$\"\"$

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 two 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 five 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 negative two times the second equation to the third equation produces a new third equation.

$\"\"$

$\"\"$

So, the new system of equations are

$\"\"$ (Equation 1)

$\"\"$ (Equation 2)

$\"\"$ (Equation 3)

The statement $\"\"$ is false , you can conclude that this system is inconsistent and has no solutions.Moreover, because this system is equivalent to the original system, you can conclude that the original system also has no solution. $\"\"$

The system has no solution.