# Solve 1/(x+y)+2/(x−y)=2 and 2/(x+y)−1/(x−y)=3 where, x + y ≠ 0 and x – y ≠ 0 - Mathematics

Sum

Solve \frac{1}{x+y}+\frac{2}{x-y}=2\text{ and }\frac{2}{x+y}-\frac{1}{x-y}=3 where, x + y ≠ 0 and x – y ≠ 0

#### Solution

Taking \frac { 1 }{ x+y } = u and \frac { 1 }{ x-y } = v

the above system of equations becomes

u + 2v = 2 ….(1)

2u – v = 3 ….(2)

Multiplying equation (1) by 2, and (2) by 1, we get;

2u + 4v = 4 ….(3)

2u – v = 3 ….(4)

Subtracting equation (4) from (3), we get

5v = 1 ⇒ v = \frac { 1 }{ 5 }

Putting v = 1/5 in equation (1), we get;

u + 2 × \frac { 1 }{ 5 } = 2 ⇒ u = 2 – \frac { 2 }{ 5 } = \frac { 8 }{ 5 }

Here, u = \frac { 8 }{ 5 } = \frac { 1 }{ x+y } ⇒ 8x + 8y = 5 ….(5)

And, v = \frac { 1 }{ 5 } = \frac { 1 }{ x-y } ⇒ x – y = 5 ….(6)

Multiplying equation (5) with 1, and (6) with 8, we get;

8x + 8y = 5 ….(7)

8x – 8y = 40 ….(8)

Adding equation (7) and (8), we get;

16x = 45 ⇒ x = \frac { 45 }{ 16 }

Now, putting the above value of x in equation (6), we get;

\frac { 45 }{ 16 } – y = 5 ⇒ y = \frac { 45 }{ 16 } – 5 = \frac { -35}{ 16 }

Hence, solution of the system of the given equations is ;

x = \frac { 45 }{ 16 } , y = \frac { -35 }{ 16 }

Concept: Equations Reducible to a Pair of Linear Equations in Two Variables
Is there an error in this question or solution?