You need to solve for x and y the given system of equations, hence, you may use substitution method, thus, you may use the top equation to write x in terms of `y` , such that:

`(1/4)x = 2 - (4/5)y => x = 4(2 - (4/5)y) => x = 8 - (16/5)y`

Replace `8 - (16/5)y` for `x` in the bottom equation, such that:

`(5/16)(8 - (16/5)y) - (1/5)y = 3`

`40/16 - (5/16)*(16/5)y - (1/5)y = 3`

`10/4 - y - (1/5)y = 3`

Isolate the terms that contain y to the left side, such that:

`-y - y/5 = 3 - 10/4 => -y - y/5 = 3 - 5/2`

You need to bring the terms to the left side to a common denominator, such that:

`(-5y - y)/5 = 3 - 5/2`

`-6y/5 = (6 - 5)/2 => -6y/5 = 1/2 => -6y = 5/2 => y = -5/12`

Replace -`5/12` for `y` in equation `x = 8 - (16/5)y` , such that:

`x = 8 - (16/5)(-5/12) => x = 8 + 16/12 => x = 8 + 4/3 => x = (24 + 4)/3 => x = 28/3`

**Hence, evaluating the solution to the given system, using substitution, method, yields **`x = 28/3, y = -5/12.`

`1/4x+4/5y=2` (i)

`5/16x-1/5y=3` (ii)

Multiply (ii) by 4 and add to (i)

`1/4x+4/5y=2`

`5/4x-4/5y=12`

`6/4x=14`

`x=(14xx4)/6`

`x=28/3`

Substitute x in (i)

(1/4)(28/3)+(4/5)y=2

(4/5)y=2-7/3

(4/5)y=-1/3

y=-5/12

Thus

**x=28/3 and y=-5/12**

**Ans.**

Addendum: In the first case, when `x=3/28` ,

`1/(4*3/28)+4/(5y)=2`

`rArr 1/(3/7)+4/(5y)=2`

`rArr 7/3+4/(5y)=2`

`rArr 4/(5y)=2-7/3=-1/3`

`rArr y=-12/5`

Again in the second case, when `x=28/3`

`1/4 * 28/3+4/5 y=2`

`rArr 7/3+4/5y=2`

`rArr 4/5y=2-7/3=-1/3`

`rArr y=-5/12`

I am not quite sure about the set of equations.

First, if the given set is:

`1/(4x)+4/(5y)=2` --- (i)

and `5/(16x)-1/(5y)=3` --- (ii)

Multiply (ii) by 4 to eliminate the second term,

`1/(4x)+4/(5y)=2`

and `5/(4x)-4/(5y)=12`

-------------------------------

(+) -> `6/(4x)=14`

`rArr x=6/56` , i.e. `3/28`

Then, if the given set is:

`1/4x+4/5y=2` --- (iii)

and `5/16x-1/5y=3` --- (iv)

Multiply (iv) by 4, again, to eliminate the second term,

`1/4x+4/5y=2`

and `5/4x-4/5y=12`

-------------------------------

(+) -> `6/4x=14`

`rArr x=56/6` , i.e. `28/3`