How do you solve #3x + 2y = 4# and #2x + y = 3#?

1 Answer
Mar 17, 2017

Answer:

See the entire solution process below:

Explanation:

Step 1) Solve the second equation for #y#:

#2x + y = 3#

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

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

#0 + y = 3 - 2x#

#y = 3 - 2x#

Step 2) Substitute #3 - 2x# for #y# in the first equation and solve for #x#:

#3x + 2y = 4# becomes:

#3x + 2(3 - 2x) = 4#

#3x + (2 xx 3) - (2 xx 2x) = 4#

#3x + 6 - 4x = 4#

#3x - 4x + 6 = 4#

#(3 - 4)x + 6 = 4#

#-1x + 6 = 4#

#-x + 6 - color(red)(6) = 4 - color(red)(6)#

#-x + 0 = -2#

#-x = -2#

#color(red)(-1) xx -x = color(red)(-1) xx -2#

#x = 2#

Step 3) Substitute #2# for #x# in the solution to the second equation at the end of Step 1 and calculate #y#:

#y = 3 - 2x# becomes:

#y = 3 - (2 xx 2)#

#y = 3 - 4#

#y = -1#

The solution is #x = 2# and #y = -1# or #(2, -1)#