Question

the denominator of a fraction is greater than the numerator by 8 if the numerator is increased by 17 and denominator is decreased by 1 the number obtained is 3/2 find the fraction​

Answer 1

Step-by-step explanation:

n+8=d

n-d=-8eq------1

n+17/d-1=3/2---------eq.2

2(n+17) =3(d-1)

2n+34=3d-3

2n+34+3-3d=0

2n+37-3d=0

2n-3d=-37

now multiply the eq.1by 2

(n-d) 2=8x2

2n-2d=16

eliminate eq1nd eq2

2n-3d=37

(2n-2d=16)x(-)

2n-3d=37

-2n+2d=-16

d=53

then n=-45

hope it may help u

Answer 2

$\large \bf \clubs \: Given :-$

• The denominator of a fraction is greater than the numerator by 8.

• The denominator of a fraction is greater than the numerator by 8.If the numerator is increased by 17 and denominator is decreased by 1, the number obtained is 3/2.

$\large \bf \clubs \: To \: Find :-$

• The Original Fraction

$\large \bf \clubs \: Solution :-$

Let,

• Numerator of Original Fraction = x .

So,

• Denominator should be = x + 8.

Hence ,

$\pink{\bf Original \: Fraction = \dfrac{x}{x + 8} }$

When the Numerator is increased by 17 and

Denominator is decreased by 1 :

$\red{ \bf New \: Fraction = \dfrac{x + 17}{x + 7} }$

According To The Given Condition :

$\begin{gathered} \sf \dfrac{\mathtt{x} + 17}{\mathtt{x} + 7} = \frac{3}{2} \\ \\ :\longmapsto \sf2(\mathtt{x} + 17) = 3(\mathtt{x} + 7) \\ \\ :\longmapsto \sf2\mathtt{x} + 34 = 3x + 21 \\ \\ :\longmapsto \sf3\mathtt{x} - 2\mathtt{x} = 34 - 21 \\ \\ \purple{ \Large :\longmapsto \underline {\boxed{{\bf x = 13} }}}\end{gathered}$

Hence ,

$\begin{gathered} {\bf Original \: Fraction = \dfrac{13}{13 + 8} }\\ \\ \underline{ \underline{\pink{:\longmapsto\bf Original \: Fraction = \dfrac{13}{21} }}}\end{gathered}$