Question

in a certain fraction on the denominator is 4 then the numerator if 3 is added to both the numerator and the denominator the resulting fraction is equal to 9 by 7 find the original number​

Answer 1

CORRECT QUESTION :-

In a certain fraction the denominator is 4 less than the numerator. If 3 is

added to both the numerator and the denominator the resulting fraction

is equal to 9 by 7. Find the original number.

SOLUTION :-

Let,

Numerator = x

Denominator = y

According to the first condition,

$\longrightarrow\sf y = x - 4 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ..........................(1)$

According to the first condition,

$\longrightarrow \sf \dfrac{x+3}{y+3}=\dfrac{9}{7} \\\\\longrightarrow 7(x+3)=9(y+3) \\\\\longrightarrow 7x+21 = 9y + 27 \\\\\longrightarrow 7x-9y = 6 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ..........................(1)$

Substitute y = x - 4 in equation (2),

$\longrightarrow\sf 7x-9(x-4)=6\\\\\longrightarrow 7x-9x+36 = 6 \\\\\longrightarrow 7x-9x = 6-36\\\\\longrightarrow -2x= -30\\\\\longrightarrow \bf x =15$

Numerator = 15

Denominator = 15 - 4 = 11

$\bf FRACTION =\dfrac{15}{11}$

Answer 2

Answer:

Let the Numerator be x and Denominator be x - 4

☯ $\underline{\boldsymbol{According\: to \:the\: Question\:now :}}$

$:\implies\sf \dfrac{Numerator + 3}{Denominator + 3} = \dfrac{9}{7}$

$:\implies\sf \dfrac{x + 3}{x - 4 + 3} = \dfrac{9}{7}$

$:\implies\sf \dfrac{x + 3}{x - 1} = \dfrac{9}{7}$

$:\implies\sf (x + 3)7 = 9(x - 1)$

$:\implies\sf 7x + 21 = 9x - 9$

$:\implies\sf 21 + 9 = 9x - 7x$

$:\implies\sf 30 = 2x$

$:\implies\sf x = \dfrac{30}{2}$

$:\implies \underline{ \boxed{\textsf{ \textbf{x = 15}}}}$

$\boxed{\bf{\mid{\overline{\underline{\bigstar\: Therefore :}}}}\mid}$

$\bullet\:\:\textsf{Numerator = x = \textbf{15}}$

$\bullet\:\:\textsf{Denominator = x - 4 = 15 - 4 = \textbf{11}}$

$\therefore\:\underline{\textsf{Required \: fraction \: is \: \textbf{ {} {\text{15}}\!/{}_{\text{11}} }}}.$