How do solve the following linear system?: # -6x+6y=6 , -6x+3y=-12 #?

2 Answers
Apr 4, 2018

#(x,y) = (5,6)#

Explanation:

#6y=6x+6#
#y=x+1# solve for #y# in terms of #x# in the first equation
#-6x+3(x+1)=-12# plug into the other equation
#-6x+3x+3=-12# distribute
#3-3x=-12# combine like terms
#-3x=-15# subtract 3 from both sides
#x=5# divide both sides by -3

Now, going back to the other equation
#-6(5)+6y=6# plug in the value 5 for x
#-30+6y=6# multiply
#6y=36# subtract 30 from both sides
#y=6# divide both sides by 6

Apr 4, 2018

Subtract equation 2 from equation 1 to solve for #y#, and then back-solve for #x# to find that #x=5# and #y=6#

Explanation:

Let's use linear algebra to solve the system of equations. This will involve subtracting whole equations from each other:

# -6x+6y=6#
#ul(-(-6x+3y=-12))#

#-6x+6y=6#
#ul(+6x-3y=12)#
#cancel(0x)+3y=18#

Now that #x# is eliminated, we can solve for #y#:

#3y=18#

#rArr (cancel(3)y)/cancel(color(red)(3))=18/color(red)(3)#

#rArr color(blue)(y=6)#

Finally, we can plug our solution for #y# back into either equation to solve for #x#:

#-6x+3y=-12#

#rArr -6x+3(color(blue)(6))=-12#

#rArr -6x+18=-12#

#rArr -6xcancel(+18color(red)(-18))=-12color(red)(-18)#

#rArr -6x=-30#

#rArr (cancel(-6)x)/cancel(color(red)(-6))=(-30)/color(red)(-6)#

#rArr color(green)(x=5)#