Question 564321
<pre>
{{{(matrix(4,6,
1,    1,    1,    1, "|",   4,
0,    1,    4,    3, "|",   8,
0,    0,    1,  3/5, "|", 9/5, 
0,    0,    0,    1, "|",  -2))}}}

This matrix is in "row echelon" form because each
row's left-most non-zero element is 1 and they
move to the right as we go down the matrix.  So the
matrix is an abbreviation for this system:
{{{system(
1w   +1x  +  1y +   1z =   4,
0w   +1x  +  4y +   3z =   8,
0w   +0x  +  1y + expr(3/5)z = 9/5, 
0w   +0x  +  0y +   1z =  -2)}}}

Removing all the 1 coefficients and the 0 terms:
{{{system(

w   +x  +  y +   z =   4,
x  +  4y +   3z =   8,
y + expr(3/5)z = 9/5, 
z =  -2)}}}

We use "back substitution".

The bottom, fourth, equation is already solved for z,

so we substitute -2 for z in the third equation:

{{{y + expr(3/5)z = 9/5}}}

{{{y + expr(3/5)(-2) = 9/5}}}

{{{y - 6/5 = 9/5}}}

{{{y = 9/5 + 6/5}}}

{{{y = 15/5}}}

{{{y = 3}}}

Now we substitute 3 for y and -2 for z in the
second equation:

{{{x  +  4y +   3z = 8}}}
{{{x  +  4(3) +  3(-2) = 8}}}
{{{x + 12 - 6 = 8}}}
{{{x + 6 = 8}}}
{{{x = 2}}}


Finally we substitute 2 for x, 3 for y 
and -2 for z in the first equation:

{{{w   +x  +  y +   z =   4}}}
{{{w+2 + 3 - 2 = 4}}}
{{{w +3 = 4}}}
{{{w=1}}}

Solution (w,x,y,z) = (1,2,3,-2)

Edwin</pre>