Question

Solve for x and y
1/3x+y +1/3x-y= 3/4

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

Plzz.,give me the answer fast tomorrow is my xam ...plzzz

Answer 1

hope this helps you ........

Answer 2

Answer: The value of x =1 and y = 1 by solving the given equations.

Given

$\begin{array}{l}{\frac{1}{3 x+y}+\frac{1}{3 x-y}=\frac{3}{4} \ldots \ldots(1)} \\ {\frac{1}{2(3 x+y)}-\frac{1}{2(3 x-y)}=\frac{-1}{8} \ldots \ldots(2)}\end{array}$

Solution:

Multiply eqn (2) by 2, it becomes,

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

Add eqn (1) and (3),

$\frac{2}{3 x+y}=\frac{2}{4}$

Hence 3x + y = 4 ….. (4)

Subtract eqn (1) and (3)

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

3x – y = 2 …… (5)

Add eqn (4) and (5), we get 6x = 6,

Hence x = 1, substitute in eqn (4)

3(1) + y = 4

y = 4 – 3 = 1

x=1,y=1