Question

Solve the following pair of linear equations:

`10/(x + 1) - 2/(y - 1) = 3/2, 5/(x + 1) + 8/(y - 1) = 3`

Answer 1

Given:

`10/(x + 1) - 2/(y - 1) = 3/2, 5/(x + 1) + 8/(y - 1) = 3`

Step 1: Introduce substitutions

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

Rewriting the system with these substitutions:

`10u - 2v = 3/2` 

5u + 8v = 3

Step 2: Solve the linear system in (u) and (v)

Multiply the first equation by 2 to clear the fraction:

20u – 4v = 3 

Second equation remains:

5u + 8v = 3

Multiply the second equation by 4:

20u + 32v = 12

Now subtract the first equation from this result:

(20u + 32v) – (20u – 4v) = 12 – 3 

20u + 32v – 20u + 4v = 9 

36v = 9

⇒ `v = 9/36`

⇒ `v = 1/4`

Put `(v = 1/4)` in the first original equation:

`10u - 2 xx 1/4 = 3/2` 

`10u - 1/2 = 3/2` 

`10u = 3/2 + 1/2`

10u = 2 

`u = 2/10`

`u = 1/5`

Step 3: Find values of (x) and (y)

Recall the substitutions:

`u = 1/(x + 1)`

`u = 1/5`

⇒ x + 1 = 5

⇒ x = 4

`v = 1/(y - 1)`

`v = 1/4`

⇒ y – 1 = 4

⇒ y = 5