Algebra help!!??!?!?!please!!

i need help with this problem. please show every step, because i am clueless, thx.
|1-y| is less than or equal to 2

|1-y| ≤ 2

IMPLIES

1 - y ≤ 2 and (-1)(1 - y) ≤ 2

Just like |x| implies -x & +x

1 - y ≤ 2
-y ≤ 1
y ≥ -1 (oh & don't forget, multiplying by a (-) number switches less/greater than sign)

(-1)(1 - y) ≤ 2
-1 + y ≤ 2
y ≤ 3

ANSWER --> -1≤ y ≤3

so how would i do this: |8x+2|= 14
would it be -14= 8x +2 = 14 ?

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|1-y| <= 2

-2 <= 1 - y <= + 2..................Subt 1 from all three parts

-3 <= - y <= 1........................Divide all three parts by -1.....RTeverse the < sign

3 >= y >= -1..........................Write it in reverse order

-1 <= y <= 3.........ANSWER

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|1-y| is less than or equal to 2

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-Break it down into 2 cases and remove the absolute value bars. so it would be 1-y is less than or equal to 2 *OR* 1-y is greater than or equal to -2 then find the value of y by adding 1 to each side (which eliminates 1 from the equation)

i) I suppose you wish to solve the equation |1 - y| ≤ 2

ii) In the first case, let (1 - y) ≤ 2; ==> -y ≤ 1
==> y ≥ -1 [Multiplying by negative quantity, the inequality changes] ------- (1)

iii) Now let (1 - y) ≥ - 2; ==> -y ≥ -3
So, y ≤ 3 ----- (2)

Hence from (1) & (2), y in [-1, 3]

|1 - y| <= 2

This means that (1 - y) will be within a circle with radius of 2.
Or rather: http://www.wolframalpha.com/input/?i=%7C…

so:
-2 < 1 - y < 2

subtract 1:
-3 < -y < 1

multiply with -1 (and remember to change '<' to '>')
3 > y > -1

So you're left with:
-1 < y < 3

|1 - y| ≤ 2
-2 ≤ 1 - y ≤ 2
-2 - 1 ≤ -y ≤ 2 - 1
-3 ≤ -y ≤ 1
3 ≥ y ≥ -1