1 = 1/3

y= -2/3

Answer 2

$\huge \boxed{ \bf{ Answer}}$


$\frac{1}{x + 1} + \frac{1}{y + 1} = 10 \: \: and \: \: \frac{1}{x + 1} - \frac{1}{y + 1}$ = 4


Let,

$\frac{1}{x + 1} = a \: \: and \: \: \frac{1}{y + 1} = b$


So, replacing the value we get,

=> 1(a) + 1(b) = 10 -------->[1]

=> 1(a) - 1(b) = 4 -------->[2]

Now, [1] + [2]

=> a + b + a - b = 10 + 4

=> 2a = 14

=> a = 7


Now, putting the value of 'a' in equation [1] we get,

=> a + b = 10

=> 7 + b = 10

=> b = 10 - 7

=> b = 3

So we got the value of 'a' and 'b


Now, as we have taken earlier

$= > \: \frac{1}{x + 1} = a \: \: and \: \: \frac{1}{y + 1} = b$


so by putting the value of 'a' and 'b' we get,

$= > \: \frac{1}{x + 1} = 7$

$= > \: 7 + 7x = 1$

$= > x = \frac{ - 6}{7}$


and


$= > \frac{1}{y + 1} = 3$

$= > \: 1 = 3y + 3$

$= > y = \frac{ - 2}{3}$



Hence , the value is

$\boxed{ \bf{x = \frac{ - 6}{7} \: \: and \: \: y = \frac{ - 2}{3} }}$