Question

if 1 is substracted from the numerator of a fraction it becomes 2/3 but if 5 is added to denominator of the fraction it becomes 1/2 find the fraction​

Answer 1

AnswEr :

$\sf{Let\: the\: Numerator\: be \: x \:\& Denominator\: be \: y.}$

$\sf{Then,\: fraction\: is \: \dfrac{x}{y}}$

$\bigstar\:\:\bold{\underline{\sf{According\:to\: Question\;Now}}}$

$\sf{If\: 1 \; is \: Subtracted\; from\; Numerator}$

$\longrightarrow\sf\: \dfrac{x \:-\: 1}{y} = \dfrac{2}{3}$

$\longrightarrow\sf\: 3(x - 1) = 2y$

$\longrightarrow\sf\: 3x - 3 = 2y$ ----------Eq(1)

$\rule{150}2$

$\sf{If\: 5 \: is \: added\: to \: Denominator}$

$\longrightarrow\sf\: \dfrac{x}{y + 5} = \dfrac{1}{2}$

$\longrightarrow\sf\: 2x = y + 5$

$\longrightarrow\sf\:y = 2x - 5$ ----------Eq(2)

$\rule{150}2$

Now, Substituting the Value of y in Eq(1)

$\longrightarrow\sf\: 3x - 3 = 2y$

$\longrightarrow\sf\: 3x - 3 = 2(2x - 5)$

$\longrightarrow\sf\: 3x - 3 = 4x - 10$

$\longrightarrow\sf\red{x\: = \: 7}$

Substituting the Value of x in Eq(2)

$\longrightarrow\sf\: y = 2(7) - 5$

$\longrightarrow\sf\: y = 14 - 5$

$\longrightarrow\sf\red{y\:=\: 9}$

Here, x = 7 & y = 9

So, Required Fraction is $\sf\dfrac{7}{9}.$

$\rule{150}2$

Answer 2

$\large{\underline{\red{\rm{AnswEr:\frac{7}{9}}}}}$

$\bold\green{\underline{\underline{Let:}}}$

The numerator be X

The denominator be y

$\bold\orange{The\: fraction=\frac{X}{Y}}$

$\bold\green{\underline{\underline{Given:}}}$

If 1 is subtracted from numerator

it becomes 2/3

⟹X-1/Y=2/3

⟹3X-3=2Y

⟹3X-2Y=3

$\bold\orange{\boxed{3X-2Y=3.....(1)}}$

Given in the question again if 5 is added

to denominator it becomes 1/2

⟹X/Y+5=1/2

⟹2X=Y+5

⟹2X-Y=5

$\bold\purple{\boxed{2X-Y=5.....(2)}}$

By solving equations 1 and 2

⟹3X-2Y=3

⟹2X-Y=5

• Here multiplying and subtracting

⟹6x-4y=6-----(3)

⟹6x-3y=15----(4)

⟹-y=-9

⟹y=9

Now, putting the value of y inEQ1

⟹3X-2Y=3

⟹3X-18=3

⟹3X=21

⟹X=7

Hence,

$\bold\red{\boxed{The\: fraction=\frac{7}{9}}}$