Answer :
To determine the solution to the system of inequalities:
[tex]\[ \left\{ \begin{array}{l} y < -\frac{1}{2} x \\ y \geq 2 x + 3 \end{array} \right. \][/tex]
we’ll analyze each inequality step-by-step, plot them, and then identify the region where both conditions are satisfied.
### 1. Graph the Inequalities
#### First Inequality: [tex]\( y < -\frac{1}{2}x \)[/tex]
This inequality represents a region below the line [tex]\( y = -\frac{1}{2}x \)[/tex].
- The line [tex]\( y = -\frac{1}{2}x \)[/tex] has a slope of [tex]\(-\frac{1}{2}\)[/tex] and passes through the origin (0, 0). It slopes downward to the right.
- The region [tex]\( y < -\frac{1}{2}x \)[/tex] is the area below this line.
#### Second Inequality: [tex]\( y \geq 2x + 3 \)[/tex]
This inequality represents a region above or on the line [tex]\( y = 2x + 3 \)[/tex].
- The line [tex]\( y = 2x + 3 \)[/tex] has a slope of 2 and a y-intercept of 3. It slopes upward to the right.
- The region [tex]\( y \geq 2x + 3 \)[/tex] is the area above or on this line.
### 2. Plot the Lines
We'll need to identify important points to plot each of these lines:
- For [tex]\( y = -\frac{1}{2}x \)[/tex]:
- Use points such as (0, 0) and (2, -1) to plot the line.
- For [tex]\( y = 2x + 3 \)[/tex]:
- Use points such as (0, 3) and (-1, 1) to plot the line.
### 3. Determine the Feasible Region
Now we need to find the intersection of the regions described by each inequality.
- Since [tex]\( y < -\frac{1}{2}x \)[/tex] represents the region below that line, and [tex]\( y \geq 2x + 3 \)[/tex] represents the region above or on that line, we need to find possible intersection points if any.
#### Finding Intersection:
Set the equations [tex]\( y = -\frac{1}{2}x \)[/tex] and [tex]\( y = 2x + 3 \)[/tex] equal to find their intersection:
[tex]\[ -\frac{1}{2}x = 2x + 3 \][/tex]
[tex]\[ -\frac{1}{2}x - 2x = 3 \][/tex]
[tex]\[ -\frac{5}{2}x = 3 \][/tex]
[tex]\[ x = -\frac{6}{5} \][/tex]
By substituting [tex]\( x = -\frac{6}{5} \)[/tex] back into either equation (we'll use [tex]\( y = -\frac{1}{2}x \)[/tex]):
[tex]\[ y = -\frac{1}{2} \left(-\frac{6}{5}\right) = \frac{3}{5} \][/tex]
Intersection point: [tex]\(\left(-\frac{6}{5}, \frac{3}{5}\right)\)[/tex]
### 4. Shading and Final Region
- Below the line [tex]\( y = -\frac{1}{2}x \)[/tex]: This includes all points where the y-value is less than on the line.
- Above or on the line [tex]\( y = 2x + 3 \)[/tex]: This includes all points where the y-value is greater than or equal to on the line.
The only feasible region in this system is where these two intersect, but because one respective needs below [tex]\( y = -\frac{1}{2}x\)[/tex] and other needs above [tex]\( y =2x+3 \)[/tex], they have no common intersection on x-y plane.
Thus, no region satisfies both inequalities simultaneously:
[tex]\[ \boxed{\text{There is no region that satisfies both inequalities.}} \][/tex]
[tex]\[ \left\{ \begin{array}{l} y < -\frac{1}{2} x \\ y \geq 2 x + 3 \end{array} \right. \][/tex]
we’ll analyze each inequality step-by-step, plot them, and then identify the region where both conditions are satisfied.
### 1. Graph the Inequalities
#### First Inequality: [tex]\( y < -\frac{1}{2}x \)[/tex]
This inequality represents a region below the line [tex]\( y = -\frac{1}{2}x \)[/tex].
- The line [tex]\( y = -\frac{1}{2}x \)[/tex] has a slope of [tex]\(-\frac{1}{2}\)[/tex] and passes through the origin (0, 0). It slopes downward to the right.
- The region [tex]\( y < -\frac{1}{2}x \)[/tex] is the area below this line.
#### Second Inequality: [tex]\( y \geq 2x + 3 \)[/tex]
This inequality represents a region above or on the line [tex]\( y = 2x + 3 \)[/tex].
- The line [tex]\( y = 2x + 3 \)[/tex] has a slope of 2 and a y-intercept of 3. It slopes upward to the right.
- The region [tex]\( y \geq 2x + 3 \)[/tex] is the area above or on this line.
### 2. Plot the Lines
We'll need to identify important points to plot each of these lines:
- For [tex]\( y = -\frac{1}{2}x \)[/tex]:
- Use points such as (0, 0) and (2, -1) to plot the line.
- For [tex]\( y = 2x + 3 \)[/tex]:
- Use points such as (0, 3) and (-1, 1) to plot the line.
### 3. Determine the Feasible Region
Now we need to find the intersection of the regions described by each inequality.
- Since [tex]\( y < -\frac{1}{2}x \)[/tex] represents the region below that line, and [tex]\( y \geq 2x + 3 \)[/tex] represents the region above or on that line, we need to find possible intersection points if any.
#### Finding Intersection:
Set the equations [tex]\( y = -\frac{1}{2}x \)[/tex] and [tex]\( y = 2x + 3 \)[/tex] equal to find their intersection:
[tex]\[ -\frac{1}{2}x = 2x + 3 \][/tex]
[tex]\[ -\frac{1}{2}x - 2x = 3 \][/tex]
[tex]\[ -\frac{5}{2}x = 3 \][/tex]
[tex]\[ x = -\frac{6}{5} \][/tex]
By substituting [tex]\( x = -\frac{6}{5} \)[/tex] back into either equation (we'll use [tex]\( y = -\frac{1}{2}x \)[/tex]):
[tex]\[ y = -\frac{1}{2} \left(-\frac{6}{5}\right) = \frac{3}{5} \][/tex]
Intersection point: [tex]\(\left(-\frac{6}{5}, \frac{3}{5}\right)\)[/tex]
### 4. Shading and Final Region
- Below the line [tex]\( y = -\frac{1}{2}x \)[/tex]: This includes all points where the y-value is less than on the line.
- Above or on the line [tex]\( y = 2x + 3 \)[/tex]: This includes all points where the y-value is greater than or equal to on the line.
The only feasible region in this system is where these two intersect, but because one respective needs below [tex]\( y = -\frac{1}{2}x\)[/tex] and other needs above [tex]\( y =2x+3 \)[/tex], they have no common intersection on x-y plane.
Thus, no region satisfies both inequalities simultaneously:
[tex]\[ \boxed{\text{There is no region that satisfies both inequalities.}} \][/tex]
