Question

Solve the following pair of linear equations:

3x + 5y = 2xy, 21x – 15y = 4xy, x ≠ 0, y ≠ 0

Answer 1

Given equations:

3x + 5y = 2xy, 21x – 15y = 4xy, with the conditions x ≠ 0, y ≠ 0.

Step 1: Rewrite the equations

Rewrite both equations in a form involving `1/x` and `1/y` by dividing all terms by (xy):

`(3x)/(xy) + (5y)/(xy) = (2xy)/(xy)`

⇒ `3/y + 5/x = 2`

`(21x)/(xy) - (15y)/(xy) = (4xy)/(xy)`

⇒ `21/y - 15/x = 4`

Step 2: Set substitutions

Let `u = 1/x, v = 1/y`.

Then the equations become 3v + 5u = 2, 21v – 15u = 4.

Step 3: Solve the system of linear equations in (u, v)

The system is:

3v + 5u = 2   ...(1) 

21v – 15u = 4   ...(2)

Multiply equation (1) by 3:

9v + 15u = 6   ...(3)

Add equations (2) and (3):

(21v + 9v) + (–15u + 15u) = 4 + 6

30v = 10

`v = 10/30`

`v = 1/3`

Put `(v = 1/3)` into equation (1):

`3 xx 1/3 + 5u = 2`

⇒ 1 + 5u = 2

⇒ 5u = 1

⇒ `u = 1/5`

Step 4: Back substitute to find (x) and (y)

`u = 1/x`

`u = 1/5`

⇒ x = 5

`v = 1/y`

`v = 1/3`

⇒ y = 3

The solution to the system is x = 5, y = 3.