Question

$\frac{1}{3x+y}$ + $\frac{1}{3x-y}$ = $\frac{3}{4}$
$\frac{1}{2(3x+y)}$ - $\frac{1}{2(3x-y)}$ = $\frac{-1}{8}$
Find the values of X and Y with full solution.

Answer 1

(1)

2x

1

+

3y

1

=2

3x

1

+

2y

1

=

6

13

3x

1

+

2y

1

=

6

13

x

3

−4y=23

x+y

8x+7y

=15

x+y

15

−

x−y

5

=−2

(ii)

x

2

+

y

3

=2

(iv)

x−1

5

+

y

1

=−1

(iv) 6x+3y=6xy

(vii)

3x+y

1

+

3x−y

1

=

4

3

2(3x+y)

1

−

2(3x−y)

1

=

8

−1

Answer 2

$\color{orchid}\frac{1}{3x + y} + \frac{1}{3x - y} = \frac{3}{4}$

Let $\underline{\boxed{\sf\purple{1/3x + y = a ,}}}$ and $\underline{\boxed{\sf\purple{1/3x - y = b ,}}}$

$\sf\color{darkgreen}{a + b = 3/4}$ (1)

$\sf\color{darkgreen}{4a + 4b = 3}$ (1)

$\bold\color{red}{Also ,}$

$\color{purple}\frac{1}{2 ( 3x + y)} - \frac{1}{2(3x - y)} = -1/8$

$\frac{a}{2}$ - $\frac{b}{2}$= $\frac{-1}{8}$

$\frac{(a - b)}{2}$ $\frac{-1}{8}$

$\sf\color{blue}{a - b}$ =$\frac{-1}{4}$

$\underline{\boxed{\sf\purple{4a - 4b = -1 }}}$$\sf\color{red}{(2)}$

$\small\textbf\color{green}{Adding equations (1) and (2) :-}$

$\sf\color{orange}{4a + 4b = 3}$

$\sf\color{orange}{4a - 4b = -1}$

$\bold{8a = 2}$

$\sf \: a =\cancel\dfrac { 2 } { 8 }$

$\underline{\boxed{\sf\purple{a = 1/4}}}$

$\sf\color{orange}{4a - 4b = -1}$

$\color{orange} 4\times 1/4$$\sf\color{orange}{- 4b = -1}$

$\sf\color{orange}{1 - 4b = -1}$

$\sf\color{orange}{4b = 1 + 1}$

$\sf\color{orange}{4b = 2}$

$\sf \: b =\cancel\dfrac { 2 } { 4 }$

$\bold{b\: =\: 1/2}$

$\frac{1}{3x}$ + y = a

$\frac{1}{3x}$1/3x + y = $\frac{1}{4}$

$\sf\color{orange}{3x + y = 4}$$\sf\color{red}{(3)}$

$\frac{1}{3x}$ - y = b

$\frac{1}{3x} - y = 1/2$

$\underline{\boxed{\sf\purple{3x - y = 2}}}$$\sf\color{red}{(2)}$

$\small\textbf\color{orchid}{Adding equations (3) and (4) :-}$

$\sf\color{orange}{3x + y = 4}$

$\sf\color{orange}{3x - y = 2}$

$\sf\color{orange}{6x = 6}$

$\underline{\boxed{\sf\purple{x = 1}}}$

$\sf\color{orange}{3x + y = 4}$

$\color{orange}3 \times 1$$\sf\color{orange}{+ y = 4}$

$\sf\color{orange}{y = 4 - 3 = 1}$

$\bold{Hence,}$

$\color{purple}\underline{\underline{ \tt{{ \boxed {x = 1 , y = 1} }}}}$

Thankyou :)