Answer:
[tex]\begin{aligned}\textsf{1.} \quad y & \geq-\dfrac{3}{4}x+3\\y & \leq\dfrac{1}{2}x+2\end{aligned}[/tex]
Step-by-step explanation:
When a line has a positive slope, the y-values increase as the x-values increase.
When a line has a negative slope, the y-values decrease as the x-values increase.
Given equations:
[tex]y=-\dfrac{3}{4}x+3 \quad \rightarrow \textsf{negative\;slope}[/tex]
[tex]y=\dfrac{1}{2}x+2 \quad \rightarrow \textsf{positive\;slope}[/tex]
When graphing inequalities:
- < or > : dashed line.
- ≤ or ≥ : solid line.
- < or ≤ : shade under the line.
- > or ≥ : shade above the line.
Therefore, from inspection of the given graph:
- The line with the negative slope has shading above the line:[tex]\implies y\geq-\dfrac{3}{4}x+3[/tex]
- The line with the positive slope has shading below the line:
[tex]\implies y \leq\dfrac{1}{2}x+2[/tex]
Therefore, the system of inequalities that represents the given graph is:
[tex]\boxed{\begin{aligned} y & \geq-\dfrac{3}{4}x+3\\y & \leq\dfrac{1}{2}x+2\end{aligned}}[/tex]
