Can someone explain set builder notation and interval notation?!?

Question

Im sitting here doing my math homework about to cry. No matter how much cancelling out a do and stuff, both sides never seem to even up so i can write the notation.

here are 2 examples of what im having trouble with

3+2(x+5)>(greater or equal to) x+5(x+1)+1

&

x+2>(greater or equal to) or 3x<(less then or equal to)x

HELP!!!!

Answer 1

1) 3+2(x+5)≥x+5(x+1)+1

we use that we can do anyting we want, as long we do it on both side of the inequality.
Ofcause we want to have x on one side and the numbers on the other side.

1 step: add -5(x+1) on both side and get:

3+2(x+5)-5(x+1)≥x+1

2 step add -x on both side and get

3+2(x+5)-5(x+1)-x≥1

3 step reduce the expression
8-4x≥1

4 step add -8 on both side and get
-4x≥-7

5 step divide with -4 on both side and remember to turn the inequality sign because you divide with a negative number, and get:

x≤7/4

you could also add 4x+7 to bot side it would have given
7≥4x <=> 7/4≥x you see its the same
the solution is then
x={x in R l x≤7/4}
===============

2) you have forgot something i guess ist a 0

x+2≥0 or 3x≤x
exact the same procedure

x≥-2 or 3x≤x

x≥-2 or 2x≤0 <=>
x≥-2 or x≤0
.......-2...........0
--------l-----x----l---- both end point are in the solution.

the solution is then

x={x in R l -2≤x≤0 }
i think you you should check to se wheter you have been written the problem correct

Answer 2

3 + 2(x+5)≥x+5(x+1)+1

3 + 2x + 10 ≥ x + 5x + 5 +1

13 + 2x ≥ 6x +6

13 +2x -6x -6 ≥ 0

7 - 4x ≥0
-4x ≥ -7 multiply both sides by -1 and change ≥ into ≤

4x ≤ 7

x ≤ 7/4

The solution includes 7/4 and all real numbers less than 7/4
x belongs to the interval (-infinity,7/4]

Answer 3

Yes I am capable of answering your question but I refuse to do your homework for you !!!