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

1/2(3x+y)-1/2(3x-y)=-1/8

Solve for x and y

Open in App

Solution

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

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

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

also ,

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

41/4 - 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
31 + y = 4

y = 4 - 3 = 1

Hence,

x = 1 , y = 1

Suggest Corrections

Similar questions

Q. Evaluate :

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

Solve the following pairs of equations by reducing them to a pair of linear equations:
$\frac{1}{3 x + y} + \frac{1}{3 x - y} = \frac{3}{4}$
$\frac{1}{2 (3 x + y)} - \frac{1}{2 (3 x - y)} = \frac{- 1}{8}$

Question 1 (viii)
Solve the following pairs of equations by reducing them to a pair of linear equations:
$\frac{1}{3 x + y} + \frac{1}{3 x - y} = \frac{3}{4}$
$\frac{1}{2 (3 x + y)} - \frac{1}{2 (3 x - y)} = \frac{- 1}{8}$

Q. Evaluate:
(i)

\frac{4}{x - 1} + \frac{5}{y - 1} = 2, \frac{8}{x - 1} + \frac{15}{y - 1} = 3, x \neq 1, y \neq 1

(ii)

\frac{1}{3 x + y} + \frac{1}{3 x - y} = \frac{3}{4}, \frac{1}{2 (3 x + y)} - \frac{1}{2 (3 x - y)} = \frac{- 1}{8}, 3 x + y \neq 0, 3 x - y \neq 0

Q. Solve the following simultaneous equations:

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

Join BYJU'S Learning Program

1/3y+ y + 1/3x-y=3/4 1/2(3x+y)-1/2(3x-y)=-1/8 Solve for x and y

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

1/2(3x+y)-1/2(3x-y)=-1/8

Solve for x and y