Question

x/3-2/x+y=1
X/4+3/X+y=3

Answer 1

$\dfrac{x}{3} - \dfrac{2}{x+y}=1.$



$\mathsf{Let \: \: \dfrac{1}{x + y}= a}$



$\dfrac{x}{3} - (2 \times \dfrac{1}{x+y})=1 \\ \\ \\ \dfrac{x}{3} - 2a=1$

$\dfrac{x -6a}{3} = 1 \\ \\ \\ x - 6a = 3 \\ \\x = 3 + 6a \: \: \: \: \: \: \: \: \: \bold{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ...( \: 1 \: )}$




$\mathsf{ \text{And doing same in } \: \dfrac{x}{4} + \dfrac{3}{x+y} = 3}$


$\dfrac{x}{4} + 3a = 3 \\ \\ \\ \dfrac{x + 12a}{4} = 3 \\ \\ \\ x + 12a =12\\ \\ x = 12 - 12a\: \: \: \: \: \: \: \: \: \bold{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ...( \: 2\: )}$



Comparing both the values of x from ( 1 ) and ( 2 ) ,



3 + 6a = 12 - 12a

12a + 6a = 12 - 3

18a = 9

$\dfrac{9}{18} = a \\ \\ \dfrac{1}{2} = a$



Putting the value of a in ( 1 ),



$x = 3 + 6( \dfrac{1}{2} ) \\ \\ \\ x = 3 + 3 \\ \\ x = 6$




In the beginning, we assumed 1 / x + y as a ,so


$\dfrac{1}{x + y} = a \\ \\ \\ \frac{1}{x+ y} = \frac{1}{2} \\ \\ \\ \frac{1}{y + 6} = \frac{1}{2} \\ \\ y + 6 = 2 \\ \\ y = - 4$

Answer 2

x / 3 - 2 / x + y = 1


Let 1 / x + y = z


Therefore, x / 3 - 2 / x +y = x / 3 - 2z

And, x / 4 + 3 / x + y = x / 4 + 3z



Then,

x / 3 - 2z = 1 ---- 1

x / 4 + 3z = 3 ----2



In 1 ,

x / 3 - 2z = 1

x - 6z = 3

x = 3 + 6z ----3



In 2,

x / 4 + 3z = 3

x + 12z = 12

x = 12 - 12z --- 4




eq 3 = eq 4

x = x

3 + 6z = 12 - 12z

1 + 2z = 4 - 4z

- 6z = - 3

z = 1 / 2

1 / x + y = 1 / 2

x + y = 2



substituting the value of x + y or z in 3

x = 3 + 6( 1 / 2 )

x = 3 + ( 6 × 1 / 2 )

x = 3 + 3

x = 6



x + y = 2

6+ y = 2

y = - 4


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