Question

Solve the following system of equations.

3x – 2y + z = –6
x – 3y + z = –3
2x + y – 2z = –7

(1, 5, 1)
(–2, 1, 2)
(–3, –6, –7)
(1, –3, 3)

Answer 1

do you remember how to solve a 2by2 system using elimination?

Answer 2

or addition method

Answer 3

the short easy way is by using matrix algebra and a calculator...but i doubt you have learned that yet

Answer 4

no i haven't learned that

Answer 5

ok there are multiple steps...

first you must eliminate a variable, meaning the goal is to reduce this to a system of 2 equations
lets eliminate z, so pick any 2 equations and use elimination method to make the z's cancel out
3x -2y +z = -6
x -3y +z = -3

multiply bottom equation by -1 so that the "z" coefficients will add to zero
then add the equations
3x -2y +z = -6
-x +3y -z = 3
--------------
2x +y = -3

Now pick 2 other equations and do the same thing, eliminate z
x -3y +z = -3
2x + y -2z = -7

multiply top equation by 2 so "z" coefficients add to zero
2x -6y +2z = -6
2x +y -2z = -7
--------------
4x -5y = -13

Answer 6

Next step, take those 2 new equations and try to eliminate either x or y
lets do y

2x +y = -3
4x -5y = -13

multiply top by 5 so that "y" coefficients add to zero
10x +5y = -15
4x -5y = -13
---------------
14x = -28

Leaving a simple single equation to solve for x

Answer 7

Oh ok and then you would plug x in to get y and z. Thank you so much for the help!

Answer 8

no problem