Question

5m-n=3m+4n then find the values of m²+n²/m²-n²

Answer 1

See the attachment for detail solution
Hope it will help you

Answer 2

Answer:

$\dfrac{29}{21}.$

Step-by-step explanation:

Given

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

On solving,

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

Putting this value of m in $\rm \dfrac{m^2+n^2}{m^2-n^2}$, we get,

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