The denominator of a fraction is one more than twice the numerator. If the sum of the fraction and its reciprocal is 2 16/21, find the fraction.
Answers
SOLUTION : Let the numerator be 'x' then denominator is 2x+1 Fraction = numerator /denominator Fraction = x /2x+1 Reciprocal of fraction = 2x+1/x A.T.Q x / 2x+1+ 2x+1/x=2 16/21 x / 2x+1+ 2x+1/x =58/21 x(x)+2x+1(2x+1) / 2x+1(x) =58/21 x² +(2x+1)² / 2x² + x = 58/21 x² + 4x² + 4x+1 / 2x² + x =58/21 [(a+b) ² = a² +b² +2ab] 5x² + 4x+1 / 2x² + x = 58/21 21 (5x² + 4x+1) = (2x² + x)58 105x² + 84x +21= 116x² + 58x 105x² - 116x² + 84x - 58x +21= 0 - 11x² + 26x + 21= 0 -(11x² - 26x -21)= 0 11x² - 26x -21 = 0 11x² -33x +7x -21 = 0 [By Middle term splitting] 11x (x - 3 ) +7(x -3) = 0 (11x +7) (x -3) = 0 x = 3 and x = - 7/3 (can't be negative) Numerator (x) = 3 Denominator = 2x +1 = 2×3 +1= 7 Denominator = 7 Fraction = 3/7 Hence, the fraction is 7/3. HOPE THIS ANSWER WILL HELP YOU...
Let the numerator of the fraction = x
denominator = 2x + 1
Now ,
fraction = x/(2x+1) ---( 1 )
Reciprocal of the fraction =(2x+1)/x ---( 2 )
according to the problem given ,
x/(2x+1) + (2x+1)/x = 2 16/21
=> [ x² + ( 2x + 1 )² ]/[(2x+1)x] = 58/21
=> ( x²+4x²+4x+1 )/( 2x²+x ) = 58/21
=> ( 5x² + 4x + 1 )/( 2x² + x ) = 58/21
=> 21( 5x²+4x+1 ) = 58( 2x² + x )
=> 105x² + 84x + 21 = 116x² + 58x
=> 0 = 116x² + 58x - 105x² - 84x - 21
=> 0 = 11x² - 26x - 21
=> 11x² - 26x - 21 = 0
Splitting the middle term , we get
=> 11x² - 33x + 7x - 21 = 0
=> 11x( x - 3 ) + 7 ( x - 3 ) = 0
=> ( x - 3 )( 11x + 7 ) = 0
Therefore ,
x - 3 = 0 or 11x + 7 = 0
x = 3 or x = -7/11
x is not negative .
So , x = 3
Now ,
Required fraction ,
fraction = x/(2x+1)
= 3/[2×3+1]
= 3/7
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