How do you graph and solve #2-abs(x+3)>1 #?

To graph and solve the inequality ( 2 - \lvert x + 3 \rvert > 1 ), first, we'll isolate the absolute value term and then solve for ( x ).

[ 2 - \lvert x + 3 \rvert > 1 ]

Subtract 2 from both sides:

[

• \lvert x + 3 \rvert > -1 ]

Now, multiply both sides by -1 (which reverses the inequality):

[ \lvert x + 3 \rvert < 1 ]

This inequality can be further broken down into two separate inequalities:

1. ( x + 3 < 1 ) and
2. ( x + 3 > -1 )

Solving the first inequality:

[ x + 3 < 1 \ x < 1 - 3 \ x < -2 ]

And solving the second inequality:

[ x + 3 > -1 \ x > -1 - 3 \ x > -4 ]

So, combining the solutions, we have:

[ -4 < x < -2 ]

The solution set is ( x ) such that ( -4 < x < -2 ). To graph this on a number line, plot an open circle at -4 and another open circle at -2, then draw a line segment connecting them, indicating that the solution lies between -4 and -2 but does not include these values.

To graph and solve the inequality (2 - |x + 3| > 1), follow these steps:

1. First, rewrite the inequality without absolute value: (2 - |x + 3| > 1) becomes (|x + 3| < 1).

2. Split the inequality into two cases: a) When (x + 3 \geq 0) (which leads to (|x + 3| = x + 3)). b) When (x + 3 < 0) (which leads to (|x + 3| = -(x + 3))).

3. For case (a): (x + 3 \geq 0), rewrite the inequality as (x + 3 < 1), and solve for (x). For case (b): (x + 3 < 0), rewrite the inequality as (-(x + 3) < 1), and solve for (x).

4. After solving each case, graph the solutions on the number line.

5. The solution to the original inequality is the combination of solutions from both cases.

Graphically, plot the solution set on the number line, indicating whether the points are included (solid circle) or excluded (open circle).