Question

The numerator of a fraction is 3 less than its denominator. If we add 10 to the numerator the result is 2. Find the original fraction.



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Answer 1

Answer:

Let the fraction be xx−3

By the given condition, new fraction=x+2x−3+2

                                                            =x+2x−1

xx−3+x+2x−1=2029

⇒20[(x−3)(x+2)+x(x−1)]=29(x2+2x)

⇒20[x2−x−6+x2−x]=29x2+58x

⇒20[2x2−2x−6]=29x2+58x

⇒40x2−40x−120−29x2−58x=0

⇒11x2−98x−120=0

⇒11x2−110x+12x−120=0

Step-by-step explanation:

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Answer 2

$\sf\fbox\pink{Approprite Question}$

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 the new fraction and original fraction is 29/20. Find the original fraction.

$\sf\fbox\pink{Solution}$

Let the fraction be (x-3)/x

By the given condition, new fraction=(x-3+2)/(x+2)

$= \frac{x - 1}{x + 2} \\ \\ \frac{x - 3}{x} + \frac{x - 1}{x + 2} = \frac{29}{20} \\ \\ ∴\:\:=> 20 [(x - 3)(x + 2) + x(x - 1)] = 29( {x}^{2} + 2x) \\ \\ = 20( {x}^{2} - x - 6 + {x}^{2} - x) = 29x {}^{2} + 58x \\ \\ or \: 11x {}^{2} - 98x - 120 = 0 \\ \\ or \: 11x {}^{2} - 110x + 12x - 120 = 0 \\ \\ (11x + 12)(x - 10) = 0 \\ \\ => \: x = 10$

$\boxed{∴ The\:fraction\:is \frac{7}{10}}$