Question

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

Answer 1

Step-by-step explanation:

Given:-

• Numerator of fraction is 3 less than its denominator

To find:-

• Original fraction = ....?

Find:-

Let the denominator of the require fraction be x.

Then, its numerator = (x-3)

So, the original fraction is (x-3)/x

$\frac{(x - 3) + 2}{x + 2} + \frac{x - 3}{x} = \frac{29}{20} \\ \frac{(x - 1)}{(x + 2)} + \frac{(x - 3)}{x} = \frac{29}{20} \\ \frac{x(x - 1) + (x - 3)(x + 2)}{x(x + 2)} = \frac{29}{20} \\ \frac{ ({x}^{2} - x) + ( {x}^{2} - x - 6) }{ {x}^{2} + 2x} = \frac{29}{20} \\ 20(2 {x}^{2} - 2x - 6) = 29( {x}^{2} + 2x) \\ 40 {x}^{2} - 40x - 120 = 29 {x}^{2} + 58x \\ 11 {x}^{2} - 98x - 120 = 0 \\ 11 {x}^{2} - 110x + 12x - 120 = 0 \\ 11x(x - 10) + 12(x - 10) = 0 \\ (x - 10)(11x + 12) = 0 \\ x = 10 \: or \: x = \frac{ - 12}{11} \\ x = 10$

Denominator of fraction never negative so x =10

So, denominator = 10 and numerator = (10-3)=7

Hence ,the required fraction is$\frac{7}{10}$

Hope its help uh❤

Answer 2

Answer:

Here is the correct answer mate....

Thank you ☺✌...