Solution
- Let the length of the rectangle be x
- Let the width of the rectangle be y.
- Thus, we can interpret the lines of the question as follows:
[tex]\begin{gathered} \text{ The length is 30 less than 6 times the width can be written as} \\ x=6y-30\text{ (Equation 1)} \\ \\ \text{The area of the rectangle is 4836. This is written as:} \\ xy=4836\text{ (Equation 2)} \end{gathered}[/tex]
- Now, let us solve these two equations simultaneously.
- We shall proceed by solving the system of equations graphically.
- Wherever the graphs of Equation 1 and Equation 2 intersect represents the solution to the system of equations
- The plot of the equations is given below
- Observe that the graphs cross at two points. The first point is positive and the other, negative.
- Since we cannot have negative lengths (x) or width (y), we can discard the negative coordinates.
- Thus, the length (x) and width (y) are given below:
[tex]\begin{gathered} \text{length(x)}=156 \\ \text{width(y)}=31 \end{gathered}[/tex]
Final Answer
The length of the rectangle is 156 feet
