An exam consists of two parts, Section X and Section Y. There can be a maximum of 80 questions. There must be at least 20 more questions in Section Y than in Section X. Write a system of inequalities to model the number of questions in each of the two sections. Then solve the system by graphing.
y=x+20
x+y=80
Let's assume the number of questions in Section X is represented by the variable "x", and the number of questions in Section Y is represented by the variable "y".
To model the given conditions, we can create inequalities.
1. The total number of questions cannot exceed 80:
x + y ≤ 80
2. There must be at least 20 more questions in Section Y than in Section X:
y ≥ x + 20
Now, let's graph these inequalities to find the possible solutions.
First, let's graph the equation x + y = 80. To do this, we need to convert it into slope-intercept form:
y = -x + 80
Now, let's plot this line:
- Let's choose some x-values and solve for y:
For x = 0, y = 80.
For x = 80, y = 0.
So, we have two points, (0, 80) and (80, 0), and we can draw a line passing through these points.
Next, let's graph the inequality y ≥ x + 20. To do this, we need to convert it into slope-intercept form:
y ≥ x + 20
Now, let's plot this inequality:
- Let's choose some x-values and solve for y:
For x = 0, y ≥ 20.
For x = 80, y ≥ 100.
We can draw a line with a slope of 1 passing through (0, 20), and we should shade the region above this line to represent the inequality.
Now, the feasible region that satisfies both conditions is the region below the line x + y = 80 and above the line y ≥ x + 20.
Graphically, the feasible region is the region bounded by these two lines.
Note: Due to the limitations of this text-based platform, I am unable to provide an actual graph. I recommend using graphing software or a graphing calculator to plot these lines and find the feasible region.