How do you solve the system of equations #x+3y = 12# and #-4x + 3y = -3#?

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Answer 1 · 2016-03-14T07:35:35

#x=15/4#
#y=11/4#

Given -

#x+3y=12# ---------------(1)
#-3x+3y=-3# -------------(2)

Solve the first equation for #x#

#x+3y=12#
#x=12-3y#

Substitute #x=12-3y# in equation (2)

#-3(12-3y)+3y=-3#
#-36+9y +3y=-3#
#12y=-3+36#
#y=33/12#
#y=11/4#

Substitute #y=11/4# in equation (1)

#x+3(11/4)=12#
#x+33/4=12#
#x=12-33/4#
#x=(48-33)/4#
#x=15/4#