Determine the number of integer values of x that satisfies the inequality |1/2-x|< 3?

Question

Please show your work. Thanks =)

Answer 1

1/2-x<3
adding x-1/2
0 adding -/2
-5/2 -1/2+x<3
adding -x+1/2
0<-x+7/2
-7/2<-x
7/2>x
x<7/2
so answer is
-5/2 integer values are -2,-1,0,1,2,3

Answer 2

Absolute value:
| a | < b

is the same as:
-b < a < b

In your case:
| ½ - x | < 3

This is the same as:
-3 < ½ - x < 3

Now subtract ½ everywhere:
-3½ < -x < 2½

Multiply through by -1 and reverse the signs (don't forget to do this when multiplying through by a negative number.)
3½ > x > -2½

Now reverse the order so the smaller number comes first:
-2½ < x < 3½

Since you want integers, the answers are: { -2, -1, 0, 1, 2, 3 }

You can double check by putting each of these number into the equation.
| ½ - (-2) | = 2½ which is less than 3
| ½ - (-1) | = 1½ which is less than 3
| ½ - 0 | = ½ which is less than 3
| ½ - 1 | = ½ which is less than 3
| ½ - 2 | = 1½ which is less than 3
| ½ - 3 | = 2½ which is less than 3

The answer is there are six integer values that satisfy the inequality:
{ -2, -1, 0, 1, 2, 3 }

Answer 3

Abs val less than 3 means the number must be between -3 and 3.
-3 < 1/2-x < 3
Add -1/2 to each:

-7/2 < -x < 5/2

Mult each by -1, so reverse direction:
7/2 > x > -5/2, or
-5/2 < x < 7/2

Sorry, didn't read the question properly at first:
So there are six integer values satisfying this:

-2, -1, 0, 1, 2, 3

Answer 4

(2-x)(4x-3) = 0 8x -6 -4x^2 +3x = 0 -4x^2 + 11x -6 = 0 It crosses at 2 and three/4. In between that, that's above 0. subsequently your second selection is nice, x<=3/4 or x >= 2

Answer 5

-3<1/2-x
-3,5<-x
x<3.5

1/2-x<3
-x<2,5
x>-2,5

-2.5
solution: {-2,-1,0,1,2,3}

Answer 6

six integer values... {-2,-1,0,1,2,3}