How do you solve the system of equations algebraically #x-3z=7, 2x+y-2z=11, -x-2y+9z=13#?

1 Answer
Jan 14, 2017

Answer:

#x = 10#; #y = -7#; #z = 1#

Explanation:

Write out the three equations to line up the unknown values. This will give you a clue how to proceed.

Always be careful of the signs #(-) (+)# as they move across the #(=)#. Always write the sign with the value to which it refers. #(-3z)#

# x -3z = 7 #

#2x +y -2z = 11#

# -x -2y +9z = 13#

Already we can see that it is easy to solve for #x# in the first equation:

#x -3z = 7 >> x = 3z +7#

Insert the value for #x# into the second equation;

#2(3z +7) +y -2z = 11#

#6z +14 + -2z = 11#

#4z +y = -3#

#y = -4z -3#

Substitute the values of #x# and #y# into the third equation:

#-(3z+7) -2(-4z-3) +9z = 13#

#-3z -7 +8z +6 +9z = 13#

#14z = 14#

#z = 1#

Substitute the value of #z# into the first equation:

#x -3(1) = 7#

#x = 10#

Substitute the values of #x# and #z# into the second equation:

#2(10) +y -2(1) = 11#

#20 +y -2 = 11#

#y = -20 +2 +11#

#y = -7#

Use either equation that includes all three unknowns to check:

#-x -2y +9z = 13#

#-(10) -2(-7) +9(1) = 13#

#13 = 13#