For what values of m and n the following system of linear equation has

1.	For what values of m and n the following system of linear equation has infinetly many solution 3x +4y=12 : (m+n)x +2(m-n)y=5m-1
Answer» Given :- $\rm\:3x+4y=12$ $\rm\:(m+<klux>N</klux>)x+2(m-n)=(5m-<klux>1</klux>)$ To find :- Value of m and n SOLUTION :- The equations is WRITTEN as:- $\rm\:3x+4y-12=0\:.........................(i)$ $\rm\:(m+n)x+2(m-n)y-(5m-1)=0\:......(ii)$ These equations are of the form $\rm\:{a_1}x+{b_1}y+{c_1}=0,$ $\rm\:{a_2}x+{b_2}y+{c_2}=0$ where $\rm\:a_1=3\:,b_1=4\:,c_1=-12$ and $\rm\:a_2=(m+n)\:,b_2=2(m-n)\:,c_1=-(5m-1)$ For infinitely many Solutions,we must have, $\rm\:\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}$ This holds only when, $\rm\:\dfrac{3}{(m+n)}=\dfrac{4}{2(m-n)}=\dfrac{-12}{-(5m-1)}$ $\rm\longrightarrow\:\dfrac{3}{(m+n)}=\dfrac{2}{(m-n)}=\dfrac{12}{(5m-1)}$ $\rm\longrightarrow\:\dfrac{3}{(m+n)}=\dfrac{12}{(5m-1)}\:and\:\:\dfrac{2}{(m-n)}=\dfrac{12}{(5m-1)}$ $\rm\longrightarrow\:\dfrac{1}{(m+n)}=\dfrac{4}{(5m-1)}\:and\:\:\dfrac{1}{(m-n)}=\dfrac{6}{(5m-1)}$ $\rm\longrightarrow\:4(m+n)=(5m-1)\:and\:\:(5m-1)=6(m-n)$ $\rm\longrightarrow\:m-4n=1\:and\:\:m-6n=-1$ $\rm\longrightarrow\:m=5\:and\:\:n=1$ [on solving the above equation] Hence,for infinitely many solutions the value of m is 5 and n is 1.