How do you solve the following system: #3x - 2y = 9, 2x + 3y = 12#?

#9x-6y=27#
#4x-6y=24#

when you expand the first equation (3) and the second (2)
Combine the above equations:

#9x+4x=51#
#x=51/13#

Put this value in any equation to get y

#2*51/13 + 3y=12#

#3y=(156-102)/13#

#y=54/(13*3)#

#y=18/13#

Your #x=51/13# and your #y=18/13#

Step 1) Solve the first equation for #x#:

#3x - 2y = 9#

#3x - 2y + color(red)(2y) = 9 + color(red)(2y)#

#3x - 0 = 9 + 2y#

#3x = 9 + 2y#

#(3x)/color(red)(3) = (9 + 2y)/color(red)(3)#

#(color(red)(cancel(color(black)(3)))x)/cancel(color(red)(3)) = 9/3 + (2y)/3#

#x = 3 + 2/3y#

Step 2) Substitute #3 + 2/3y# for #x# in the second equation and solve for #y#:

#2x + 3y = 12# becomes:

#2(3 + 2/3y) + 3y = 12#

#(2 * 3) + (2 * 2/3y) + 3y = 12#

#6 + 4/3y + 3y = 12#

#6 + 4/3y + (3/3 * 3y) = 12#

#6 + 4/3y + 9/3y = 12#

#6 + 13/3y = 12#

#-color(red)(6) + 6 + 13/3y = -color(red)(6) + 12#

#0 + 13/3y = 6#

#13/3y = 6#

#color(red)(3)/color(blue)(13) xx 13/3y = color(red)(3)/color(blue)(13) xx 6#

#cancel(color(red)(3))/cancel(color(blue)(13)) xx color(blue)(cancel(color(black)(13)))/color(red)(cancel(color(black)(3)))y = 18/13#

#y = 18/13#

Step 3) Substitute #18/13# for #y# in the solution to the first equation at the end of Step 1 and calculate #x#:

#x = 3 + 2/3y# becomes:

#x = 3 + (2/3 xx 18/13)#

#x = 3 + (2/color(red)(cancel(color(black)(3))) xx (color(red)(cancel(color(black)(18)))6)/13)#

#x = 3 + 12/13#

#x = (13/13 xx 3) + 12/13#

#x = 39/13 + 12/13#

#x = 51/13#

The solution is: #x = 51/13# and #y = 18/13# or #(51/13, 18/131)#