Question

$\fcolorbox{red}{lime}{PLEASE SOLVE THIS QUESTION}$
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Answer 1

Explanation:

Given :

5m - n = 3m + 4n

therefore ,

5m - 3m = 4n + n

Therefore,

2m = 5n

1. m² / n² = 25/4 [ squaring both sides ]

= m² + n² / m² - n² = 25+4/ 25-4 [ By componendo dividends ]

m² + n² / m² - n² = 29/21

m² + n² / m² - n² = 29:21

Hõpê ìt hêlps yóú

Answer 2

*Question:—

if 5m-n = 3m+4n then to find the value of m²+n²/m²-n² by complete the following activity.

*Answer:—

$\dfrac{29}{21}.$

Step-by-step explanation:—

Given,

$\rm 5m-n=3m+4n.$

On solving,

$\begin{gathered}\rm 5m-n=3m+4n\\\\\text{Subtracting 3m from both the sides},\\5m-n-3m=3m+4n-3m\\2m-n=4n\\\\\text{Adding n on both the sides,}\\2m-n+n=4n+n\\2m=5n\\\\\text{Dividing 2 on both the sides,}\\\dfrac{2m}{2} = \dfrac{5n}{2}\\m=\dfrac{5n}{2}.\end{gathered}$

Putting this value of m in :—

$\rm \dfrac{m^2+n^2}{m^2-n^2}$

We get,

$\begin{gathered}\rm \dfrac{m^2+n^2}{m^2-n^2}= \dfrac{\left( \dfrac{5n}{2}\right )^2+n^2}{\left (\dfrac{5n}{2}\right )^2-n^2}\\= \dfrac{\left( \dfrac{25n^2}{4}\right )+n^2}{\left (\dfrac{25n^2}{4}\right )-n^2}\\=\dfrac{25n^2+4n^2}{25n^2-4n^2}\\=\dfrac{29n^2}{21n^2}\\=\dfrac{29}{21}.\end{gathered}$

$@vikramjeeth$