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| PAGE: 686 | SET: Exercises | PROBLEM: 30 |
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 the system of linear equations rewrite as follows:
(Equation 1)
(Equation 2)
(Equation 3)
The leading of coefficient of the first equation is one, you can begin by saving the y at the upper left and eliminating the other y - terms from the first column.So, adding negative three 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)
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 true, you can conclude that this system will have infinitely many solutions.
In the above system, multiply the equation 2 and next equation 1 and 2 are interchange as follows:
(Equation 1)
(Equation 2)
In the second equation, solve for x in terms of z to obtain 
.
By back - substituting 
into equation 1, you can solve for y, as follows:
(Write equation 1)
(Substitute )
(Simplify)
(Subtract 
from each side)
Finally let 
, where a is a real number, the solutions of the system
.
The solution in ordered triple form is 
.
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