Solve the system of inequalities by graphing y< 2x +4 and 2x -y less than or equal to 5, Show or explain on the graph the region where the solution to the inequalities would be.
To graph the system of inequalities, let's start by graphing each individual inequality.
1) y < 2x + 4:
This is a linear inequality in slope-intercept form. To graph it, follow these steps:
- Plot the y-intercept, which is 4 (the point (0, 4)).
- Determine the slope, which is 2. Since the inequality is y < 2x + 4, the slope is positive, so you can interpret it as "up 2, right 1".
- From the y-intercept, move up 2 units and right 1 unit. Plot another point there, then draw a line through both points.
Note: Since the inequality is y < 2x + 4, the line should be dashed to indicate that the points on the line itself are not included in the solution set.
2) 2x - y ≤ 5:
This is also a linear inequality. To graph it, follow these steps:
- Rearrange the inequality to y ≥ 2x - 5, so that it is in slope-intercept form.
- Plot the y-intercept, which is -5 (the point (0, -5)).
- Determine the slope, which is 2. Since the inequality is y ≥ 2x - 5, the slope is positive, so you can interpret it as "up 2, right 1".
- From the y-intercept, move up 2 units and right 1 unit. Plot another point there.
- Draw a line through both points.
Note: Since the inequality is y ≤ 2x - 5, the line should be solid to indicate that the points on the line itself are included in the solution set.
Now, let's analyze the graph to find the region where the solution to the system of inequalities would be.
The region where the solution to the system would be is the intersection between the shaded regions of the two individual inequalities.
Since the first inequality has a dashed line, the region below the line is shaded.
Since the second inequality has a solid line, the region above the line is shaded.
By analyzing the graph, we can see that the region where the solution to the inequalities would be lies in the area above the dashed line and below the solid line.
Graphically, this region would be the triangular area between the two lines.