We need to graph on the number line the solution to the compounded inequality
[tex]\begin{gathered} y+6<10 \\ \text{or } \\ 2y-3>9 \end{gathered}[/tex]
In order to do so, let's work with each inequality separately. The final solution will be the union of the two solutions since it can be one "or" the other.
Step 1
Subtract 6 from both sides of the first inequality:
[tex]\begin{gathered} y+6<10 \\ \\ y+6-6<10-6 \\ \\ y<4 \end{gathered}[/tex]
So, the solution to the first inequality is all real numbers less than 4 (not included). Therefore, we graph this solution using an empty circle:
Step 2
Add 3 to both sides of the second inequality, and then divide both sides by 2:
[tex]\begin{gathered} 2y-3+3>9+3 \\ \\ 2y>12 \\ \\ \frac{2y}{2}>\frac{12}{2} \\ \\ y>6 \end{gathered}[/tex]
Thus, the solution to this inequality is all the real numbers greater than 6 (not included: empty circle):
Answer
Therefore, the solution to the compounded inequalities is the union of both solutions:
