Question

In a fraction, twice the numerator is $2$ more than the denominator.

If $3$ is added to the numerator and to the denominator the new fraction is $\frac{2}{3}$.

Find the original number.

Answer 1

Step 1: Express the problem statement in the mathematical expression

Let the numerator of the fraction is $x$ and the denominator is $y$.

It is given that, twice the numerator is $2$ more than the denominator.

Thus, $2x=y+2$

$\Rightarrow y=2x-2$.

Now, $3$ is added to the numerator and to the denominator.

Thus, the new fraction $\frac{x+3}{y+3}=\frac{2}{3}$.

Step 2: Solve the equations

Consider the equation: $\frac{x+3}{y+3}=\frac{2}{3}$

$$\begin{gathered} \Rightarrow 3\left(x+3\right)=2\left(y+3\right) \\ \Rightarrow 3x+9=2y+6 \end{gathered}$$

Substitute $y$ with $2x-2$.

$$\begin{gathered} \Rightarrow 3x+9=2(2x-2)+6 \\ \Rightarrow 3x+9=4x-4+6 \\ \Rightarrow 3x-4x=2-9 \\ \Rightarrow -x=-7 \\ \Rightarrow x=7 \end{gathered}$$

Thus,

$$\begin{gathered} y=2x-2 \\ =2\left(7\right)-2 \\ =14-2 \\ =12 \end{gathered}$$.

Hence, the original fraction is $\frac{x}{y}=\frac{7}{12}$.

Therefore, the original number is $\frac{7}{12}$.