Question

find the graph of this system of linear inequalities. \\(\

$$\begin{cases} y \\leq 3x - 4 \\\\ y \\geq -x + 3 \\end{cases}$$

\\)

Explanation:

Step1: Analyze \( y \leq 3x - 4 \)

The line \( y = 3x - 4 \) has a slope of \( 3 \) and a y - intercept of \( - 4 \). Since the inequality is \( y\leq3x - 4 \), we draw a solid line (because the inequality is non - strict) and shade the region below the line.

Step2: Analyze \( y\geq - x + 3 \)

The line \( y=-x + 3 \) has a slope of \( - 1 \) and a y - intercept of \( 3 \). Since the inequality is \( y\geq - x+3 \), we draw a solid line (because the inequality is non - strict) and shade the region above the line.

Step3: Find the intersection region

The solution to the system of inequalities is the region that is shaded for both inequalities. We can also find the intersection point of the two lines \( y = 3x-4 \) and \( y=-x + 3 \) by setting them equal:
\( 3x-4=-x + 3 \)
\( 3x+x=3 + 4 \)
\( 4x=7 \)
\( x=\frac{7}{4}=1.75 \)
Substitute \( x = 1.75 \) into \( y=-x + 3 \), we get \( y=-1.75 + 3=1.25 \). So the intersection point is \( (1.75,1.25) \).

Now, let's analyze the graphs:

• For \( y\leq3x - 4 \), the line goes up with a slope of 3, and we shade below. For \( y\geq - x + 3 \), the line goes down with a slope of - 1, and we shade above.
• Looking at the three graphs, the third graph (the right - most one) has the region that is below \( y = 3x-4 \) (the orange region) and above \( y=-x + 3 \) (the purple region) overlapping correctly, with the intersection point around \( (1.75,1.25) \) and the correct shading for both inequalities.

Answer:

The third graph (the right - most one among the three given graphs) is the graph of the system of linear inequalities \(

$$\begin{cases}y\leq3x - 4\\y\geq - x + 3\end{cases}$$

\)