Substitution is an interesting way to solve systems of equations. The first step with substitution is to isolate a variable. We are given two equations, #2x+4y=32# and #-2x+6y=-44#. I'm going to isolate #x# in the first equation, for no particular reason except that I don't want to mess with the negatives.
Anyways, if we start with #2x+4y=32#, then I am first going to subtract #4y# on both sides, until I have #2x=32-4y#. From here I just divide by #2#, and congratulations people, we have our #x#, which is #(32-4y)/2#, or #16-2y#.
Now let's plug that into the second equation, #-2x-6y=-44#, so that #x# is replaced by #16-2y#, like so: #-2(16-2y)-6y=-44#. If we distribute the #-2#, we have #-32+4y-6y=-44#. Adding #32# to both sides gives us #-2y=-12#, and if we divide by #-2# on both sides, we are left with #y=6#.
Now that we have #y#, let's go find #x#. I'm going to use the second equation, but you can use either one. That gives us #-2x-6(6)=-44# or #-2x-36=-44#. If we add #36# on both sides, we have #-2x=-8#, and by dividing by #-2# we are left with #x=4#.
Now we have #x# and #y#, and we could just stop here, but I want to double check our work. We can do this by plugging in our values for #x# and #y# into one of the equations. Let's try it with the first equation, #2x+4y=32#. This becomes #2(4)+4(6)=32#. If we simplify this, we have #8+24=32#, and #32=32#, so we were right!
If we want to triple check, just to be sure, we could plug in #4# and #6# for #x# and #y#, respectively, into the other equation. That gives us #-2(4)-6(6)=-44#. This becomes #-8-36=-44#, and #-44=-44#, so we were right! Again! But, it's good to be sure.
Hope this helps!
