How do you solve this system of equations: # 4x + 3y = 1 and x = 1- y#?

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Answer 1 · 2017-12-06T04:11:30

See a solution process below:

Step 1) Because the second equation is already solved for #x# we can substitute #(1 - y)# for #x# in the first equation and solve for #y#:

#4x + 3y = 1# becomes:

#4(1 - y) + 3y = 1#

#(4 xx 1) - (4 xx y) + 3y = 1#

#4 - 4y + 3y = 1#

#4 + (-4 + 3)y = 1#

#4 + (-1)y = 1#

#4 - 1y = 1#

#4 - y = 1#

#4 - color(red)(4) - y = 1 - color(red)(4)#

#0 - y = -3#

#-y = -3#

#color(red)(-1) xx -y = color(red)(-1) xx -3#

#y = 3#

Step 2) Substitute #3# for #y# in the second equation and calculate #x#:

#x = 1 - y# becomes:

#x = 1 - 3#

#x = -2#

The Solution Is: #x = -2# and #y = 3# or #(-2, 3)#