If the numerator of a fraction is increased by $2$ and the denominator by $1$, it becomes $5/8$.... If the numerator of a fraction is increased by $2$ and the denominator by $1$, it becomes $\displaystyle \frac{5}{8}$ and if the numerator and the denominator of the same fraction are each increased by $1$, the fraction becomes equal to $\displaystyle \frac{1}{2}$. Find the fraction.
I tried,
Let the numerator of the fraction be $x$,
Let the denominator of the fraction be $y$
Therefore, Original fraction is $\displaystyle \frac{x}{y}$.
$$
\left\{
\begin{eqnarray*}
\frac{x+2}{y+1}=\frac{5}{8} \\[2mm]
\frac{x+1}{y+1}=\frac{1}{2} \\
\end{eqnarray*}
\right.
$$
What should I do now?
 A: That's a fine start. Now you've got two equations in two unknowns. $$\frac{x+2}{y+1} = \frac 58\tag{1}$$
$$\frac{x+1}{y+1} = \frac 12 \tag{2}$$
Now, we have to assume that $y+1 \neq 0$, given the information that's impossible, so we can take $(1)$ for example, and "cross multiply" to get: $$ 8(x+2)  = 5(y+1) \iff \color{blue}{y + 1} = \frac{8(x+2)}{5}\tag{1}$$
From the second equation, we have
$$2(x+1) = \color{blue}{y+1}\tag{2}$$
That means $\color{blue}{y+1 = y+1}$, so that $$\frac {8(x+2)}5 = 2(x +1)\tag{3}$$
Now solve for $x$: $$\frac{8(x + 2)}5 = 2(x+1)\iff 5(x+1) = 4(x+2) \iff 5x+5 = 4x + 8 \iff x =  3$$ Then use $x=3$ to solve for $y$, using the fact that $y + 1 = 2(x + 1) \iff y = 2x + 1\implies y = 2(3) + 1 = 7$.
So the original fraction $$\frac xy = \frac 37$$
A: We have the equations
$$\frac{x+2}{y+1}=\frac{5}{8}, \ \frac{x+1}{y+1}=\frac{1}{2}=\frac{4}{8}$$
Hence
$$x+1=4, \ y+1=8 \ \implies x=3, \ y=7$$
We are allowed to do so since $(x+2)-(x+1)=5-4$
A: $\dfrac{x+2}{y+1}=\dfrac{5}{8}$ and $\dfrac{x+1}{y+1}=\dfrac{1}{2}=\dfrac{4}{8}$
Dividing first equation by second equation you would get
$$\dfrac{x+2}{x+1}=\dfrac{5}{4}=\dfrac{3+2}{3+1}$$
This should be easy now..
