## anonymous one year ago Which of the following points (x, y, z) is in the solution set of the system of inequalities: x + 3y − z ≤ 5 x − 2z ≥ 0 x + y + z ≤ 10 (a) (1, 1, 3) (b) (2, 2, 1) (c) (6, −2, 2) (d) (8, 3, 1) (e) (−1, 8, −2) (f) None of these please explain as well :) Thank you

1. freckles

you could enter in the choices and see which satisfies the inequalities

2. anonymous

3. freckles

let's check (6,-2,2) x=6 y=-2 z=2 $6+3(-2)-2 \le 5 \\ 6-6-2 \le 5 \\ -2 \le 5 \text{ the first inequality is true } -2 \text{ is less than } 5 \\ 6-2(2) \ge 0 \\ 6-4 \ge 0 \\ 2 \ge 0 \text{ the second inequality is true 2 is greater than 0 } \\ 6-2+2 \le 10 \\ 6 \le 10 \text{ the third inequality is true 6 is less than 10 }$ so yes (6,-2,2) seems to check out

4. freckles

can there be more than one option?

5. anonymous

I think that's the only one..

6. freckles

checking (1,1,3) $1+3-3 \le 5 \\1 \le 5 \text{ first inequality true }$ $1-2(3) \ge 0 \\ 1-6 \ge 0 \\ -5 \ge 0 \text{ \not true }$ no need to check third inequality (1,1,3) is not a solution checking (2,2,1) $2+6-1 \le 5 \text{ is false }$ no need to check the other inequalities (2,2,1) is not a solution checking (8,3,1) $8+9-1 \le 5 \text{ is false } \\ (8,3,1) \text{ doesn't work }$ checking (-1,8,-2) $-1+3(8)+2 \le 5 \text{ is false } \\ (-1,8,-2) \text{ is \not a solution }$ most of those choices fail at the first inequality

7. freckles

good job

8. anonymous

Thank you :)