First, we have to find the equation of the blue and red lines.
Blue line.
Let's use the points (-3, 0) and (0, 1). We use the slope formula.
![m=(y_2-y_1)/(x_2-x_1)](https://img.qammunity.org/2023/formulas/mathematics/high-school/78uaqhwt0aws3qfwxigaftpihnmb1gzxtp.png)
Replacing the points, we have.
![m=(1-0)/(0-(-3))=(1)/(3)](https://img.qammunity.org/2023/formulas/mathematics/college/6t6si3m8xdaxxppsozgxllch3eu25xw44j.png)
The equation of the blue line is
![y=(1)/(3)x+1](https://img.qammunity.org/2023/formulas/mathematics/high-school/ql5jthrrm6xz1j9w5g45k8lg3q6v5r9jew.png)
Red line.
Let's use the points (1,1) and (2,3).
![m=(3-1)/(2-1)=(2)/(1)=2](https://img.qammunity.org/2023/formulas/mathematics/college/jey90btd1y9r2p3ab4d892mxcnnkttdjrj.png)
The equation of the red line is
![y=2x-1](https://img.qammunity.org/2023/formulas/mathematics/college/opq4zhveiafyly8dg4vkzfh38hux18vdwt.png)
Then, we find the inequality signs for each equation. Notice that the area of the solution is above the blue line and on top of the red line.
So, the inequalities are
![\begin{gathered} y\leq(1)/(3)x+1 \\ y\ge2x-1 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/2wctbx7fs1on9fevksmntq1dt3uqh77nw1.png)