Question

match each compound inequality on the left to the graph that represents its solution on the right. -5x + 9 < -6 or -3x + 1 ≥ 7 -6x > -18 and 1 ≤ 2x + 5 -16 ≤ 6x + 2 < 14

Explanation:

Step1: Solve $-16\leq6x + 2\lt14$

Subtract $2$ from all parts: $-16-2\leq6x+2 - 2\lt14 - 2$, so $-18\leq6x\lt12$.
Divide by $6$: $\frac{-18}{6}\leq\frac{6x}{6}\lt\frac{12}{6}$, which gives $-3\leq x\lt2$.

Step2: Solve $-6x\gt - 18$ and $1\leq2x + 5$

For $-6x\gt - 18$, divide by $-6$ (reverse inequality): $x\lt3$.
For $1\leq2x + 5$, subtract $5$: $1-5\leq2x+5 - 5$, so $-4\leq2x$. Divide by $2$: $-2\leq x$.
The solution is $-2\leq x\lt3$.

Step3: Solve $-5x + 9\lt - 6$ or $-3x+1\geq7$

For $-5x + 9\lt - 6$, subtract $9$: $-5x+9 - 9\lt - 6 - 9$, so $-5x\lt - 15$. Divide by $-5$ (reverse inequality): $x\gt3$.
For $-3x + 1\geq7$, subtract $1$: $-3x+1 - 1\geq7 - 1$, so $-3x\geq6$. Divide by $-3$ (reverse inequality): $x\leq - 2$.
The solution is $x\gt3$ or $x\leq - 2$.

Answer:

The compound inequality $-16\leq6x + 2\lt14$ corresponds to the first graph (with closed - circle at $-3$ and open - circle at $2$); the compound inequality $-6x\gt - 18$ and $1\leq2x + 5$ corresponds to the second graph (with closed - circle at $-2$ and open - circle at $3$); the compound inequality $-5x + 9\lt - 6$ or $-3x+1\geq7$ corresponds to the third graph (with closed - circle at $-2$ and no overlap in the middle, with part going to the left of $-2$ and part to the right of $3$).