Question 1 of 10

On a piece of paper, graph this system of inequalities. Then determine which region contains the solution to the system.

[tex]\[
\begin{array}{l}
y = \frac{1}{4}x + 3 \\
y \geq -5
\end{array}
\][/tex]

Text description for graph:



Answer :

To solve and graph this system of inequalities, follow these steps:

### Step 1: Understand and Rewrite the Inequalities
1. The first line given is:
[tex]\[ y = \frac{1}{4} + 3 \][/tex]

This simplifies to:
[tex]\[ y = \frac{1}{4}x + 3 \][/tex]

2. The second inequality is:
[tex]\[ y \geq -5 \][/tex]

### Step 2: Plot the First Equation

We plot the equation [tex]\( y = \frac{1}{4}x + 3 \)[/tex]:
- Identify the y-intercept (where [tex]\( x = 0 \)[/tex]):
[tex]\[ y = 3 \][/tex]
So the point (0, 3) is on the graph.

- Identify the slope, which is [tex]\(\frac{1}{4}\)[/tex]:
For every increase of 1 unit in [tex]\( x \)[/tex], [tex]\( y \)[/tex] increases by [tex]\(\frac{1}{4}\)[/tex].

- Plot a few more points:
- When [tex]\( x = 4 \)[/tex], [tex]\( y = \frac{1}{4} \cdot 4 + 3 = 1 + 3 = 4 \)[/tex], so the point (4, 4) is on the graph.
- When [tex]\( x = -4 \)[/tex], [tex]\( y = \frac{1}{4} \cdot (-4) + 3 = -1 + 3 = 2 \)[/tex], so the point (-4, 2) is on the graph.

Connect these points with a straight line, extending it in both directions.

### Step 3: Represent the Inequality [tex]\( y \geq -5 \)[/tex]

To graph [tex]\( y \geq -5 \)[/tex]:
- Draw a horizontal line at [tex]\( y = -5 \)[/tex].
- Shade the region above this line including the line itself (because [tex]\( y \geq -5 \)[/tex] means all values of [tex]\( y \)[/tex] that are larger than or equal to -5).

### Step 4: Identify the Feasible Region

The feasible region is where the shaded area of [tex]\( y \geq -5 \)[/tex] overlaps with the line [tex]\( y = \frac{1}{4}x + 3 \)[/tex]:

- Start from the line [tex]\( y = \frac{1}{4}x + 3 \)[/tex].
- The graph of [tex]\( y \geq -5 \)[/tex] covers everything above the line [tex]\( y = -5 \)[/tex] up to continuing indefinitely upwards, while the equation [tex]\( y = \frac{1}{4}x + 3 \)[/tex] is a line extending diagonally.

### Text Description for the Graph

1. Line Plot: Draw the line from the equation [tex]\( y = \frac{1}{4}x + 3 \)[/tex] starting at the y-intercept (0, 3). As this is a linear function with a positive slope, the line will rise from left to right.

2. Horizontal Line and Shading: Draw a horizontal line at [tex]\( y = -5 \)[/tex] and shade the region above and including this line.

3. Overlap Region: Indicate that the feasible region (solution set) lies both above the [tex]\( y = -5 \)[/tex] line and on the [tex]\( y = \frac{1}{4}x + 3 \)[/tex] line.

This gives us the area where any point will satisfy both inequalities. The solution to the system is the region on and above the line [tex]\( y = -5 \)[/tex] that also lies on the line [tex]\( y = \frac{1}{4}x + 3 \)[/tex].