Question

For the following system of equations in echelon form, tell how many solutions there are in nonnegative integers.

2x+2y + 4z=70
y-3z=10
Select the correct choice below and, if necessary, fill in the answer box to complete your choice.
OA. There are nonnegative solutions.
OB. There are infinitely many solutions.
OC. There is no solution.

GUYS PLEASE HELP I DONT UNDERSTAND THIS

Answer 1

A. there are nonnegative solutions

Explanation:

sorry, my bad with the other answer, I missed somehow that we were looking for integer only solutions (and nonnegative ones at that).

when we have 3 variables and 2 equations (the echelon form is just irrelevant information, we would need to transform and solve the equations in any way. the echelon form just means that some of the necessary transformations had been done already), we have no single solution. we have in fact infinitely many solutions. but not with pure integer solutions.

there we might have only a few possible solutions.

and there is a little trick in the line of thinking :

let's say, y can have any value.

then the second equation tells us about z

z = (y - 10)/3

but we know z must be nonnegative.

so,

(y - 10)/3 >= 0

y/3 -10/3 >= 0

y/3 >= 10/3

y >= 10

or

10 <= y

and because z (= (y - 10)/3) has to be a nonnegative integer

(y - 10) has to be a nonnegative multiple of 3, so that the division by 3 delivers an integer.

so, y can be 10, 13, 16, 19, ...

now we use the z identity in the first equation

2x + 2y + 4(y - 10)/3 = 70

6x + 6y + 4(y - 10) = 210

6x + 6y + 4y - 40 = 210

6x + 10y = 250

3x + 5y = 125

3x = 125 - 5y

x = (125 - 5y)/3

now, also x has to be a nonnegative integer

(125 - 5y)/3 >= 0

125/3 - 5y/3 >= 0

125/3 >= 5y/3

125 >= 5y

25 >= y

or

y <= 25

so, we have 10 <= y <= 25

and as (y - 10) must be a multiple of 3, we have the possible y values :

10, 13, 16, 19, 22, 25

for each of these y values there is a valid solution for the system.

so there are only a limited number of nonnegative integer solutions.