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

1 Answer

Answer:

#(x, y, z)=(-3/2,11/4,19/4)#

Explanation:

I'm going to label these equations:

#E1: x+y+z=6#
#E2: x-2y=-7#
#E3:4x+3y+z=7#

There's no #z# term in E2, so let's solve it for #x# and then substitute into E1 and E3:

#E2: x=2y-7#

#E1: (2y-7)+y+z=6#
#E3: 4(2y-7)+3y+z=7#

I'm going to simplify them:

#E1: 3y+z=13#
#E3: 11y+z=35#

We can subtract E1 from E3 to eliminate the #z# terms:

#E3-E1: (11y-3y)+(z-z)=(35-13)#

#E3-E1: 8y=22=>y=22/8=>color(blue)(ul(bar(abs(color(black)(y=11/4))))#

I'm going to plug this into E2 for to solve for #x#, and then will substitute the #x, y# values into E1 and E3 to solve for #z# and verify my work.

#E2: x-2y=-7#

#E2: x-2(11/4)=-7#

#E2: x-(11/2)=-14/2#

#E2: x=-14/2+11/2=-3/2=>color(blue)(ul(bar(abs(color(black)(x=-3/2))))#

~~~~~

#color(brown)(E1: x+y+z=6#
#color(red)(E3:4x+3y+z=7#

#color(brown)(E1: -3/2+11/4+z=6#
#color(red)(E3:4(-3/2)+3(11/4)+z=7#

#color(brown)(E1: z=24/4+6/4-11/4=19/4#
#color(red)(E3:z=28/4+24/4-33/4=19/4#

And so this shows we've got it right and that #z=38/8=>color(blue)(ul(bar(abs(color(black)(z=19/4))))#