The sum of the numerator and denominator of a fraction is 3 less than twice the denominator.
Question:

The sum of the numerator and denominator of a fraction is 3 less than twice the denominator. If the numerator and denominator are decreased by 1, the numerator becomes half the denominator. 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 3 less than twice the denominator. Thus, we have

$x+y=2 y-3$

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

$\Rightarrow x-y+3=0$

If the numerator and denominator are decreased by 1, the numerator becomes half the denominator. Thus, we have

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

$\Rightarrow \frac{x-1}{y-1}=\frac{1}{2}$

$\Rightarrow 2(x-1)=y-1$

$\Rightarrow 2 x-2=y-1$

 

$\Rightarrow 2 x-y-1=0$

So, we have two equations

$x-y+3=0$

 

$2 x-y-5=0$

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

By using cross-multiplication, we have

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

$\Rightarrow \frac{x}{1+3}=\frac{-y}{-1-6}=\frac{1}{-1+2}$

$\Rightarrow \frac{x}{4}=\frac{-y}{-7}=\frac{1}{1}$

$\Rightarrow \frac{x}{4}=\frac{y}{7}=1$

$\Rightarrow x=4, y=7$

Hence, the fraction is $\frac{4}{7}$.

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