Problem in solving questions with Algebra I was trying to solve a question:
Income of A and B is 3:5 respectively and their expenditures are 1:5. If both saves $100 each. Then find their income?
$$\frac{3x-y}{5x-5y}= \frac{100}{100}$$
$$300x - 100y = 500x - 500y$$
$$400y = 200x$$
$$\frac xy = \frac{200}{400}$$
400 x 3 = 1200
400 x 5 = 2000
But that was wrong, and I asked my friend he said that two variables can't be in only single side.
For eg. 2x + 4y =16 //can't be solved
But  2x+4y=8y can be solved
The Answer is:
$$\frac {3x - 100}{5x - 100} = \frac {1}{5}$$
$$15x - 500 = 5x - 100$$
$$10x = 400$$
$$x = 40$$
40 x 3 = 120
40 x 5 =200
I can now solve problem using the correct method, but i remember I have solved many questions which had two variables on one side and RHS no variable at all.
For eg. a question:
Find the number that must be added to the terms of the ratio 7:13 to make it equal to 2:3
How I solved it:
$$\frac{7x+y}{13x+y} = \frac{2}{3}$$
$$21x +3y = 26x +2y$$
$$Y = 5x$$
$$\frac yx = 5$$
$$y=5$$
And the answer is correct. 
Now, clearly I used the same method in the first problem but it didn't work why?
EDIT: I added the answer using the correct method
 A: You need two equations to solve for two unknowns.  The two equations are 
$$3x-y=100\\5x-5y=100$$
You divided them, making only one equation.  You should solve them simultaneously
$$15x-5y=500\\5y=15x-500\\5x-(15x-500)=100\\10x=400$$
So $B$ makes $200$ while $A$ makes $120$.  $B$'s expenses are $100$ and $A$'s are $20$
A: Based on the provided information, the original $2$ equations are
$$3x-y = 100 \tag{1}\label{eq1}$$
$$5x-5y= 100 \tag{2}\label{eq2}$$
where $x$ is the ratio of the incomes and $y$ is the ratio of the expenditures. These are $2$ linear equations in $2$ unknowns. You can solve them by multiplying \eqref{eq1} by $5$ and subtracting \eqref{eq2} multiplied by $3$ to get
$$10y = 200 \iff y = 20 \tag{3}\label{eq3}$$
Substituting this into \eqref{eq1} gives $x = 40$. Note in your equation, you have the reciprocal of the correct ratio, i.e., it should be $\frac{x}{y} = \frac{400}{200} = 2$. As you can see, $x = 40$ and $y = 20$ gives this same ratio. Also, incomes can be determined to be $A = 3x = 120$ and $B = 5x = 200$.
When you took the ratio of the $2$ equations, you combined them so there was then only $1$ equation in $2$ unknowns, so you can't answer for specific $x$ and $y$, only something like their ratio that you derived.
With the second equation, you're only given that one equation with $2$ unknowns, so you again can only get the ratio. As you are only asked to determine one of the possible values, this is how you can determine that $y = 5$ is one of them. However, note that any $y = 5k$, for any non-zero integer $k$, will work, with this giving $x = k$.
Update: What you've done in your just updated question text is use an alternate way to eliminate a variable from \eqref{eq1} and \eqref{eq2} so you get one equation in $1$ unknown. In particular, moving $100$ to the left and the $y$ terms to the right gives
$$3x - 100 = y \tag{4}\label{eq4}$$
$$5x - 100 = 5y \tag{5}\label{eq5}$$
Dividing \eqref{eq4} by \eqref{eq5} gives the equation you've added of
$$\frac{3x-100}{5x-100} = \frac{y}{5y} = \frac{1}{5} \tag{6}\label{eq6}$$
where it's assumed $y \neq 0$.
