Step-by-step explanation:
Let
{1/(x+2y)} = u -----Eqn(1)
{1/(x-2y)} = v ------Eqn(2)
So, given equation can be rewritten as
⇛6u + 5v = -3 ---Eqn(3)
⇛3u + 7v = -6 -----Eqn(4)
On multiply equation (4) by 2, we get
⇛6u + 14v = -12 -----Eqn(5)
On subtracting equation (3) from equation (5), we get
⇛9v = -9
⇛v = -9/9
⇛v = -1 ----Eqn(6)
Put value of v in equation (3), we get
6u + 5v = -3
⇛6u + 5 * (-1) = -3
⇛6u -5 = -3
⇛6u = -3 + 5
⇛6u = 2
⇛u = 2/6
⇛u = (2÷2)/(6÷2)
⇛u = 1/3 -----Eqn(7)
Now, substituting value of u and v in equation (1) and equation (2), we get
{1/(x+2y)} = 1/3 and {1/(x-2y)} = -1
⇛x + 2y = 3 -----Eqn(8)
⇛x - 2y = -1 -------Eqn(9)
Now, add equation (8) and equation (9) , we get
⇛2x = 2
⇛x = 2/2
<h3>⇛x = 1 -----Eqn(10)</h3>
On substituting x = 1 in equation (8), we get
x + 2y = 3
⇛ 1 + 2y = 3
⇛2y = 3 - 1
⇛2y = 2
⇛y = 2/2
<h3>⇛y = 1 ------Eqn(11)</h3>
Therefore, x = 1 and y = 1
<u>Answer</u><u>:</u> Hence the value of x and y is 1 and 1.
