Question

If x is an integer, what is the set of solutions to |3x + 2| < 11?

A. {-4, -3, -2, -1, 0, 1, 2, 3}

B. {-4, -3, -2, -1, 0, 1, 2}

C. {-1, 0, 1, 2, 3}

D. {-3, -2, -1, 0, 1, 2, 3}

Answer 1

can u see the file i attached

Answer 2

yes

Answer 3

i basically graphed it, the straight line is y=11 and i have also graphed the absolute value. can u see the x values that are in the shaded region?

Answer 4

Here is the rule:
________________________________________________________
To solve an absolute value inequality of the form:

|X| < k, where
X is an expression in x, and
k is a number,

solve the following compound inequality

-k < X < k
_________________________________________________________

Now let's look at your problem

|3x + 2| < 11

Your absolute value inequality becomes:

-11 < 3x + 2 < 11

Which can be solved by subtracting 2 from all three sides:
-13 < 3x < 9

Then divide all three sides by 3:
-4.333... < x <3

Since x is an integer, the first integer greater than -4.333... is -4.
Also, x is an integer less than 3, so the largest x can be is 2.
That means x can be all integers from -4 to 2, inclusive.

Answer 5

Thanks! @mathstudent55

Answer 6

You're welcome.