Question

the numerator of a fraction is 8 less than the denominator if the numerator is increased by 17 and the denominator is decreased by 1 the new fraction obtained is 3.2 find the fraction​

Answer 1

Given:

According to the question

Let the fraction be x/y

x=y-8

x-y=-8         _(1)

x-17/y+1=3/2

2(x-17)=3(y+1)

2x-34=3y+3

2x-3y=3+34

2x-3y=37      -(2)

by solving equation

2 × {x-y=-8 }

1 × { 2x-3y=37}

2x-2y=-16

2x-3y=37

__________

y=-53

put value of y in equation (1)

x-y=-8

x-(-53)=8

x+53=8

x=8-53

x=-45

The fraction is x/y=-53/-45

Hence the fraction is x/y=53/45

Answer 2

Let, $\frac{x}{y}$ be the fraction,

So, for equation (i),

$=>x=y-8$

$=>x-y=-8...(i)$

Now, for equation (ii)

$=>x-\frac{17}{y+1}=\frac{3}{2}$

By simplifying it we get,

$=>2x-34=3y-3$

$=>2x-3y=3+34$

$=>2x-3y=37...(ii)$

By solving equation (i) and (ii), we get

Multiplying equation (i) by 2 and (ii) by 1 we get,

$=>2*{x-y=-8}\\1*{2x-3y=37}$

$=>2x-2y=-16\\=>2x-3y=37$

By simplifying it we get,

$y=-53$

Now, put the value of y in equation (i),

$=>x-y=-8\\=>x-(-53)=8\\=>x+53=8\\=>x=8-53\\x=-45$

The fraction we assume that is $\frac{x}{y} =\frac{-53}{-45}$

So, the required fraction is $\frac{53}{45}$