# Three-Variable Linear Systems

• January 31st 2010, 01:13 PM
jasimmons86
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

• January 31st 2010, 01:21 PM
pickslides
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?
• January 31st 2010, 01:27 PM
jasimmons86
i dont know how to find the inverse
• January 31st 2010, 01:32 PM
pickslides
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.
• January 31st 2010, 01:35 PM
Quacky
(1) X + Y + Z = 18
(2) X - Y - Z = 12
(3) 3X - 2Y + 4Z = 4

Here's my simplistic and basic method :o

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
$Y=8.8333$
• January 31st 2010, 01:43 PM
Soroban
Hello, jasimmons86!

What methods are you allowed to use?

Quote:

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

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