Question

can anyone help me understand this problem Solve using the addition principle. Graph and write set-builder notation and interval notation for each answer.

5(t + 3) + 9 ≥ 3(t − 2) − 10 i have solved it but not sure if i got it right
{t|t ≥ 0} or (–∞, 0]

Answer 1

what class is it for. Algebra or set theory?

Answer 2

basic algebra

Answer 3

Can you tell me how you arrived at that solution?

Answer 4

5(t + 3) + 9 ≥ 3(t − 2) − 10
5t+3+9 ≥ 3t-2-10
5t+12 ≥ 3t-12
5t-3t +12-12 ≥ 3t-3t-12+12
2t /2 ≥ 0/2
t≥0 that was the answer i got

Answer 5

I think I see what you did wrong

Answer 6

5(t+3)+9>or= 3(t-2)-10
You have to use distributive property.(Everything inside the parenthesis should be multiplied by number outside.)

Answer 7

dam it cant believe i let that one by me

Answer 8

So it is
5*t+5*3+9>0r=3*t+(3*-2) -10

Answer 9

So did you end up getting an answer?

Answer 10

working it out now

Answer 11

5(t + 3) + 9 ≥ 3(t − 2) − 10
5t+15+9 ≥ 3t-6-10
5t+24 ≥ 3t-16
5t-3t +24-24 ≥ 3t-3t-16+24
2t /2 ≥ 8/2
t≥4
{t|t ≥ 4} or (–∞, 4]

Answer 12

not sure about the last line though thats whats messing me up

Answer 13

since t is greater or equal to 4 we should start from 4 to infinity.
so it should be [4,infinity)

Answer 14

ok so i have the answer backwards
5(t + 3) + 9 ≥ 3(t − 2) − 10
5t+15+9 ≥ 3t-6-10
5t+24 ≥ 3t-16
5t-3t +24-24 ≥ 3t-3t-16+24
2t /2 ≥ 8/2
t≥4
{t|t ≥ 4} or [4,∞)

Answer 15

thank you i see im going to hate my next math class more then this one