Sketch the region enclosed by the graphs of x = 0, 6y - 5x = 0, and x + 3y = 21.
Find the area.
To sketch the region enclosed by the graphs of x = 0, 6y - 5x = 0, and x + 3y = 21, we first need to find the points where the lines intersect.
For x = 0, this is the y-axis.
To find the intersection point of 6y - 5x = 0 and x + 3y = 21, we can solve the system of equations:
6y - 5x = 0
x + 3y = 21
Solving for x in the second equation gives:
x = 21 - 3y
Substitute this into the first equation:
6y - 5(21 - 3y) = 0
6y - 105 + 15y = 0
21y = 105
y = 5
Now substitute y = 5 back into x = 21 - 3y:
x = 21 - 3(5)
x = 6
So the intersection point is (6, 5).
Now we can sketch the region enclosed by the graphs:
1. x = 0 is the y-axis.
2. 6y - 5x = 0 passes through the origin (0,0) and (6,5).
3. x + 3y = 21 passes through (0,7) and (21,0).
The enclosed region is a triangle with vertices at (0,0), (6,5), and (0,7).
To find the area of the triangle, we can calculate the base and height of the triangle:
Base = 6 units
Height = 5 units
Area = 0.5 * base * height
Area = 0.5 * 6 * 5
Area = 15 square units
Therefore, the area of the region enclosed by the graphs of x = 0, 6y - 5x = 0, and x + 3y = 21 is 15 square units.