What is the solution of the following system?:5 x + 3y + -2z = -12, x + 7y - 9z = 7, x + -y + z = -4

$x = - \frac{23}{8}$ and $y = \frac{1}{8}$ and $z = - 1$

Explanation:

From the give equations
Let
$5 x + 3 y - 2 z = - 12 \text{ }$first equation
$x + 7 y - 9 z = 7 \text{ }$second equation
$x - y + z = - 4 \text{ }$third equation

Also the equations are equivalent to the following; after multiplying the terms of second and third equations by 5
$5 x + 3 y - 2 z = - 12 \text{ }$first equation
$5 x + 35 y - 45 z = 35 \text{ }$second equation
$5 x - 5 y + 5 z = - 20 \text{ }$third equation

Now we eliminate x using the first and second equations by subtraction

$5 x + 3 y - 2 z = - 12 \text{ }$first equation
$\underline{5 x + 35 y - 45 z = 35 \text{ }}$second equation
$0 - 32 y + 43 z = 47$

$32 y - 43 z = 47 \text{ }$fourth equation

Now we eliminate x using the first and third equations by subtraction
$5 x + 3 y - 2 z = - 12 \text{ }$first equation
$\underline{5 x - 5 y + 5 z = - 20 \text{ }}$third equation
$0 + 8 y - 7 z = 8$

$8 y - 7 z = 8 \text{ }$fifth equation

Multiply the fifth equation by 4 then subtract the fourth equation

$32 y - 28 z = 32 \text{ }$fifth equation
$\underline{32 y - 43 z = 47 \text{ }}$fourth equation
$0 + 15 z = - 15$

$15 z = - 15$
$\textcolor{red}{z = - 1}$

Use the $8 y - 7 z = 8 \text{ }$fifth equation and $z = - 1$ to solve for $y$

$8 y - 7 z = 8 \text{ }$fifth equation
$8 y - 7 \left(- 1\right) = 8 \text{ }$fifth equation
$8 y + 7 = 8$
$8 y = 8 - 7$
$8 y = 1$
$y = \frac{1}{8}$
$\textcolor{red}{y = \frac{1}{8}}$

Use the third equation $x - y + z = - 4 \text{ }$ and $\textcolor{red}{z = - 1}$ and $\textcolor{red}{y = \frac{1}{8}}$ to solve for $x$

$x - y + z = - 4 \text{ }$third equation
$x - \left(\frac{1}{8}\right) + \left(- 1\right) = - 4 \text{ }$third equation
$x = \frac{1}{8} + 1 - 4$
$x = \frac{1 - 24}{8}$
$\textcolor{red}{x = - \frac{23}{8}}$

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