How do you solve the system #x-4y-z=-3#, #x+2y+z=5#, and #3x-7y-2z=-6#?

2 Answers
Feb 19, 2018

Answer:

#x=0.5; y=-0.5; z=5.5#

Explanation:

Add first two expressions to eliminate
the z
#x-4y-z+x+2y+z=-3+5#
#2x-2y=2#

Multiply the first expression by -2, then add to the third expression to eliminate the z again
#-2x+8y+2z=6#
#-2x+8y+2z+3x-7y-2z=6-6#
#x+y=0#

Multiply this answer by 2, then add to the first result to eliminate the y.
#2x+2y=0#
#2x+2y+2x-2y=0+2#
#4x=2#
#x=0.5#

Substitute the x-value into one of the result formulae and solve for y
#0.5+y=0#
#y=-0.5#

Substitute the x-value and the y-value into one of the original formulae and solve for z
#0.5-1+z=5#
#z=5.5#

Substitute your answers into BOTH of the other original formulae to check your results
#0.5+2-5.5=-3. True#
#1.5+3.5-11=-6. True#

Feb 19, 2018

Answer:

#x=1/2#, #y=-1/2# and #z=11/2#

Explanation:

Perform the Gauss Jordan elimination on the augmented matrix

#A=((1,-4,-1,|,-3),(1,2,1,|,5),(3,-7,-2,|,-6))#

I have written the equations not in the sequence as in the question in order to get #1# as pivot.

Perform the folowing operations on the rows of the matrix

#R2larrR1-R2#; #R3larrR3-3R1#

#A=((1,-4,-1,|,-3),(0,6,2,|,8),(0,5,1,|,3))#

#R2larr(R2)/2#

#A=((1,-4,-1,|,-3),(0,3,1,|,4),(0,5,1,|,3))#

#R2larrR2-R3#

#A=((1,-4,-1,|,-3),(0,-2,0,|,1),(0,5,1,|,3))#

#R2larr(R2)/(-2)#

#A=((1,-4,-1,|,-3),(0,1,0,|,-1/2),(0,5,1,|,3))#

#R1larrR1+4R2; R3-5R2#

#A=((1,0,-1,|,-5),(0,1,0,|,-1/2),(0,0,1,|,11/2))#

#R1larrR1+R3#

#A=((1,0,0,|,1/2),(0,1,0,|,-1/2),(0,0,1,|,11/2))#

Thus, #x=1/2#, #y=-1/2# and #z=11/2#