Question

solve this


Find x and y

$\frac{1}{3x + y} + \frac{1}{3x - 1} = \frac{3}{4}$
$\frac{1}{2(3x + y)} - \frac{1}{2(3x - y)} = \frac{ - 1}{8}$

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Answer 1

Answer:

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Refers with attachments

Step-by-step explanation:

Answer:-

$x = 1 \\ \\ and \ \\ y = 1$

Answer 2

Answer:

+

3x−y

1

=

4

3

Let 1/3x + y = a , and 1/3x - y = b ,

a + b = 3/4 -----------------> (1)

4a + 4b = 3 -------------> (1)

also ,

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

2(3x+y)

1

−

2(3x−y)

1

=−1/8

a/2 - b/2 = -1/8

(a - b)/2 = -1/8

a - b = -1/4

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

Adding equations (1) and (2) :-

4a + 4b = 3

4a - 4b = -1

--------------------

8a = 2

a = 2/8

a = 1/4

4a - 4b = -1

4*1/4 - 4b = -1

1 - 4b = -1

4b = 1 + 1

4b = 2

b = 2/4

b = 1/2

1/3x + y = a

1/3x + y = 1/4

3x + y = 4 -----------------> (3)

1/3x - y = b

1/3x - y = 1/2

3x - y = 2 -----------------> (2)

Adding equations (3) and (4) :-

3x + y = 4

3x - y = 2

--------------------------

6x = 6

x = 1

3x + y = 4

3*1 + y = 4

y = 4 - 3 = 1

Hence,

x = 1 , y = 1