Solve the following absolute value inequalities using operations.
\left| 3x+2 \right|≤2
The inequality:
|3x+2| \leq 2
Can be rewritten as:
-2 \leq 3x+2 \leq 2
Subtracting 2 from both all sides gives:
-2-2 \leq 3x + 2-2 \leq 2-2
-4 \leq 3x \leq 0
Dividing everything by 3 gives (note that 3 is positive):
\dfrac{-4}{3} \leq \dfrac{3x}{3} \leq \dfrac{0}{3}
The set of solutions is:
\dfrac{-4}{3} \leq x \leq 0
\left | 2x + 4 \right | ≤8
The inequality:
\left | 2x + 4 \right | ≤8
Can be rewritten as:
-8 \leq 2x + 4 \leq 8
Subtracting 4 from all sides gives:
-8-4 \leq 2x + 4 -4\leq 8-4
-12 \leq 2x \leq 4
Dividing everything by 3 gives (note that 3 is positive):
\dfrac {-12} {2} \leq \dfrac {2x} {2} \leq \dfrac {4} {2}
The set of solutions is:
-6 \leq x \leq 2
\left | 3x + 4 \right | ≤-2
The absolute value must always be positive. Therefore \left | 3x + 4 \right | ≤-2 is not possible and there is no solution.
There is no solution.
\left | x + 4 \right | \geq 8
\left | x + 4 \right | \geq 8 can be rewritten as:
x+4 \geq 8 or x + 4 \leq -8
Solve the first inequality
x+4 \geq 8
Subtracting 4 from both sides:
x+4-4 \geq 8-4
x\geq 4
Solve the second inequality
x+4 \leq -8
Subtracting 4 from both sides:
x+4-4 \leq -8-4
x \leq -12
The set of solutions is:
x \geq4 \text{ or } x\leq -12
\left | -2x + 6 \right | \geq 12
\left | -2x + 6 \right | \geq 12 can be rewritten as:
-2x+6 \geq 12 or -2x+6 \leq -12
Solve the first inequality
-2x+6 \geq 12
Subtracting 6 from both sides:
-2x+6 -6\geq 12-6
-2x \geq 6
Dividing both sides by -2 and interchanging the direction of the inequality:
\dfrac{-2x}{-2} \leq \dfrac{6}{-2}
x \leq -3
Solve the second inequality
-2x+6 \leq -12
Subtracting 6 from both sides:
-2x+6-6 \leq -12-6
-2x \leq -18
Dividing both sides by -2 and interchanging the direction of the inequality:
\dfrac{-2x}{-2} \geq \dfrac{-18}{-2}
x \geq 9
The set of solutions is:
x\leq -3 \text{ or } x\geq 9
\left | -2x + 4 \right | \lt 20
\left | -2x + 4 \right | \lt 20 can be rewritten as:
-20 \lt -2x+4 \lt 20
Subtracting 4 from all sides:
-20-4 \lt -2x+4-4 \lt 20-4
-24 \lt -2x \lt 16
Dividing both sides by -2 and interchange the direction of both inequalities:
\dfrac{-24}{-2} \gt \dfrac{-2x}{-2} \gt \dfrac{16}{-2}
Equivalently:
12 \gt x \gt -8
The set of solutions is:
12 \gt x \gt -8
\left | 2x + 7 \right | \lt 0
The absolute value must always be positive. Therefore \left | 2x + 7 \right | \lt 0 is not possible and there is no solution.
There is no solution.