How do you solve the following system using substitution?: #[(2x,+, y, -, z,=,3), (-x, + ,2y, +, 4z,=,-3), (x,-,2y,-,3z,=,4)]#

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Answer 1 · 2018-03-16T13:04:49

See explanation below

From third equation we can express #x=2y+3z+4#

Lets substitute in second equation

#-2y-3z+4+2y+4z=-3# adding similar terms we have

#z=-3-4=-7#

Using this value (and expresion for #x#) in first equation

#2(2y+3·(-7)+4)+y+7=3#

#4y-42+6+y+7=3#
#5y=32#
#y=32/5# an finally #x=2·32/5-21+4=-21/5#

The only solution is #x=-21/5; y=32/5; z=-7#