Question

Solve the following pairs of linear equation:
(3/x+y)+(2/x-y)=3
(2/x+y)+(3/x-y)=11/3

Answer 1

Step-by-step explanation:

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Answer 2

$\frac{3}{x+y}+\frac{2}{x-y}=3\\ \\\frac{2}{x+y}+\frac{3}{x-y} =\frac{11}{3}$

Let 1/(x + y) be a and 1/(x - y) be b.

$3a+2b=3--(1)\\\\2a+3b=\frac{11}{3}\\ \\6a+9b=11--(2)$

Multiplying (1) equation by 2 and then subtracting both the equations :

$6a+4b=6\\6a+9b=11\\-------\\-5b=-5\\b=1$

Putting value of b in (1) equation :

$3a+2(1)=3\\3a=3-2\\a= \frac{1}{3}$

Now we know that :

$\frac{1}{x+y}=a\\ \\\frac{1}{x+y}=\frac{1}{3} \\ \\x+y=3----(3)\\\\also,$

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

Adding equations (3) and (4) :

$x + y=3\\x-y=1\\-------\\2x =4\\x=2$

Putting value of x in (4) equation :

$x-y=1\\2-y=1\\-y=-1\\y=1$

Hence, value of x is 2 and y is 1.