Question

Solve the following pair of linear equations:

`4/(x + 2) - 6/(y - 1) = -1, 2/(x + 2) + 3/(y - 1) = 1`

Answer 1

Given the pair of linear equations:

`4/(x + 2) - 6/(y - 1) = -1, 2/(x + 2) + 3/(y - 1) = 1`

Step 1: Introduce substitutions

Let: `u = 1/(x + 2), v = 1/(y - 1)`

Rewrite the equations:

4u – 6v = –1   ...(1) 

2u + 3v = 1   ...(2)

Step 2: Solve for (u) and (v)

From equation (2),

Multiply both sides by 2:

4u + 6v = 2   ...(3)

Step 3: Add equations (1) and (3)

(4u – 6v) + (4u + 6v) = –1 + 2

8u = 1

`u = 1/8`

Step 4: Substitute `(u = 1/8)` into equation (2)

`2 xx 1/8 + 3v = 1 1/4 + 3v`

`2 xx 1/8 + 3v = 1`

`3v = 1 - 1/4`

`3v = 3/4`

`v = 1/4`

Step 5: Back-substitute for (x) and (y)

`u = 1/(x + 2)`

`u = 1/8`

⇒ x + 2 = 8

⇒ x = 6

`v = 1/(y - 1)`

`v = 1/4`

⇒ y – 1 = 4

⇒ y = 5