Let's see what we have:
#4x - 2y = 6#
#3x + 2y = 8#
Hmm... what can we do with these? Since we're looking for a solution, we're looking for a pair of numbers #(x, y)# that, when substituted, will make both equations true. To do that, we need to solve for one first, then the other.
It seems like we could try to... add them together? The #y#s sure could cancel out!
#(4x - 2y) + (3x + 2y) = (6) + (8)#
#7x = 14#
Nice! Seems like we could simply divide by #7#:
#(7x)/7 = 14/7#
#x = 2#
Now that we've solved for #x#, we just need to solve for #y#. Let's try substituting for #x# in one of the equations... err... the second one:
#3x + 2y = 8 rarr 3(2) + 2y = 8#
#6 + 2y = 8#
Ah, we could subtract both sides by #6#:
#(6 + 2y) - 6 = 8 - 6#
#2y = 2#
And divide by #2#:
#(2y)/2 = 2/2#
#y = 1#
There we go! We have #x = 2# and #y = 1#, or more simply, #(2, 1)# as a solution. Just to make sure, let's "verify" the answers by plugging them in to the equations and seeing if we get a true statement in return:
#4x - 2y = 6 rarr 4(2) - 2(1) = 6 rarr 8 - 2 = 6 rarr 6 = 6#
#3x + 2y = 8 rarr 3(2) + 2(1) = 8 rarr 6 + 2 = 8 rarr 8 = 8#
Yep!
