#### Question

Solve the following systems of equations:

`5/(x + 1) - 2/(y -1) = 1/2`

`10/(x + 1) + 2/(y - 1) = 5/2` where `x != -1 and y != 1`

#### Solution

Let `1/(x + 1) = u and 1/(y - 1) = v`

Then, the given system of equations becomes

`=> 5u - 2v = 1/2` .....(i)

`=> 10u + 2y = 5/2 ....(2)`

Adding equation (i) equation (ii), we get

`5u + 10u = 1/2 + 5/2`

`=> 15u = (1 + 5)/2`

`=> 15u = 6/2 = 3`

`=> u = 3/15 = 1/5`

Putting `u = 1/5` in equation (i) we get

`5 xx 1/5 - 2v = 1/2`

`=> 1 - 2v = 1/2`

`=> -2v = 1/2 - 1`

`=> -2v = (1 - 2)/2`

`=> -2v = (-1)/2`

`=> v = (-1)/(-4) = 1/4`

Now `u = 1/(x + 1)`

`=> 1/(x + 1) = 1/5`

=> x + 1 = 5

=> x = 5 -1 =4

And `v = 1/(y - 1)`

`=> 1/(y -1) = 1/4 `

`=> y - 1 = 4 `

=> y = 4 + 1 = 5

Hence, solution of the give system of equation is x = 4, y = 5

Is there an error in this question or solution?

#### APPEARS IN

Solution Solve the Following Systems of Equations: `5/(X + 1) - 2/(Y -1) = 1/2` `10/(X + 1) + 2/(Y - 1) = 5/2` Where `X != -1 and Y != 1` Concept: Algebraic Methods of Solving a Pair of Linear Equations - Substitution Method.