Question

Solve the following pair of linear equations:

`8/x + 1/y = 3, 16/x - 3/y = 1, x ≠ 0, y ≠ 0`

Answer 1

Given:

`8/x + 1/y = 3, 16/x - 3/y = 1, x ≠ 0, y ≠ 0`

Step-wise calculation:

1. Put `(1/x = u)` and `(1/y = v)`.

Then the system becomes:

8u + v = 3   ...(1) 

16u – 3v = 1   ...(2)

2. Multiply equation (1) by 3 to align the coefficients of (v):

3 × (8u + v) = 3 × 3 

⇒ 24u + 3v = 9   ...(3)

3. Add equations (2) and (3):

(16u – 3v) + (24u + 3v) = 1 + 9

`40u = 10u`

`40u = 10/40`

`40u = 1/4`

4. Substitute `(u = 1/4)` into equation (1):

`8 xx 1/4 + v = 3`

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

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

`8 xx 1/4 + v = 1`

5. Recall `(u = 1/x)` and `(v = 1/y)`:

`1/x = 1/4`

⇒ `x = 4 1/y`

⇒ x = 1

⇒ y = 1