The system of linear 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)
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 second equation to the 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
(Equation 1)
(Equation 2)
(Equation 3)
To solve for y, the value of substitute in equation 2 to obtain as follows:
(Substitute
)
(Add 3 to each side)
(Apply LCM rule:
)
To solve for x, the values of substitute in equation 1 to obtain as follows:
(Substitute
)
(Subtract
from each side)
(Simplify)
(Apply LCM rule:
)
The solution is , which can be write as the ordered triple
.
The solution is .