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.​

Answer 1

Answer:

Step-by-step explanation:

let the value of numerator be x

let the value of denominator be y

x + y = 2y-3

x - 1/y - 1 = (1 / 2) * y /  y  (As numerator becomes half of denominator)

x + y = 2y - 3

x - y = -3......... 1

x - 1/y - 1 = (1/2) * y / y

y in numerator and denominator cancels out

x - 1/y - 1 = 1 / 2

2x - 2 = y - 1

2x -y = 1........ 2

multiplying eq 1 by 2 we get

2x - 2y = -6

and therefore

     2x   -     y  =       1

( - ) 2x ( + ) 2y = ( + )6

__________________

     0   + y = 7

__________________

y = 7

substituting in 1

we get

x - y = -3

x - 7 = -3

x = 4

therefore the fraction is

x / y = 4 / 7

any doubts just ask me

Answer 2

Let Numerator = x and Denominator = y

Then, fraction = x/y

⋆ According to the first condition,

$\sf{Numerator + Denominator = twice \: of\: den. - 3}$

$: \implies \: x + y = 2y - 3$

$: \implies \:2 y - y = 3 + x$

$: \implies \: y = 3 + x \: .......(i)$

⋆ According to the second condition,

$\sf{Decreased \: numerator \: by \: 1 = \frac{1}{2} (Decreased \: den.) }$

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

$: \implies \: 2(x - 1) = y - 1$

$: \implies \: 2x - 2 = y - 1$

$: \implies \: 2x - y = - 1 + 2$

$: \implies \: 2x - y = 1 \: .......(ii)$

✪Substituting y = 3 + x in equation ( ii ) ,We have

$2x - (3 + x) = 1$

$\rightarrow \: 2x - x = 1 + 3$

$\rightarrow \: x = 4$

✪ Substituting x = 4 in equation ( i ) , We get

$y = 3 + 4$

$\rightarrow \: y = 7$

Hence, the fraction = x/y = 4/7