Question

alg. 2 1.6 - 1.9 test review
graph the inequality:

1. x≤2
2. -8<x<4
3. |x|≥9

solve and graph the following inequalities:

1. x - 2≤13
2. 4x > -20
3. 6x - 3≥9
4. -5<x + 7<17

solve and graph these absolute - value inequalities:

1. |2x - 8|<-10
2. |x + 6|≤9
3. |-3x + 6|≥15

Explanation:

Step1: Solve $x - 2\leq13$

Add 2 to both sides: $x\leq13 + 2$, so $x\leq15$.

Step2: Solve $4x>-20$

Divide both sides by 4: $x>\frac{-20}{4}$, so $x > - 5$.

Step3: Solve $6x-3\geq9$

Add 3 to both sides: $6x\geq9 + 3$, then $6x\geq12$. Divide both sides by 6, we get $x\geq2$.

Step4: Solve $-5

Subtract 7 from all parts: $-5-7

Step5: Solve $|2x - 8|<-10$

Since the absolute - value of any real number is non - negative, $|2x - 8|\geq0$ for all real $x$. So the solution set is the empty set $\varnothing$.

Step6: Solve $|x + 6|\leq9$

Rewrite as $-9\leq x + 6\leq9$. Subtract 6 from all parts: $-9-6\leq x\leq9 - 6$, so $-15\leq x\leq3$.

Step7: Solve $|-3x + 6|\geq15$

Rewrite as two inequalities: $-3x+6\geq15$ or $-3x + 6\leq - 15$.
For $-3x+6\geq15$, subtract 6 from both sides: $-3x\geq15 - 6$, then $-3x\geq9$. Divide both sides by $-3$ and reverse the inequality sign: $x\leq - 3$.
For $-3x + 6\leq - 15$, subtract 6 from both sides: $-3x\leq-15 - 6$, then $-3x\leq-21$. Divide both sides by $-3$ and reverse the inequality sign: $x\geq7$.

Answer:

1. The graph of $x\leq2$ has a closed - circle at 2 and the line extends to the left.
2. The graph of $-8
3. The graph of $|x|\geq9$ has closed - circles at $-9$ and 9 and the lines extend to the left of $-9$ and to the right of 9.
4. The solution of $x - 2\leq13$ is $x\leq15$, and its graph has a closed -[SSE Completed, Client Connection Error][SSE Completed, Client Connection Error][SSE Completed, Client Connection Error][LLM SSE On Failure]