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 three equations, we first need to solve for the points where they intersect.
1. x = 0 represents the y-axis. So one of the boundaries is the y-axis. This line is vertical and passes through the origin.
2. 6y - 5x = 0 can be rearranged to y = (5/6)x. This line will intersect the y-axis at (0,0) and the x-axis at (6,5). It has a positive slope, so it will be the line that forms the upper boundary of the region.
3. x + 3y = 21 can be rearranged to y = (21-x)/3. This line will intersect the y-axis at (0,7) and the x-axis at (21,0). It has a negative slope, so it will be the line that forms the lower boundary of the region.
Now, we sketch the region enclosed by these three lines:
The vertices of the region are: (0,0), (0,7), (6,5), (3,4)
To find the area of the region, we can split it into two triangles and calculate the area of each.
Area of triangle 1 (formed by x-axis, y-axis and the line y = (5/6)x):
base = 6
height = 5
Area = 0.5 * base * height = 0.5 * 6 * 5 = 15
Area of triangle 2 (formed by the lines y = (5/6)x, y = (21-x)/3, and the y-axis):
base = 3
height = 4
Area = 0.5 * base * height = 0.5 * 3 * 4 = 6
Total area = Area of triangle 1 + Area of triangle 2 = 15 + 6 = 21
Therefore, the area of the region enclosed by x=0, 6y-5x=0, and x+3y=21 is 21 units squared.