# The numerator of a fraction is 4 less than the denominator.

Question:

The numerator of a fraction is 4 less than the denominator. If the numerator is decreased by 2 and denominator is increased by 1, then the denominator is eight times the numerator. Find the fraction.

Solution:

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

The numerator of the fraction is 4 less the denominator. Thus, we have

$x=y-4$

$\Rightarrow x-y=-4$

If the numerator is decreased by 2 and denominator is increased by 1, then the denominator is 8 times the numerator. Thus, we have

$y+1=8(x-2)$

$\Rightarrow y+1=8 x-16$

$\Rightarrow 8 x-y=1+16$

$\Rightarrow 8 x-y=17$

So, we have two equations

$x-y=-4$

$8 x-y=17$

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

Subtracting the second equation from the first equation, we get

$(x-y)-(8 x-y)=-4-17$

$\Rightarrow x-y-8 x+y=-21$

$\Rightarrow-7 x=-21$

$\Rightarrow 7 x=21$

$\Rightarrow x=\frac{21}{7}$

$\Rightarrow x=3$

Substituting the value of in the first equation, we have

$3-y=-4$

$\Rightarrow y=3+4$

$\Rightarrow y=7$

Hence, the fraction is $\frac{3}{7}$