Question

a fraction becomes 9/11 when 2 is added to both the numerator and the denominator. if 3 I'd added to both num. and deno. it becomes 5/6. find the fraction​

Answer 1

Given :-

If 2 is added to the both Numerator and Denominator the fraction becomes 9/11 .

If 3 is added to the both Numerator and Denominator the fraction becomes 5/6

To find :-

The fraction

Solution:-

$\bf \: Let \: the \: Fraction \: be \: \red {\dfrac{x}{y} }$

So,

If 2 is added to both Numerator and Denominator it becomes 9/11

The equation is :-

$\bf \: \red{\dfrac{x + 2}{y + 2} = \dfrac{9}{11} } \:$

By cross multiplication

$\rm \: (x + 2)11 = (y + 2)9$

$\rm \: 11x + 22 = 9y + 18$

$\rm \: 11x + 22 - 9y - 18 = 0$

$\bf \red{\: 11x - 9y + 4 = 0} - - \: 1$

If 3 is added to both Numerator and Denominator it becomes 5/5

The equation is :-

$\bf \: \red{\dfrac{x + 3}{y + 3} = \dfrac{5}{6} } \:$

By cross multiplication

$\rm \: (x + 3)6 = (y + 3)5$

$\rm \: 6x + 18 = 5y + 15$

$\rm \: 6x + 18 - 5y - 15 = 0$

$\bf \red{ \: 6x - 5y + 3 = 0} \: - - 2$

Now ,

We have the two equations.

11x -9y + 4 = 0 and 6x-5y + 3 =0 .By solving these two equations we get the value of x,y .

We cannot solve by elimination method So, we can use by substituting method .

$\rm \: 11x - 9y + 4 = 0$

$\rm \: 11x - 9y \: = \: - 4$

Making subject 'x'

$\rm \: 11x = \: - 4 + 9y$

$\bf \red{x = \bigg( \dfrac{9y - 4}{11} \bigg)}$

Now,

Substituting the value of 'x' in equation-2 in order to get the value of 'y'

$\bf \red{ \: 6x - 5y + 3 = 0} \: - - 2$

$\rm \: 6 \bigg( \dfrac{9y - 4}{11} \bigg) - 5y + 3 = 0$

$\rm \: \dfrac{54y - 24}{11} - 5y + 3 = 0$

$\rm \: \dfrac{54y - 24 - 5y(11) + 3(11)}{11} = 0$

$\rm \: \dfrac{54y - 24 - 55y + 33}{11} = 0$

By cross multiplication

$\rm \: 54y - 55y - 24 + 33 = 0$

$\rm \: - y + 9 = 0$

$\rm \: - y = - 9$

$\boxed{ \underline{ \bf{ \red { \bigstar{y = 9}}}}}$

Substituting the value of y in equation 2

$\bf \red{ \: 6x - 5y + 3 = 0} \: - - 2$

$\rm \: 6x \: - 5(9) + 3 = 0$

$\rm \: 6x - 45 + 3 = 0$

$\rm \: 6x - 42 = 0$

$\rm \: 6x = 42$

$\boxed{ \underline{ \bf{ \red { \bigstar{x = 7}}}}}$

We got the values of x,y So, the fraction is

$\bf \red {\dfrac{x}{y} = \dfrac{7}{9} }$

$\: \: \: \: \: \boxed{ \underline{ \bf{ \red { \bigstar \: So, \: the \: Required \: Fraction \: is \: \pink{ \frac{7}{9} }}}}}$