How do you solve the system #a + 2b = -2#, #-a + b + 4c = -7#, #2a + 3b -c =5#?

1 Answer

Answer:

#(a, b, c)=(52,-27,18)#

Explanation:

#a+2b=-2#
#-a+b+4c=-7#
#2a+3b-c=5#

I'm going to use the first equation to substitute for a in the other two equations:

#a=-2b-2#

~~~~~

#-(-2b-2)+b+4c=-7#

#2b+2+b+4c=-7#

#3b+4c=-9#

~~~~~

#2(-2b-2)+3b-c=5#

#-4b-4+3b-c=5#

#-b-c=9#

And we can now use this last equation to substitute into the one prior:

#b=-c-9#

~~~~~

#3(-c-9)+4c=-9#

#-3c-27+4c=-9#

#c=18#

And now let's find b:

#-b-c=9#

#-b-18=9#

#b=-27#

And check it:

#3b+4c=-9#

#3b+4(18)=-9#

#3b+72=-9#

#3b=-81#

#b=-27#

And now let's find the last variable:

#a+2b=-2#

#a+2(-27)=-2#

#a-54=-2#

#a=52#

and check it:

#-a+b+4c=-7#

#-52-27+4(18)=-7#

#-79+72=-7#

#-7=-7#

To summarize,

#(a, b, c)=(52,-27,18)#