Given that the rectangle has dimensions a and b where a is less than or equal to b
Let b represent the length of the rectangle.
Let a represent the width of the rectangle.
Given that the area of the rectangle is 198 square feet, it means that
ba = 198 equation 1
We know that one side is 15 ft longer than three times the other side. It means that
b = 15 + 3a
Substituting b = 15 + 3a into equation 1, it becomes
a(15 + 3a) = 198
15a + 3a^2 = 198
3a^2 + 15a - 198 = 0
Dividing through the equation by 3, it becomes
a^2 + 5a - 66 = 0
We would solve the quadratic equation by applyling the method of factorisation. We would find two numbers such that their sum or difference is 5a and their product is - 66a^2. The numbers are 11a and 6a. The equation becomes
a^2 + 11a - 6a - 66 = 0
a(a + 11) - 6(a + 11) = 0
(a - 6)(a + 11) = 0
a - 6 = 0 or a + 11 = 0
a = 6 or a = - 11
Since the dimension of the rectangel cannot be negative, then the value of a is 6
Substituting a = 6 into equation 1, it becomes
6b = 198
b = 198/6
b = 33
The shorter side is 6 feet
The longer side is 33 feet