Considering:
#Eq_1: 3x+2y=18#.
#Eq_2: 8x+4y=41#.
When we multiply the entire equation by a #k# number, the equality is maintained:
#2*(3x+2y)=2*(18)#
#rArr6x+4y=36#
Since the equality is maintained, its possible to sum each part of it on the second equation. But first, its interesting that we multiply again by #-1#: its interesting because we are finding a number that takes #y# out of the equation. Then, we find #x#'s value and solve it:
#-1*(6x+4y) = -1*(36)#
#rArr-6x-4y=-36#
Now, we sum the result of the manipulation that we did using the #Eq_1# on #Eq_2#:
#8x+4y + (-6x-4y) = 41 + (-36)#
#rArr8x - 6x + cancel(4y) cancel(-4y) = 41 - 36#
#rArr2x = 5#
#rArr x = 5/2#
We now have the value of #x# and need to find the value of #y#:
To do this, is simple: put #x = 5/2# into one of the equations.
I'm going to be doing this in #Eq_2# because the coefficient of #x# can be divided by #2" "# (since #x=5/2#):
#8*(5/2) + 4y = 41 #
#rArr4 * 5 + 4y = 41#
#rArr4y = 41-20#
#rArr4y = 21#
#rArr y = 21/4#
