The sum of the numerator and denominator of a fraction is 4 more than twice the numerator.
Question:

The sum of the numerator and denominator of a fraction is 4 more than twice the numerator. If the numerator and denominator are increased by 3, they are in the ratio 2 : 3. Determine the fraction.

Solution:

Let the numerator and denominator of the fraction be $x$ and $y$ respectively. Then the fraction is $\frac{x}{y}$

The sum of the numerator and denominator of the fraction is 4 more than twice the numerator. Thus, we have

$x+y=2 x+4$

$\Rightarrow 2 x+4-x-y=0$

$\Rightarrow x-y+4=0$

If the numerator and denominator are increased by 3, they are in the ratio 2:3. Thus, we have

$x+3: y+3=2: 3$

$\Rightarrow \frac{x+3}{y+3}=\frac{2}{3}$

$\Rightarrow 3(x+3)=2(y+3)$

$\Rightarrow 3 x+9=2 y+6$

$\Rightarrow 3 x-2 y+3=0$

So, we have two equations

$x-y+4=0$

$3 x-2 y+3=0$

Here x and y are unknowns. We have to solve the above equations for x and y.

By using cross-multiplication, we have

$\frac{x}{(-1) \times 3-(-2) \times 4}=\frac{-y}{1 \times 3-3 \times 4}=\frac{1}{1 \times(-2)-3 \times(-1)}$

$\Rightarrow \frac{x}{-3+8}=\frac{-y}{3-12}=\frac{1}{-2+3}$

$\Rightarrow \frac{x}{5}=\frac{-y}{-9}=\frac{1}{1}$

$\Rightarrow \frac{x}{5}=\frac{y}{9}=1$

$\Rightarrow x=5, y=9$

Hence, the fraction is $\frac{5}{9}$.

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