Question

match each compound inequality with its number line solution. show your work on a separate sheet of paper. 12 < k + 2 ≤ 18 2c < -10 or -3c < -12 3n + 11 ≤ 13 or 3x ≥ 12 3 < 2x - 3 < 15

Explanation:

Step1: Solve $12 < k + 2\leq18$

Subtract 2 from all parts: $12-2 < k+2 - 2\leq18 - 2$, so $10 < k\leq16$. Matches the first number - line (open - circle at 10, closed - circle at 16).

Step2: Solve $2c < - 10$ or $-3c < - 12$

For $2c < - 10$, divide by 2: $c < - 5$. For $-3c < - 12$, divide by - 3 and reverse the inequality sign: $c>4$. Matches the fourth number - line (open - circle at - 5, open - circle at 4).

Step3: Solve $3n + 11\leq13$ or $3x\geq12$

For $3n + 11\leq13$, subtract 11 and then divide by 3: $3n\leq13 - 11$, $3n\leq2$, $n\leq\frac{2}{3}$. For $3x\geq12$, divide by 3: $x\geq4$. Matches the third number - line (closed - circle at $\frac{2}{3}$, closed - circle at 4).

Step4: Solve $3 < 2x - 3 < 15$

Add 3 to all parts: $3 + 3<2x-3 + 3<15 + 3$, $6 < 2x<18$. Divide by 2: $3 < x<9$. Matches the second number - line (open - circle at 3, open - circle at 9).

Answer:

$12 < k + 2\leq18$ matches the first number - line (open - circle at 10, closed - circle at 16).
$2c < - 10$ or $-3c < - 12$ matches the fourth number - line (open - circle at - 5, open - circle at 4).
$3n + 11\leq13$ or $3x\geq12$ matches the third number - line (closed - circle at $\frac{2}{3}$, closed - circle at 4).
$3 < 2x - 3 < 15$ matches the second number - line (open - circle at 3, open - circle at 9).