How do you solve the system of equations: #2x + 6y + 3z = 5, - 2x + 2y + z = 15 , and - 3y + 4z = 9#?

Question

1364 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-13T07:58:51

#x=-5#
#y=1#
#z=3#

The process used is involves row reductions/matrices

#2x+6y+3z=5#
#-2x+2y+z=15#
#0x+−3y+4z=9#

Take the three equations and put them into a matrix to make working easier

#([2,6,3,5],[-2,2,1,15],[0,-3,4,9])#

Add row 1 to row 2

#([2,6,3,5],[0,8,4,20],[0,-3,4,9])#

Then subtract row 3 from row 2

#([2,6,3,5],[0,11,0,11],[0,-3,4,9])#

Now converting the rows back to equations,

#2x+6y+3z=5#
#0x+11y+0z=11#
#0x+−3y+4z=9#

Since #11y=11#, you know #y=1#
Then, substituting that into the third equation,

#-3(1)+4z=9#
#4z=12#
#z=3#

Finally, substituting both y and z values into the first equation,

#2x+6(1)+3(3)=5#
#2x+6+9=5#
#2x=-10#
#x=-5#