Find the area in sq units bounded by |3x| + |y| = 36.
The modulus function takes the positive value of whatever is inside the brackets. Whatever is inside the brackets can be either positive or negative. Since we have two modulus functions, this means that we actually have 4 equations, not just one.
These equations are,
3x + y = 36
3x - y = 36
-3x + y = 36
-3x - y = 36
The intersection of these four lines can be found by pairing off any two equations (provided they don't have the same gradient) and solving for x and y.
The intersection points are: A(0, -36), B(12, 0), C(0, 36), D(-12, 0)
These four points define a parallelogram shape which is the graph of |3x| + |y| = 36.
The area enclosed by this graph, |3x| + |y| = 36, is the area of the parallelogram ABCD.
This area is 4 times the area of triangle ABO, where O is the origin (0, 0).
area = (1/2)*base*height = (1/2)*12*36 = 216
Answer: enclosed area is 216 sq units