Question

The numerator of a fraction is 3 less than its denominator. If 2 is added to both the
and the denominator, then the sum of the new fraction and original fraction is 2930. Find the
original fraction
​

Answer 1

Solution (Question Error):

$\bf{\red{\underline{\bf{Given\::}}}}$

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

$\bf{\red{\underline{\bf{To\:find\::}}}}$

The original fraction.

$\bf{\red{\underline{\bf{Explanation\::}}}}$

Let the denominator be r

Let the numerator be (r-3)

So;

$\boxed{\bf{The\:orginal\:fraction=\frac{r-3}{r} }}}}}$

A/q

$\longrightarrow\sf{\dfrac{r-3}{r} +\dfrac{r-3+2}{r+2} =\dfrac{29}{20} }\\\\\\\longrightarrow\sf{\dfrac{r-3}{r} +\dfrac{r-1}{r+2} =\dfrac{29}{20} }\\\\\\\longrightarrow\sf{\dfrac{(r+2)(r-3)+(r)(r-1)}{r(r+2)} =\dfrac{29}{20} }\\\\\\\longrightarrow\sf{\dfrac{r^{2}-3r+2r-6 +r^{2} -r}{r^{2} +2r} =\dfrac{29}{20} }\\\\\\\longrightarrow\sf{\dfrac{2r^{2}-2r-6 }{r^{2}+2r } =\dfrac{29}{20} }\\\\\\\longrightarrow\sf{20(2r^{2} -2r-6)=29(r^{2} +2r)}\\\\\\\longrightarrow\sf{40r^{2}-40r-120=29r^{2}+58}$

$\longrightarrow\sf{40r^{2} -29r^{2} -40r-58r-120=0}\\\\\\\longrightarrow\sf{11r^{2} -98r-120=0}$

$\underline{\underline{\bf{Using\:Quadratic\:formula\::}}}}}$

As the given polynomial as compared with ax² + bx + c

• a = 11
• b = -98
• c = -120

Now;

$\boxed{\bf{x=\frac{-b\pm\sqrt{b^{2}-4ac } }{2a}}}}}$

$\longrightarrow\sf{x=\dfrac{-(-98)\pm\sqrt{(-98)^{2} -4\times 11\times (-120)} }{2\times 11}} \\\\\\\longrightarrow\sf{x=\dfrac{98\pm\sqrt{9604-(-5280)} }{22} }\\\\\\\longrightarrow\sf{x=\dfrac{98\pm\sqrt{9604+5280} }{22} }\\\\\\\longrightarrow\sf{x=\dfrac{98\pm\sqrt{14884} }{22} }\\\\\\\longrightarrow\sf{x=\dfrac{98\pm122}{22} }\\\\\\\longrightarrow\sf{x=\dfrac{98+122}{22} \:\:Or\:\:x=\dfrac{98-122}{22} }\\\\\\\longrightarrow\sf{x=\cancel{\dfrac{220}{22}} \:\:\:Or\:\:\:x=\dfrac{-24}{22} }$

$\longrightarrow\sf{\orange{x=10\:\:\:Or\:\:\:x\neq \dfrac{-24}{22} }}$

Thus;

r = 10

$\boxed{\bf{The\:orginal\:fraction=\frac{10-3}{10} ={\boxed{\sf{\frac{7}{10}}}}}}}$

Answer 2

Answer:

Step-by-step explanation:

☆ Let the denominator be x

           Numerator  is x-3

The fraction is (x-3)/x

when 2 is added to numerator and denominator

                             

 we have fraction as (x-1)/(x+2) 

A/Q  (x-3)/x +(x-1)/(x+2)= 29/20

  ⇒    (x-3)((x+2) +x(x-1)=x(x+2)*29/20

  ⇒    (x²-6-x + x²-x)20= 29x²+58x

  ⇒    40x²-120-40x=29x²+58x

  ⇒    11x²-98x- 120=0

  ⇒    11x² - 110x +12x -120=0

  ⇒     (11x+12)(x-10)=0

Canceling the negative fraction value  of x we have x=10

Then the fraction (x-3)/x=7/10

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