Question

Give a MEDAL who solve find how many nonnegative integer solutions of the system are here
x+y<8
x+z<8
t+y<8
z+t<8

Answer 1

try graphing all these regions on a single page on your calculator. FInd the intersection of these regions.

Answer 2

Do you have a TI-NSpire CX

Answer 3

I'll try, wait a minute

Answer 4

sorry I was wrong, you have to graph it by hand, TI sucks on this one

Answer 5

@amistre64 @Ashleyisakitty @ahoward79 @Falco276 @goformit100 @genius12

Answer 6

4 equations with 4 unknowns

Answer 7

x+y<8 ; y < 8-x
x+z<8 ; z < 8-x

well, y = z

t+8-x<8

x = t

Answer 8

8-x >= 0 , when x <= 8

Answer 9

seems to me that the solution set is:

x,y,z,t = (0,8]

Answer 10

@amistre64 not obligatory y=z

Answer 11

x+y < 8
0+7 ,1+6, 2+5, 3+4, 4+3, 5+2, 6+1

x + z < 8
0+7 ,1+6, 2+5, 3+4, 4+3, 5+2, 6+1

t+y<8
0+7 ,1+6, 2+5, 3+4, 4+3, 5+2, 6+1

z+t<8
0+7 ,1+6, 2+5, 3+4, 4+3, 5+2, 6+1

Answer 12

forgot to tack onthe (7+0)s

Answer 13

when x=0, y=7, z=7, t=0

when x=1, y=6, z=6, t=1

Answer 14

at any rate, the set is the cross of {0,1,2,3,4,5,6,7}^4

Answer 15

no when x+y<8 you also can have 1,2 1,3 1,4 1,5 1,0 1,1 1,6

Answer 16

you are making x+y=7

Answer 17

but the other cases?

Answer 18

im simply defining a limit to it

Answer 19

i beleive there are 7^4 elements of the set (x,y,z,t)

Answer 20

or is that 8^4

Answer 21

at any rate, its a lot of ordered pairs

Answer 22

This is a tricky counting problem. You could try to devise some case-by-case counting scheme, or you could enumerate all the possibilities (that satisfy the constraints) using a computer programming language. It's actually pretty straightforward if you know how to program.

numSolutions = 0
for x in [0, 7]
..for y in [0, 7]
....for z in [0, 7]
......for t in [0, 7]
........if x + y < 8 and x + z < 8 and t + y < 8 and z + t < 8 then
..........numSolutions = numSolutions + 1

print numSolutions

This will be extremely quick, since there are only 8^4 iterations.

Answer 23

Btw, it's 876.

Answer 24

yes its corect