We begin with #2m+3n=4# and #-m+2n=5#. I prefer to solve these kinds of problems by elimination. We could also solve this by substitution or by graphing.
The way we set up elimination is like so:
#color(white)(.)2m+3n=4#
#-m+2n=5#
Now, the goal of elimination is to get rid of a variable. Right now, nothing would subtract or add to zero, so let's tinker with the problem. I'm going to multiply #-m+2n=5# by #2# so that I can add away the #m#s. I'll show you what I mean:
#color(white)(..............)2m+3n=4#
#+#
#color(white)(.....)2*(-m+2n=5)#
Which becomes
#color(white)(.....)2m+3n=4#
#+#
#color(white)(.)-2m+4n=10#
#-------#
#color(white)(........)0+7n=14#
Divide by #7# on both sides, and we have #n=2#
To find #m#, we just plug in #n=2# to one of the equations. I'm just going to use #-m+2n=5#. We have #-m+2(2)=5#, or #-m+4=5#, which we can simplify to #-m=1#. If we divide by #-1# on both sides, we have #m=-1#.
To double check our work, let's plug our values for #m# and #n# into the other equation, #2m+3n=4#. Once we plug in #m# and #n# for #-1# and #2#, respectively, we have #2(-1)+3(2)=4#. This becomes #-2+6#, which equals #4#. #4=4#, and we got it right! Nice job!
