How do you solve the simultaneous equations #x+y+z=-2#, #2x+5y+2z=-10#, #-x+6y-3z=-16# ?

2 Answers
Mar 25, 2017

Answer:

#(x, y, z) = (-2, -2, 2)#

Explanation:

Given:

#{ (x+y+z=-2),(2x+5y+2z=-10),(-x+6y-3z=-16) :}#

Subtracting twice the first equation from the second, we get:

#3y = -6#

Dividing both sides by #3# we find:

#y = -2#

Adding the first and third equation together, we get:

#7y-2z = -18#

Substituting #y=-2# into this equation, we get:

#-14-2z = -18#

Add #14# to both sides to get:

#-2z = -4#

Divide both sides by #-2# to get:

#z = 2#

Then putting #y=-2# and #z = 2# in the first equation, we find:

#x-color(red)(cancel(color(black)(2)))+color(red)(cancel(color(black)(2)))=-2#

Hence:

#x = -2#

Mar 25, 2017

Answer:

Use the 3 equations to write an Augmented Matrix and then perform elementary row operations until you obtain an identity matrix.

Explanation:

Write the augmented matrix:

#[ (1,1,1,|,-2), (2,5,2,|,-10), (-1,6,-3,|,-16) ]#

Perform elementary row operations.

#-2R_1+R_2toR_2#

#[ (1,1,1,|,-2), (0,3,0,|,-6), (-1,6,-3,|,-16) ]#

#R_2/3#

#[ (1,1,1,|,-2), (0,1,0,|,-2), (-1,6,-3,|,-16) ]#

#R_1+R_3toR_3#

#[ (1,1,1,|,-2), (0,1,0,|,-2), (0,7,-2,|,-18) ]#

#-7R_2+R_3toR_3#

#[ (1,1,1,|,-2), (0,1,0,|,-2), (0,0,-2,|,-4) ]#

#R_3/-2#

#[ (1,1,1,|,-2), (0,1,0,|,-2), (0,0,1,|,2) ]#

#R_1-R_2toR_1#

#[ (1,0,1,|,0), (0,1,0,|,-2), (0,0,1,|,2) ]#

#R_1-R_3toR_1#

#[ (1,0,0,|,-2), (0,1,0,|,-2), (0,0,1,|,2) ]#

We have obtained an identity matrix and the right column contains the solution set:

#x = -2, y = -2, and z = 2#