Algebra help please!

Question

1) Identify the solution of the inequality |x + 3| – 3 ≥ 1.
x ≤ –7 or x ≥ 1
–7 ≤ x ≤ 1
x ≤ –7
x ≥ 1
2) If f(x) = –3x2 – 5x + 7 , find f(–2).
–15
–10
7
5
3) Identify the standard form of the equation x = 3/4y – 5/8.
x – 6y = –5
8x – 6y = –5
4x – 3y = –5
x – 3y = –5
4) Find the slope of the line containing the points (–3, 1) and (3, 4).

m = 1/2
m = –1/2
m = 2
m = –2

Answer 1

1) |x + 3| – 3 ≥ 1.

    x + 3 – 3 ≥ 1.

    x   ≥ 1.

Option 4 is correct answer.

2) If f(x) = –3x2 – 5x + 7 , 

f(–2) = -3(-2)^2-5(-2)+7

       = -3*4+10+7

       = -12+17

       = 5  .
Option 4 is correct answer.

3 ) x = 3/4 y - 5/8

   x - 3/4 y = -5/8

   8(x-3/4 y) = -5/8

   8x-8*3/4 y = -5

   8x-6y = -5

Option 2 is correct answer.

4) Given points are (-3,1) = (x1,y1)

                           (3,4) = (x2,y2)

Slope m = y2-y1/x2-x1

m = 4-1/3-(-3)

m = 3/6

m = 1/2

Option 1 is correct answer.

Comments

|x + 3| - 3 ≥ 1 solution is wrong.

See correct solution below, answered by david.

Answer 2

1) The absolute inequality is  |x + 3| - 3 ≥ 1

|x + 3| ≥ 1 + 3

|x + 3| ≥ 4

|x| ≥ a Then x ≥ a or  x ≤ -a

x + 3 ≥ 4 or  x + 3 ≤ -4

x ≥ 4 - 3 or  x ≤ - 4 -3

x  ≥ 1 or  x ≤ -7

Solution of the inequality is x  ≥ 1 or  x ≤ -7.

Option 1 is correct choice.