For each of the following systems of equations in echelon fo

For each of the following systems of equations in echelon form, tell how many solutions there are in nonnegative integers. x + 2y + 3z = 90 x - 7y + 4z = 75 3y + 4z = 36 2y + 7z = 60

Solution

23.

Multiplying first by 3 and second by 2 and subtracting gives

2x-3z=198

2x=3z+198

And

3y=-4z+36

So z must be multiple of 2 and 3

Non negative means

-4z+36>=0

4z<=36

0=<z<=9

So ten possible values of z but z must be a multiple of 6 so only two non trivial solution for

z=6,0

24

Multiplying first equation by 2 and second by 7 and adding gives

2x+57z=570

So

2x=-57z+570

So z must be multiple of 2

-57z+570>=0

z<=10

2y=60-7z

So

60-7z>=0

7z<=60

So

z<=8

So ,z=0,2,4,6,8

So 5 possible solutions