$\"\"$

The system of linear equations are

$\"\"$ (Equation 1)

$\"\"$ (Equation 2)

$\"\"$ (Equation 3)

In this system leading of coefficient of the first equation does not one, so 1st and 3rd equations are interchange.So, the new system of equations are

$\"\"$ (Equation 1)

$\"\"$ (Equation 2)

$\"\"$ (Equation 3)

The leading of coefficient of the first equation is one, you can begin by saving the x at the upper left and eliminating the other x - terms from the first column.So, adding negative one times the first equation to the second equation produces a new second equation.

$\"\"$ $\"\"$

So, the new system of equations are

$\"\"$ (Equation 1)

$\"\"$ (Equation 2)

$\"\"$ (Equation 3)

Adding negative two times the first equation to the third equation produces a new third equation.

$\"\"$ $\"\"$

So, the new system of equations are

$\"\"$ (Equation 1)

$\"\"$ (Equation 2)

$\"\"$ (Equation 3)

Now that all but the first x have been eliminated from the first column, go to back on the second column.(You need to eliminate y from the third equation.)So, adding two times the second equation to three times third equation produces a new third equation.

$\"\"$ $\"\"$

So the new system of equations are

$\"\"$ (Equation 1)

$\"\"$ (Equation 2)

$\"\"$ (Equation 3)

Finally, you need a coefficient of 1 for y in the second equation and coefficient of 1 for z in the third equation, so multiply the second equation by $\"\"$ produces a new second equation and so multiply the third equation by $\"\"$ produces a new third equation respectively.The new system in row - echelon form of equations are