Question 506518
<pre>
2x + 3y - 4z = 7
3x + 4y - 2z = 9
5x + 7y - 6z = 20


{{{(matrix(3,5,

2,  3, -4, "|", 7,
3,  4, -2, "|", 9,
5,  7, -6, "|", 20))}}}

Use Row 1 to get a 0 where the 3 is in Row 2:
Multiply Row 1 by -3 and the Row 2 by 2

{{{(matrix(3,5,

-6, -9, 12, "|", -21,
6,  8, -4, "|", 18,
5,  7, -6, "|", 20))}}}

Add Row 1 to Row 2

{{{(matrix(3,5,

6,  9, -12, "|", 21,
0, -1, 8, "|", -3,
5,  7, -6, "|", 20))}}}

Restore Row 1 as it was

{{{(matrix(3,5,

2,  3, -4, "|", 7,
0, -1, 8, "|", -3,
5,  7, -6, "|", 20))}}}

Use Row 1 to get a 0 where the 5 is in Row 3
Multiply Row 1 by -5 and Row 3 by 2

{{{(matrix(3,5,

-10,  -15, 20, "|", -35,
0, -1, 8, "|", -3,
10,  14, -12, "|",40))}}}

Add Row 1 to Row 3

{{{(matrix(3,5,

-10,  -15, 20, "|", -35,
0, -1, 8, "|", -3,
0,  -1, 8, "|",5))}}}

Restore Row 1 as it was:

{{{(matrix(3,5,
2,  3, -4, "|", 7,
0, -1, 8, "|", -3,
0,  -1, 8, "|",5))}}}

Use Row 2 to get a 0 where the -1 is in Row 3
Multiply Row 2 by -1 and keep Row 3 as it is

{{{(matrix(3,5,
2,  3, -4, "|", 7,
0, 1, -8, "|", 3,
0,  -1, 8, "|",5))}}}

Add Row 2 to Row 3

{{{(matrix(3,5,
2,  3, -4, "|", 7,
0, 1, -8, "|", 3,
0,  0, 0, "|",8))}}}

Oh oh!  We can see there is no solution because
the bottom Row is an abbreviation for the equation:

0x + 0y + 0z = 8

There are no values of x,y, or z that we can substitute
for any of them that will make the left side equal to 8,
because the left side will always be 0 no matter what
numbers you substitute for the letters.  Therefore there
is no solution and this is called an "inconsistent" system.

Edwin</pre>