How do you solve the following system: #-x+6y=12, x-4y=2 #?

1 Answer
Dec 14, 2015

Answer:

#color(white)(xx)x=30#, #y=7#

Explanation:

#color(white)(xx)-x+6y=12, x-4y=2#

#color(white)(xx)-x+6y=12<=>y=(x+12)/6#
#color(white)(xx)x-4y=2<=>y=(x-2)/4#

#color(white)(xx)y=(x+12)/6#
#=>color(blue)((x-2)/4)=(x+12)/6#

Multiply both side by #color(red)12#:
#color(white)(xx)color(red)(12xx)(x-2)/4=color(red)(12xx)(x+12)/6#
#=>3x-6=2x+24#

Add #color(red)(-2x+6)# to both side:
#color(white)(xx)3x-6color(red)(-2x+6)=2x+24color(red)(-2x+6)#

#color(white)(xx)x=30#

#color(white)(xx)y=(x-2)/4#

#color(white)(xxx)=(color(red)30-2)/4#
#color(white)(xxx)=(color(red)28)/4#
#color(white)(xxx)=7#