Question

Solve the following system of equations:
x + 4y - z = 6
2x - y + z = 3
3x + 2y + 3z = 16

Answer 1

Do you prefer the substitution method or elimination? or does it not matter.

Answer 2

matrix ?

Answer 3

Elimination, but it doesn't really matter.

Answer 4

it really looks like a matrix problem.

Answer 5

What do you mean matrix problem? I need to solve for x y z.

Answer 6

Yep. I'm in precalc. 10th Grade.

Answer 7

Well, I am saying that I would prefer matrix, but you don't really have to use it if you don't like it.

Answer 8

How would you do that?

Answer 9

\(\large\color{black}{ \displaystyle \left[\begin{matrix}? & ? & ?~~~|~~~? \\ ? & ? & ?~~~|~~~? \\ ? & ? & ?~~~|~~~? \end{matrix}\right] }\)

on the right on each row, after the " | " you have the constant that the equation is equivalent to, and inside you will in the variable coefficients.

\(\large\color{black}{ \displaystyle x + 4y - z = 6 \\
2x - y + z = 3 \\
3x + 2y + 3z = 16 }\)

\(\large\color{black}{ \displaystyle \left[\begin{matrix}1 & 4 & -1~~~|~~~6 \\ 2 & -1 & 1~~~|~~~3 \\ 3 & 2 & 3~~~|~~~16 \end{matrix}\right] }\)

Answer 10

BUT, this is only I am allowed to do when variables in each row correspond to appropriate column.

like first column (if you look at 3 equations) is X's, 2nd column is Y's and etc...

Answer 11

then by manipulating row (multiplying adding subtracting and all that)

we must get to

\(\large\color{black}{ \displaystyle \left[\begin{matrix}1 & 0 & 0~~~|~~~a \\ 0 & 1 & 0~~~|~~~b \\ 0 & 0 & 1~~~|~~~c \end{matrix}\right] }\)

where as we are done, we will have
x=a
x=b
x=c

(for whatever numbers you have as your a b c )

Answer 12

if you haven't done those before though, lets use some more usual approaches.

Answer 13

How do you manipulate the rows to get there?

Answer 14

I mean
x=a, y=b, z=3
excuse me

Answer 15

by adding and subtracting row... just the same way as by doing so with regular equations.

Answer 16

except that this is at times more convenient when you have a lot of variables.
but again, if this is not something you are well familiar with, then lets go with other approaches (like substitution).

Answer 17

How do you do that with 3?

Answer 18

how to do what with 3 variables ?

Answer 19

the substitution you will need to sub into the other 2 (remaining) equations.

Answer 20

3 equations

Answer 21

x + 4y - z = 6
2x - y + z = 3
3x + 2y + 3z = 16

rearranging the first equation for z

x + 4y - z = 6
x + 4y = 6 + z
`x + 4y - 6 = z`

so the new system we will use is:
x + 4y - 6 = z
2x - y + z = 3
3x + 2y + 3z = 16

NOW, the substitution comes in handy.

2x - y + (x + 4y - 6) = 3
3x + 2y + 3(x + 4y - 6) = 16

Answer 22

then you have a regular system of equations.

After you find the x and y, substitute the values of x and y (that you have found) into any of the ORIGINAL (from very beginning) equations to find z.

Answer 23

So you just sub in one equation into the others.