Question

review: solve and graph numbers 1 to 4. they are normal solution sets for compound inequalities and absolute - value inequalities. 1. $x + 6<4$ or $3x>3$ 2. $3leq2x + 1leq11$ 3. $|x + 2|<3$ 4. $|4x|geq8$

Explanation:

Step1: Solve $x + 6<4$

Subtract 6 from both sides: $x<4 - 6$, so $x<-2$.

Step2: Solve $3x>3$

Divide both sides by 3: $x > 1$.
The solution of $x + 6<4$ or $3x>3$ is $x<-2$ or $x>1$.

Step3: Solve $3\leq2x + 1\leq11$

Subtract 1 from all parts: $3-1\leq2x+1 - 1\leq11 - 1$, so $2\leq2x\leq10$.

Step4: Divide all parts by 2

$\frac{2}{2}\leq\frac{2x}{2}\leq\frac{10}{2}$, so $1\leq x\leq5$.

Step5: Solve $|x + 2|<3$

Rewrite as $-3

Step6: Solve $|4x|\geq8$

Rewrite as two inequalities: $4x\geq8$ or $4x\leq - 8$.
For $4x\geq8$, divide by 4: $x\geq2$. For $4x\leq - 8$, divide by 4: $x\leq - 2$.

Answer:

1. Solution: $x<-2$ or $x>1$. Graph: Open - circle at $x = - 2$ and shade left, open - circle at $x = 1$ and shade right.
2. Solution: $1\leq x\leq5$. Graph: Closed - circle at $x = 1$ and closed - circle at $x = 5$, shade between.
3. Solution: $-5
4. Solution: $x\leq - 2$ or $x\geq2$. Graph: Closed - circle at $x=-2$ and shade left, closed - circle at $x = 2$ and shade right.