Question

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

Answer 1

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) 

ATQ  (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

Answer 2

AnswEr :

7/10.

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

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

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

The original fraction.

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

Let the Denominator be r

Let the Numerator be (r-3)

$\therefore\underline{\bf{The\:original\:fraction\:=\:\green{\dfrac{r-3}{r} }}}}}$

A/q

$\mapsto\sf{New\:fraction=\dfrac{r-3+2}{r+2} =\dfrac{r-1}{r+2} }$

The sum of the new fraction and original fraction is 29/20.

$\mapsto\tt{\dfrac{r-3}{r} +\dfrac{r-1}{r+2} =\dfrac{29}{20} }\\\\\\\mapsto\tt{\dfrac{(r-3)(r+2)+r(r-1)}{r(r+2)} =\dfrac{29}{20}}\\\\\\\mapsto\tt{\dfrac{r^{2}+2r-3r-6+r^{2} -r }{r^{2}+2r } =\dfrac{29}{20} }\\\\\\\mapsto\tt{\dfrac{2r^{2}-2r-6 }{r^{2}+2r } =\dfrac{29}{20} }\\\\\\\mapsto\tt{20(2r^{2} -2r-6)=29(r^{2} +2r)}\\\\\\\mapsto\tt{40r^{2} -40r-120=29r^{2} +58r}\\\\\\\mapsto\tt{40r^{2} -29r^{2} -40r-58r-120=0}\\\\\\\mapsto\tt{11r^{2} -98r-120=0}}$

$\blacksquare\bf{\blue{\underline{\underline{\bf{Using\;quadratic\:formula\::}}}}}$

We have quadratic equation as we compared with ax² + bx + c = 0

• a = 11
• b = -98
• c = -120

$\leadsto\tt{\blue{x=\dfrac{-b\pm\sqrt{b^{2}-4ac } }{2a} }}\\\\\\\leadsto\tt{x=\dfrac{-(-98)\pm\sqrt{(-98)^{2}-4*11*(-120) } }{2*11} }\\\\\\\leadsto\tt{x=\dfrac{98\pm\sqrt{9604-44*(-120)} }{22} }\\\\\\\leadsto\tt{x=\dfrac{98\pm\sqrt{9604+5280} }{22} }\\\\\\\leadsto\tt{x=\dfrac{98\pm\sqrt{14884} }{22} }\\\\\\\leadsto\tt{x=\dfrac{98\pm122}{22} }\\\\\\\leadsto\tt{x=\dfrac{98+122}{22} \:\:Or\:\:x=\dfrac{98-122}{22} }\\\\\\\leadsto\tt{x=\cancel{\dfrac{220}{22}} \:\:Or\:\:x=\dfrac{-24}{22} }$

$\leadsto\tt{\red{x=10\:\:\:Or\:\:\:x=\dfrac{-24}{22} }}$

∴ We know that negative value isn't acceptable.

Thus;

r = 10

$\underbrace{\sf{The\:original\:fraction\:is\:\frac{r-3}{r} =\frac{10-3}{10} =\pink{\frac{7}{10}}}}}$