# Math Help - Three-Variable Linear Systems

1. ## Three-Variable Linear Systems

I've tried solving this problem for a couple hours and I am stumped
a little help would be much appreciated

X + Y + Z = 18
X - Y - Z = 12
3X - 2Y + 4Z = 4

2. You need to solve

$\left[ \begin{array}{c}
x\\
y\\
z\end{array}\right]=\left[ \begin{array}{ccc}
1 & 1 & 1 \\
1 & -1 & -1 \\
3 & -2 & 4 \end{array} \right]^{-1} \left[ \begin{array}{c}
18\\
12\\
4\end{array}\right]$

You you know how to find this inverse?

3. i dont know how to find the inverse

4. You can use this Matrix calculator

Assuming you know simple matrix operations like mulitplication and finding determinants,

To solve the system consider cramer's Method.

Cramer's rule - Wikipedia, the free encyclopedia

If you don't know these elementary ideas I suggest this problem should be solved using subtitution and elimination.

5. (1) X + Y + Z = 18
(2) X - Y - Z = 12
(3) 3X - 2Y + 4Z = 4

Here's my simplistic and basic method

Rearranging (1) to isolate X
(4) $X=18-Z-Y$
Substituting into (2)
$18-Z-Y-Y-Z=12$
$18-2Z-2Y=12$
Rearranging this gives:
$2Y+2Z=6$
Taking out (and then dividing through by) a common factor of 2 gives
$Y+Z=3$
Substituting this into equation 1 gives
$X+3=18$
$X=15$

Substituting that into all equations leaves:
$Y+Z=3$
$2Y-4Z=41$
Which can then be solved easily simultaneously to give
$Y=3-Z$
$2(3-Z)-4Z=41$
$6-6Z=41$
$-6Z=35$
$Z=-5.833$
Then substituting this back into the above equations will give
$Y=8.8333$

6. Hello, jasimmons86!

What methods are you allowed to use?

$\begin{array}{cccc}x + y + z &=& 18 & [1] \\
x - y - z &=& 12 & [2] \\
3x - 2y + 4z &=& 4 & [3] \end{array}$

Add [1] and [2]: / $2x \:=\:30 \quad\Rightarrow\quad \boxed{x \;=\;15}$

$\begin{array}{ccccccc}\text{Substitute into [2]:} & 15 - y - z \:=\:12 & \Rightarrow & y + z \:=\:3 & [3] \\
\text{Substitute into [3]:} & 45 - 2y + 4z \:=\:4 & \Rightarrow & 2y - 4z \:=\:41 & [4] \end{array}$

. . $\begin{array}{ccccc}4 \times [3]: & 4y + 4z \:=\:12 \\
\text{Add [4]:} & 2y - 4z \:=\:41 \\ \hline \\[-3mm]
\text{We have: }& 6y\qquad \:=\:53 \end{array}$

Hence: . $\boxed{y \:=\:\frac{53}{6}}$

Substitute into [3]: . $\frac{53}{6}+ z \:=\:3 \quad\Rightarrow\quad\boxed{ z \:=\:-\frac{35}{6}}$