Question

match each compound inequality on the left to the graph that represents its solution on the right.
-14 ≤ 6x + 4 < 16
-4x + 3 > - 9 and -6x ≤ 12
-4x + 3 < - 9 or 6x ≤ - 12
click and hold an item in one column, then drag it to the matching item in the other column. be sure your cursor is over the target before releasing. the target will highlight or the cursor will change. need help? watch this video.

Explanation:

Step1: Solve $-14\leq6x + 4<16$

Subtract 4 from all parts: $-14 - 4\leq6x+4 - 4<16 - 4$, so $-18\leq6x<12$. Then divide by 6: $\frac{-18}{6}\leq\frac{6x}{6}<\frac{12}{6}$, which gives $- 3\leq x<2$.

Step2: Solve $-4x + 3>-9$ and $-6x\leq12$

For $-4x + 3>-9$, subtract 3: $-4x>-9 - 3=-12$, then divide by - 4 and reverse the inequality sign: $x < 3$. For $-6x\leq12$, divide by -6 and reverse the sign: $x\geq - 2$. So the solution is $-2\leq x<3$.

Step3: Solve $-4x + 3<-9$ or $6x\leq - 12$

For $-4x + 3<-9$, subtract 3: $-4x<-9 - 3=-12$, divide by -4 and reverse the sign: $x>3$. For $6x\leq - 12$, divide by 6: $x\leq - 2$. So the solution is $x\leq - 2$ or $x>3$.

Answer:

Match $-14\leq6x + 4<16$ with the graph having a closed - circle at - 3 and an open - circle at 2.
Match $-4x + 3>-9$ and $-6x\leq12$ with the graph having a closed - circle at - 2 and an open - circle at 3.
Match $-4x + 3<-9$ or $6x\leq - 12$ with the graph having a closed - circle at - 2 and an open - circle at 3 (the third graph in the order presented here as it represents the union of two intervals).