# How do you solve the following linear system: 3x + 4y - z =1, 3x - y - 4z = 0, x + 3y - 3z = 9?

$x = - \frac{138}{55}$
$y = \frac{86}{55}$
$z = - \frac{25}{11}$

#### Explanation:

From the given equations
$3 x + 4 y - z = 1$first equation
$3 x - y - 4 z = 0$second equation
$x + 3 y - 3 z = 9$third equation

Let us eliminate x first using first and second equations by subtraction

$3 x + 4 y - z = 1$first equation
$3 x - y - 4 z = 0$second equation
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$0 \cdot x + 5 y + 3 z = 1$
$5 y + 3 z = 1$ fourth equation

Let us eliminate x first using first and third equations by subtraction

$3 x + 4 y - z = 1$first equation
$x + 3 y - 3 z = 9$third equation is also

$3 x + 9 y - 9 z = 27$third equation, after multiplying each term by 3

perform subtraction using the new third equation and the first equation

$3 x + 4 y - z = 1$first equation
$3 x + 9 y - 9 z = 27$third equation
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$0 \cdot x - 5 y + 8 z = - 26$
$- 5 y + 8 z = - 26$ fifth equation

Solve for y and z simultaneously using fourth and fifth equations using addition

$5 y + 3 z = 1$ fourth equation
$- 5 y + 8 z = - 26$ fifth equation
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$0 \cdot y + 11 z = - 25$
$11 z = - 25$
$z = - \frac{25}{11}$

Solve y using $5 y + 3 z = 1$ fourth equation and $z = - \frac{25}{11}$

$5 y + 3 z = 1$ fourth equation
$5 y + 3 \left(- \frac{25}{11}\right) = 1$ fourth equation
$5 y - \frac{75}{11} = 1$
$5 y = \frac{75}{11} + 1$
$5 y = \frac{86}{11}$
$y = \frac{86}{55}$

Solve for x using $3 x + 4 y - z = 1$first equation and $y = \frac{86}{55}$ and $z = - \frac{25}{11}$

$3 x + 4 y - z = 1$first equation
$3 x + 4 \left(\frac{86}{55}\right) - \left(- \frac{25}{11}\right) = 1$first equation
$3 x + \frac{344}{55} + \frac{25}{11} = 1$
$3 x = 1 - \frac{344}{55} - \frac{25}{11}$
$3 x = \frac{55 - 344 - 125}{55}$
$3 x = - \frac{414}{55}$
$x = - \frac{138}{55}$

Check using the original equations

$3 x + 4 y - z = 1$first equation
$3 \left(- \frac{138}{55}\right) + 4 \left(\frac{86}{55}\right) - \left(- \frac{25}{11}\right) = 1$first equation

$- \frac{414}{55} + \frac{344}{55} + \frac{25}{11} = 1$
$\frac{- 414 + 344 + 125}{55} = 1$
$\frac{55}{55} = 1$
$1 = 1$

$3 x - y - 4 z = 0$second equation
$3 \left(- \frac{138}{55}\right) - \left(\frac{86}{55}\right) - 4 \left(- \frac{25}{11}\right) = 0$second equation
$- \frac{414}{55} - \frac{86}{55} + \frac{100}{11} = 0$
$- \frac{500}{55} + \frac{100}{11} = 0$
$0 = 0$

$x + 3 y - 3 z = 9$third equation
$\left(- \frac{138}{55}\right) + 3 \left(\frac{86}{55}\right) - 3 \left(- \frac{25}{11}\right) = 9$third equation
$- \frac{138}{55} + \frac{258}{55} + \frac{75}{11} = 9$
$\frac{120}{55} + \frac{75}{11} = 9$
$\frac{24}{11} + \frac{75}{11} = 9$
$\frac{99}{11} = 9$
$9 = 9$

The solution set is
$x = - \frac{138}{55}$
$y = \frac{86}{55}$
$z = - \frac{25}{11}$

God bless....I hope the explanation is useful.

Mar 18, 2016

$\left(x , y , z\right) = \left(- \frac{138}{55} , \frac{86}{55} , - \frac{25}{11}\right)$

#### Explanation:

We have the three equations:

$\left\{\begin{matrix}3 x + 4 y - z = 1 \text{ "" "" "" "" ""eq. 1" \\ 3x-y-4z=0" "" "" "" "" ""eq. 2" \\ x+3y-3z=9" "" "" "" "" ""eq. 3}\end{matrix}\right.$

Multiply $\text{eq. 3}$ by $- 3$ and add it to $\text{eq. 1}$:

{:(3x+4y-z=1),(ul(-3x-9y+9z=-27" "+)),(-5y+8z=-26" "" "" "" "" ""eq. 4"):}

Multiply $\text{eq. 1}$ by $- 3$ and add it to $\text{eq. 3}$:

{:(-9x-12y+3z=-3),(ul(x+3y-3z=9" "" "" "+)),(-8x-9y=6" "" "" "" "" "" "" ""eq. 5"):}

Multiply $\text{eq. 2}$ by $3$ and add it to $\text{eq. 3}$:

{:(9x-3y-12z=0),(ul(x+3y-3z=9" "" "+)),(10x-15z=9" "" "" "" "" "" "" ""eq. 6"):}

Multiply $\text{eq. 5}$ by $5$ and $\text{eq. 6}$ by $4$ and add the two:

{:(-40x-45y=30),(ul(40x-60z=36" "" "+)),(-45y-60z=66" "" "" "" "" "" ""eq. 7"):}

Multiply $\text{eq. 4}$ by $- 9$ and add it to $\text{eq. 7}$:

{:(45y-72z=234),(ul(-45y-60z=66" "" "+)),(-132z=300" "" "" "" "" "" "" "" ""eq. 8"):}

From $\text{eq. 8}$, we can deduce that

color(red)z=300/(-132)=-150/66=-50/22color(red)(=-25/11

Use this value of $z$ in $\text{eq. 6}$ to solve for $x$:

$10 x - 15 z = 9 \text{ "=>" } 10 x - 15 \left(- \frac{25}{11}\right) = 9$

$\textcolor{w h i t e}{s l} \implies \text{ } 10 x + \frac{375}{11} = 9$

$\textcolor{w h i t e}{s l} \implies \text{ } 10 x + \frac{375}{11} = \frac{99}{11}$

$\textcolor{w h i t e}{s l} \implies \text{ } 10 x = - \frac{276}{11}$

color(white)(sl)=>" "color(red)x=-276/110color(red)(=-138/55

We can also use the value of $z$ we found to solve for $y$ by plugging into $\text{eq. 4} :$

$- 5 y + 8 z = - 26 \text{ "=>" } - 5 y + 8 \left(- \frac{25}{11}\right) = - 26$

$\textcolor{w h i t e}{s l} \implies \text{ } - 5 y - \frac{200}{11} = - 26$

$\textcolor{w h i t e}{s l} \implies \text{ } - 5 y - \frac{200}{11} = - 26$

$\textcolor{w h i t e}{s l} \implies \text{ } - 5 y - \frac{200}{11} = - \frac{286}{11}$

$\textcolor{w h i t e}{s l} \implies \text{ } - 5 y = - \frac{86}{11}$

color(white)(sl)=>" "color(red)(y=86/55

This gives us the solution set of $\left(x , y , z\right)$ as:

color(blue)((-138/55,86/55,-25/11)