The numerator of a fraction is 3 less than its denominator. if 2 is ad

1.	The numerator of a fraction is 3 less than its denominator. if 2 is added to both the numerator and the denominator, then the sum of new fraction and orignal fraction is 29/20. find the original fraction.
Answer» Topic :- Linear Equations Given :- The numerator of a fraction is 3 less than its DENOMINATOR. If 2 is added to both the numerator and the denominator, then the sum of new fraction and original fraction is 29/20. To Find :- Original Fraction Solution :- Let denominator of fraction be 'x'. Then numerator will be ' x - 3 ' as it is 3 less than denominator. So, original fraction is $\dfrac{x-3}{x}$ Add 2 to both numerator and denominator, New fraction $\dfrac{x-3+2}{x+2}$ $\dfrac{x-1}{x+2}$ Add both fractions, $\dfrac{x-3}{x}+\dfrac{x-1}{x+2}=\dfrac{29}{20}$ $\dfrac{(x-3)(x+2)+x(x-1)}{x(x+2)}=\dfrac{29}{20}$ $\dfrac{x^2+2x-3x-6+x^2-x}{x(x+2)}=\dfrac{29}{20}$ $\dfrac{2x^2-2x-6}{x^2+2x}=\dfrac{29}{20}$ Now, CROSS multiply, 20( 2x² - 2x - 6 ) = 29( x² + 2x ) 40x² - 40x - 120 = 29x² + 58x 40x² - 29x² - 40x - 58x - 120 = 0 11x² - 98x - 120 = 0 11x² - 110x + 12X - 120 = 0 11x( x - 10 ) + 12( x - 10 ) = 0 ( 11x + 12 )( x - 10 ) = 0 11x + 12 = 0 x = -12/11 or x - 10 = 0 x = 10 So, x = 10 or -12/11 Original Fraction $\dfrac{x-3}{x}$ Put x = 10, $\dfrac{10-3}{10}$ $\dfrac{7}{10}$ Put x = -12/11, $\dfrac{\dfrac{-12}{11}-3}{\dfrac{-12}{11}}$ $\dfrac{\dfrac{-45}{11}}{\dfrac{-12}{11}}$ Answer :- So, original fractions are $\dfrac{7}{10}$ and $\dfrac{\dfrac{-45}{11}}{\dfrac{-12}{11}}$

The numerator of a fraction is 3 less than its denominator. if 2 is added to both the numerator and the denominator, then the sum of new fraction and orignal fraction is 29/20. find the original fraction.

Answer»

Topic :-

Linear Equations

Given :-

The numerator of a fraction is 3 less than its DENOMINATOR. If 2 is added to both the numerator and the denominator, then the sum of new fraction and original fraction is 29/20.

To Find :-

Original Fraction

Solution :-

Let denominator of fraction be 'x'.

Then numerator will be ' x - 3 ' as it is 3 less than denominator.

So, original fraction is

$\dfrac{x-3}{x}$

Add 2 to both numerator and denominator,

New fraction

$\dfrac{x-3+2}{x+2}$

$\dfrac{x-1}{x+2}$

Add both fractions,

$\dfrac{x-3}{x}+\dfrac{x-1}{x+2}=\dfrac{29}{20}$

$\dfrac{(x-3)(x+2)+x(x-1)}{x(x+2)}=\dfrac{29}{20}$

$\dfrac{x^2+2x-3x-6+x^2-x}{x(x+2)}=\dfrac{29}{20}$

$\dfrac{2x^2-2x-6}{x^2+2x}=\dfrac{29}{20}$

Now, CROSS multiply,

20( 2x² - 2x - 6 ) = 29( x² + 2x )

40x² - 40x - 120 = 29x² + 58x

40x² - 29x² - 40x - 58x - 120 = 0

11x² - 98x - 120 = 0

11x² - 110x + 12X - 120 = 0

11x( x - 10 ) + 12( x - 10 ) = 0

( 11x + 12 )( x - 10 ) = 0

11x + 12 = 0

x = -12/11 or

x - 10 = 0

x = 10

So, x = 10 or -12/11

Original Fraction

$\dfrac{x-3}{x}$

Put x = 10,

$\dfrac{10-3}{10}$

$\dfrac{7}{10}$

Put x = -12/11,

$\dfrac{\dfrac{-12}{11}-3}{\dfrac{-12}{11}}$

$\dfrac{\dfrac{-45}{11}}{\dfrac{-12}{11}}$

Answer :-

So, original fractions are

$\dfrac{7}{10}$ and

$\dfrac{\dfrac{-45}{11}}{\dfrac{-12}{11}}$