How do you solve the system of equations algebraically #7x+5y+z=0, -x+3y+2z=16, x-6y-z=-18#?

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Answer 1 · 2017-12-04T12:52:23

Solution #x=-2,y=2,z=4# [Ans]

#7x+5y+z=0 (1) , -x+3y+2z=16 (2)# and

#x-6y-z=-18 (3)# from equation (1) we get #z=-7x-5y#

Putting #z=-7x-5y# in equation (2) and (3) we get

#-x+3y-14x-10y=16 or -15x -7y =16 (4)# and

#x-6y+7x+5y=-18 or 8x-y=-18(5)# Multiplying

equation (5) by #7# we get #56x-7y=-126(6)# Subtracting

equation (4) from equation (6) we get #71x=-142 or x=-2#

Putting #x=-2# in equation (1) and (2) we get

#5y+z=14 (7) and 3y+2z=14 (8)# Multiplying

equation (7) by #2# we get #10y+2z=28 (9)# Subtracting

equation (8) from equation (9) we get #7y=14 or y=2#

Putting #y=2# in equation (8) we get #3*2+2z=14 or#

#2z=8 or z=4 #. Solution #x=-2,y=2,z=4# [Ans]

Answer 2 · 2017-12-04T17:13:24

#x=-2#, #y=2# and #z=4#

Adding third equation to first one,

#7x+5y+z+x-6y-z=0-18# or #8x-y=-18#. #(1)#

Adding 2 times of third one to second one,

#-x+3y+2z+2*(x-6y-z)=16-2*18# or #x-9y=-20# #(2)#

Substracting 8 times of #(2)# from #(1)#,

#8x-y-8*(x-9y)=-18-8*(-20)#

#71y=142#, so #y=2#

Hence,

#x-9*2=-20# or #x-18=-20#, so #x=-2#

Thus,

#7*(-2)+5*2+z=0# or #z-4=0#, so #z=4#