Given the system of inequalities below, find the vertices of the feasible region. Report your vertices starting with the one which has the smallest x-value. If more than one vertex has the same, smallest x-value, start with the one that has the smallest y-value. Proceed clockwise from the first vertex.

y<-x+7
y>-7x+11
x>0 y> 0
if someone could explain it to me that would be wonderful!

where did you get 10?

x and y equals 10

y<-x+7

y>-7x+11
x>0 y> 0
Graphically, draw the four inequalities as through they are lines,
y=-x+7
y=-7x+11
x=0
y= 0
They should intersect at six distinct points.

Create the figure that represents the region of feasibility.

The goal is to find the boundaries of the region of feasibility bounded by four intersection points, and the correspoinding segments of straight line representing the inequalities.

This can be achieved (as described in the question) by starting at the origin (intersection of x>0 and y>0) and proceeding along the x-axis to the next intersection point. If this point satisfies all constraints, or at the border of it, proceed along the straight line. If not, go back to the previous point and continue with the intersecting line.

To find the vertices of the feasible region for the given system of inequalities, we need to follow these steps:

Step 1: Graph the linear inequalities.
Step 2: Identify the points where the boundary lines intersect.
Step 3: Test each intersection point to determine if it satisfies all the inequalities.
Step 4: Determine the vertices based on the satisfying intersection points.

Let's start with Step 1 and graph the linear inequalities:

1. Graph y < -x + 7:
- Note that this inequality represents a line with a negative slope of -1 and a y-intercept of 7. It has a shaded region below the line.

2. Graph y > -7x + 11:
- This inequality represents a line with a negative slope of -7 and a y-intercept of 11. The shaded region will be above the line.

3. Graph x > 0 and y > 0:
- These two inequalities represent half-planes that are above the x-axis and to the right of the y-axis, respectively. Shade the regions accordingly.

Now, onto Step 2:

Look for the points where the boundary lines intersect. The lines y < -x + 7 and y > -7x + 11 will intersect each other at a specific point. We can solve these equations to find their intersection point:

-y = -7x + 11
-y = -x + 7

By equating the right sides of the equations, we can find x and y values:

-7x + 11 = -x + 7
-6x = -4
x = 2/3

Substitute the value of x back into either equation to find y:

-y = -7(2/3) + 11
-y = -14/3 + 33/3
y = 19/3

So, the intersection point is (2/3, 19/3).

Now, onto Step 3:

Test the intersection point (2/3, 19/3) in all four inequalities to determine if it satisfies each one. If it satisfies all the inequalities, then it is a vertex of the feasible region.

The inequality y < -x + 7 is true for the point (2/3, 19/3) since 19/3 is less than -(2/3) + 7.

The inequality y > -7x + 11 is also true for the same point since 19/3 is greater than -7(2/3) + 11.

For x > 0, the point (2/3, 19/3) satisfies this condition.

For y > 0, it also satisfies the condition.

Therefore, the point (2/3, 19/3) satisfies all the inequalities and is a vertex of the feasible region.

Finally, onto Step 4:

We need to determine if there are any other vertices by examining the boundary lines and testing their intersection points. However, in this case, the point (2/3, 19/3) is the only intersection and also the only vertex for the feasible region since the inequalities limit the possibilities.

So, the vertices of the feasible region for the given system of inequalities are (2/3, 19/3).