Question: A rectangular lot has an area of 280m². What is the width of the lot if it requires 68meters of fencing materials to enclose it. (use a variable represent the unknown quantity, then write an equation from the given information)

[tex]\large\bold{\red{PROBLEM:}}[/tex]A rectangular lot has an area of 280m². What is the width of the lot if it requires 68meters of fencing materials to enclose it.

[tex]\large\bold{\red{SOLUTION:}}[/tex] Represents l and w as the length and the width. Formulated equations of the given statement. We have a width is 68 meters.

[tex] \begin{gathered} \begin{cases} \sf 68 = 2w + 2l \\ \sf280 = l \times w\end{cases} \end{gathered}[/tex]

[tex] \begin{gathered} \begin{cases} \sf 34 = w + l \\ \sf280 = l \times w\end{cases} \end{gathered}[/tex]

[tex] \begin{gathered} \begin{cases} \sf 34 = w + l \\ \sf280 = l \times w\end{cases} \end{gathered}[/tex]
[tex] \begin{gathered} \begin{cases} \sf w = 34 - l \\ \sf280 = l \times w\end{cases} \end{gathered}[/tex]

- Substitute the second equation in terms of w.

[tex] \begin{gathered} \begin{cases} \sf w = 34 - l \\ \sf280 = l \times (34 - l)\end{cases} \end{gathered}[/tex]

[tex] \begin{gathered} \begin{cases} \sf w = 34 - l \\ \sf280 = 34l - l^{2} \end{cases} \end{gathered}[/tex]

[tex] \begin{gathered} \begin{cases} \sf w = 34 - l \\ \sf l^{2} - 34l + 280 = 0 \end{cases} \end{gathered}[/tex]

- Solve the second equation using the quadratic formula. Make sure it gives the positive situation.

[tex] \begin{gathered} \sf l = \frac{ -( - 34) + \sqrt{ {34}^{2} - 4(1)(280)} }{2(1)} \end{gathered}[/tex]

[tex] \begin{gathered} \sf l = \frac{ 34 + \sqrt{ 1156 - 4(280)} }{2} \end{gathered}[/tex]

[tex] \begin{gathered} \sf l = \frac{ 34 + \sqrt{ 1156 - 1120} }{2} \end{gathered}[/tex]

[tex] \begin{gathered} \sf l = \frac{ 34 + \sqrt{ 36}}{2} \end{gathered}[/tex]

[tex] \begin{gathered} \sf l = \frac{ 34 + 6}{2} \end{gathered}[/tex]

[tex] \begin{gathered} \sf l = \frac{ 40}{2} \end{gathered}[/tex]

[tex] \begin{gathered} \sf l = 20 \end{gathered}[/tex]

- Substitute the l from the first equation to find the width.

[tex] \begin{gathered} \begin{cases} \sf w = 34 - 20 \\ \sf l = 20\\ \end{cases} \end{gathered}[/tex]

[tex] \begin{gathered} \begin{cases} \sf w = 14 \\ \sf l = 20\\ \end{cases} \end{gathered}[/tex]

- Therefore the width of the rectangle lot is:

[tex] \large \boxed{ \sf{ \green{14 \: meters}}}[/tex]

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