Question

solve the following systems of equations
x+3y+4z=14
2x-3y+2z=10
3x-y+z=9

Answer 1

As a first step,
try adding the first two equations together:
x+3y+4z=14
2x-3y+2z=10
can you do that ?

Answer 2

yes

Answer 3

3x+6z=24

Answer 4

that looks good. Notice all the terms can be divided by 3, and is always nice to simplify
you get
x + 2z = 8

that is one equation with x and z as unknowns.

now we need to use the last equation with one of the first 2 to get rid of the y and give us a second equation.

using the first, and the last equation
x+3y+4z=14
3x-y+z=9

to get the 3y in the first equation to "cancel out" we need -3y in the last equation
to get a -3y in the last equation, multiply both sides (and all terms) by 3. we get

9x -3y +3z= 27

now add the two equations
3x-y+z=9
9x -3y +3z= 27

Answer 5

oops, that last line should read
now add the two equations
x+3y+4z=14
9x -3y +3z= 27

Answer 6

10x+7z=41

Answer 7

now you have the two equations
x + 2z = 8
10x+7z=41
you could multiply the first equation by -10, and then add the two equations

Answer 8

z=-3

Answer 9

now pick one of the two equations with x and z (the simpler one is the easiest choice and solve for x)

Answer 10

except check your sign on z

Answer 11

I got -13 z = -39
z= -39/-13 = +3