When the numerator of a fraction is increased by $4$, the fraction increases by $2/3$... When the numerator of a fraction is increased by $4$, the fraction increases by $2/3$. What is the denominator of the fraction?
I tried,
Let the numerator of the fraction be $x$ and the denominator be $y$.
Accordingly, $$\frac{x+4}y=\frac xy+\frac 23$$
I am not able to find the second equation. 
 A: Again, you've got a fine start:
You wrote: 
$$\frac{x+4}y=\color{red}{\frac xy}+\color{blue}{\frac 23}\tag{1}$$
But note that $$\frac{x+4}{y} = \color{red}{\frac xy} + \color{blue}{\frac 4y}\tag{2}$$
From $(1),(2),$ it must follow that $$\color{blue}{\frac 4y = \frac 23 } \iff 2y = 4\cdot 3 = 12 \iff y = \frac{12}{2} = 6$$
So the denominator, $y$ is $6$.
A: If you add 4 to the numerator, the value of your fraction will be increased by $\frac4y$, where y is your denominator.
So $\frac4y=\frac23$ and $y=6$
A: In general you are correct: to solve for two unknowns, you would usually need two equations.  But in this case you are lucky, and the one equation gives a solution for the one unknown you are asked to find.
$\dfrac{x+4}{y}=\dfrac{x}{y}+\dfrac{2}{3}$ so multiplying both sides by $3y$ gives $3x+12 = 3x+2y$ making the denominator $y=6$.
It gives no specific solution for $x$, for example: $\dfrac{5+4}{6}=\dfrac{5}{6}+\dfrac{2}{3}$ and $\dfrac{7+4}{6}=\dfrac{7}{6}+\dfrac{2}{3}$. 
A: Given,
$$\frac{n+4}{d}=\frac{n}{d}+\frac{2}{3}$$
So,
$$\frac{n}{d}+\frac{4}{d}=\frac{n}{d}+\frac{2}{3}$$
Or,
$$\frac{4}{d}=\frac{2}{3}$$
That, gives us $d=6$ 
A: We have $\frac{x+4}y={\frac xy}+{\frac 23}=\frac{3x+2y}{3y}$ but: $$\frac{x+4}{y}=\frac{\color{red}3(x+4)}{\color{red}3y}=\frac{3x+12}{3y}$$ So $$\frac{3x+12}{3y}=\frac{3x+2y}{3y}$$ So if $y\neq 0$ then $3x+12=3x+2y$.
A: $$x=\frac{a}{b}$$
$$x+\frac{2}{3}=\frac{a+4}{b} \Rightarrow \frac{a}{b}+\frac{2}{3}=\frac{a+4}{b} \Rightarrow a+\frac{2}{3}b=a+4 \Rightarrow b=\frac{12}{2}=6$$
A: Hint $ $ linearity: $\ \ell(x) = x/d\,\Rightarrow\, \ell(x\!+\!x')=\ell(x)\color{#c00}{+\ell(x')},\,$ so $\,\ell(x)\,$ increases by  $\,\color{#c00}{\ell(x')} = x'/d$
Remark $\ $ Here linearity follows from the distributive law $\ d^{-1}(x+x') = d^{-1}x + d^{-1}x'$
