Question 1090883
<pre><font size = 4 color = "indigo"><b>

This system turns out to be a dependent system which has
infinitely many solutions.  I'll show you below how to
do it on your TI-83 or TI-84 calculator.  If you need
help on how to solve it without a calculator, just say
so in the thank-you note form below and I'll help you
do it by the matrix method without a calculator.

{{{system(

matrix(3,9,

red(2)x,
""+"",
red(3)y,
""-"",
red(1)z,
"",
""="",
"",
red(4),
red(3)x,
""-"",
red(1)y,
""+"",
red(2)z,
"",
""="",
"",
red(5),
1*x,
""-"",
red(4)y,
""+"",
red(3)z,
"",
""="",
"",
red(1)
))}}}
<pre>

----------------------------------


Put all those red numbers in a 3x4 augmented matrix
like this:

{{{red((matrix(3,5,

2,3,-1,"|",4,
3,-1,2,"|",5,
1,4,3,"|",1)))}}}

On your TI-83 or TI-84,

Press CLEAR

Press 2ND
Press x<sup>-1</sup> (MATRIX)
Prees the right arrow twice to highlight EDIT
Press ENTER
Press 2
Press ENTER
Press 3
Press ENTER
Type -1, 
Press ENTER
Type 4,
Press ENTER
Type 3,
Press ENTER
Type -1
Press ENTER
Type 2
Press ENTER
...
continue typing in all the rest of the elements 
in the matrix, pressing ENTER after each. Then
...
Press 2ND
Press MODE (QUIT)
Press 2ND
Press x<sup>-1</sup> (MATRIX)
Press right arrow to highlight MATH
Press up arrow 5 times to highlight B:rref(
Press ENTER  you should see rref( on the screen
Press 2nd
Press x<sup>-1</sup> (MATRIX)
Press ENTER  you should see rref([A]
Type )
Press MATH
Press ENTER
Press Enter
You should see  

rref([A]
    [[1 0 5/11  19/11]
     [0 1 -7/11 2/11 ]
     [0 0       0    ]]

Interpret this as the system:

{{{system(

matrix(3,9,

1*x,
""+"",
0*y,
""+"",
expr(5/11)*z,
"",
""="",
"",
19/11,
0*x,
""+"",
1*y,
""+"",
expr(-7/11)*z,
"",
""="",
"",
2/11,
0*x,
""+"",
0*y,
""+"",
0*z,
"",
""="",
"",
0
))}}}

That's the same as

{{{system(x+expr(5/11)z=19/11,
y-expr(7/11)z=2/11,
0*z=0)}}}

Notice that the third equation 0*z=0 is such that
z can be any number whatever, for any number
substituted for z in 0*z=0 will always give 0=0.
So let a = any number.
Since z = any number, we substitute a for z

{{{system(x+expr(5/11)a=19/11,
y-expr(7/11)a=2/11,
0*a=0)}}}

Solve the first for x, the second for y,
then write z = a

{{{system(x=19/11-expr(5/11)a,
y=2/11+expr(7/11)a,
z=a)}}}

So the general solution is

{{{matrix(1,3,

(matrix(1,5,x,",",y,",",z)),
""="",
(matrix(1,5,19/11-expr(5/11)a,",",2/11+expr(7/11)a,",",a))

)}}}

We can get as many solutions as we like by choosing 
different values for the number "a".  For instance,
if we choose a = -5, we have the solution

{{{matrix(1,3,

(matrix(1,5,x,",",y,",",z)),
""="",
(matrix(1,5,4,",",-3,",",-5))) }}}

and if we choose a = 6, we have the solution

{{{matrix(1,3,

(matrix(1,5,x,",",y,",",z)),
""="",
(matrix(1,5,-1,",",4,",",6))) }}}

Edwin</pre>