Question 976765
our matrix
	

{{{(matrix(2,4,1,-7,4,75,0,2,7,60))}}}


Find the pivot in the 1st column in the 1st row

	
{{{(matrix(2,4,1,-7,4,75,0,2,7,60))}}}


Make the pivot in the 2nd column by dividing the 2nd row by 2
	
{{{(matrix(2,4,1,-7,4,75,0,1,7/2,30))}}}


Eliminate the 2nd column
	

{{{(matrix(2,4,1,0,57/2,285,0,1,7/2,30))}}}


Solution set:

{{{x = 285 - (57/2)z}}}
{{{y = 30 - (7/2)z}}}
{{{z }}}- free

Therefore the solutions can be written as
({{{x}}}, {{{y}}}, {{{z}}}) = ({{{285 - (57/2)z}}},{{{30 - (7/2)z}}},{{{z}}})



Because we want to find non-negative solutions, we have three inequalities

{{{285 - (57/2)z>=0}}}
{{{30 - (7/2)z>=0}}}
{{{z>=0}}}

so, 
{{{285 >=(57/2)z}}}
{{{30 >=(7/2)z}}}

{{{285/(57/2) >=z}}}
{{{30/(7/2) >=z}}}

{{{z<=10 }}}
{{{z<=8.57 }}}

and 

{{{z>=0}}}

so,  the integer values of {{{z}}} satisfying all of those inequalities {{{z>=0}}},{{{z<=8.57 }}}, and {{{z<=10 }}} are

 {{{z}}}={{{0}}},{{{1}}},{{{2}}},{{{3}}},{{{4}}},{{{5}}},{{{6}}},{{{7}}},{{{8}}} 

so, there will be nine non-negative solutions