Question

Solve the following system of equations:

`4/x + 3y = 8`

`6/x - 4y = -5`

Answer 1

Given: `4/x + 3y = 8, 6/x - 4y = -5`

Step-wise calculation:

1. Let `u = 1/x` (so x ≠ 0). 

The system becomes: 4u + 3y = 8, 6u – 4y = –5

2. Solve the linear system.

From 4u + 3y = 8, `y = (8 - 4u)/3`. 

Substitute into 6u – 4y = –5: `6u - 4((8 - 4u)/3) = -5` 

Multiply both sides by 3:

18u – 4(8 – 4u) = –15 

18u – 32 + 16u = –15 

34u – 32 = –15 

34u = 17

`u = 1/2`

3. Back-substitute:

`1/x = u = 1/2` 

⇒ x = 2

Then `y = (8 - 4u)/3` 

= `(8 - 4 xx (1/2))/3` 

= `(8 - 2)/3`

= `6/3`

= 2

4. Check: `4/x + 3y = 4/2 + 3 xx 2` 

= 2 + 6

= 8

`6/x - 4y = 6/2 - 4 xx 2` 

= 3 – 8

= –5 

The solution is x = 2, y = 2.