Question

solve this question........​

Answer 1

Answer:       n=7

and

                    d=10          

Step-by-step explanation:

$OLet\;the\;original\;fraction=\frac{n}{d}$

But, it is given that n(numerator)=d-3

so., the original equation

$=\frac{d-3}{d}.............(i)$

Now, A.T.Q.

The new equation is

$= \frac{d-3 +2}{d+2}=\frac{d-1}{d+2}...........(ii)$

$=\frac{ (d-3)(d+2) +d(d-1) }{ d(d+2) }=\frac{ d^{2}-3d+2d-6+d^{2}-d }{ d^{2}+2d }\\\\=\frac{2d^{2}-2d-6}{d^{2}+2d}=\frac{29}{20}[Given\;in\;the\;question]\\ \\ Now,\;cross\;multiplying\\ =>20(2d^{2}-2d-6)=29(d^{2}+2d)\\ =>40d^{2}-40d-120=29d^2+58d\\=>11d^{2}-98d-120=0\\=>11d^{2}-110d+12d-120=0\\=>11d(d-10)+12(d-10)=0\\=>(11d+12)(d-10)=0\\=>d=\frac{-12}{11}\;\;\;\;or\;\;\;d=10\\d\neq \frac{-12}{11}(It\;doesn't\;satisfy\;the\;condoitions\;of\;question\\=>d=10\\and\;numerator\;\;(n)=d-3[given]$

$=>n=10-3=7$

$Learning\;\;together\;:)$