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.
clear

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 all parts 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 from both sides: $-4x+3 - 3>-9 - 3$, so $-4x>-12$. Divide both sides by -4 and reverse the inequality sign: $x < 3$. For $-6x\leq12$, divide both sides by -6 and reverse the inequality sign: $x\geq - 2$. The solution of the compound - inequality is $-2\leq x<3$.

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

For $-4x + 3<-9$, subtract 3 from both sides: $-4x+3 - 3<-9 - 3$, so $-4x<-12$. Divide both sides by -4 and reverse the inequality sign: $x>3$. For $6x\leq - 12$, divide both sides by 6: $x\leq - 2$. The solution is $x\leq - 2$ or $x>3$.

Answer:

Match $-14\leq6x + 4<16$ with the graph where the closed - circle is at $x = - 3$ and the open - circle is at $x = 2$.
Match $-4x + 3>-9$ and $-6x\leq12$ with the graph where the closed - circle is at $x=-2$ and the open - circle is at $x = 3$.
Match $-4x + 3<-9$ or $6x\leq - 12$ with the graph where the closed - circle is at $x=-2$ and the open - circle is at $x = 3$ (this is incorrect, it should be a graph with a closed - circle at $x=-2$ and values to the left of it and an open - circle at $x = 3$ and values to the right of it).