Answer:
[tex]To \: express \: this \: as \: an \: equation, \: \\ let \: L \: represents \: \: length \: and \: w \: = \: width\: \\ The \: width \: of rectangle \: is \: 3 \: feet \: less \: than \: \\ 5 \: times \: its \: length. \: This \: tells \: us \: that \: 3 \: \\ will \: be \: subtracted \: from \: 5l. \: So,
\\ w \: = 5L \: - \: 3. \: \\ \\ Next \: the \: area \: is \: 14 \: and \: since \: the \: area \: of \: a \: rectangle \: is \\ \: length \times \: width, \: it will \: be \: expressed \: as \: \\ 14 \: = L \times w \: since \: w \: = (5L - 3) \\ 14 = L \times (5L - 3) \\ 14 = 5L {}^{2} - 3L \\ 5L {}^{2} - 3L - 14 = 0 \\ factor \: the \: polynomial \\ (5L + 7)(L - 2) = 0 \\ L = \frac{ - 7}{5} \: or \: 2 \: but \: the \: measurement \: cannot be \: \\ negative \: so \: we \: will \: consider \: 2 \: instead \: \\ \\ L = 2 ft \ \\\\ to \: find \: width \: \\ w \: = \: 5L - 3 \\ w \: = \: 5(2) - 3 \\ w \: = \: 10 - 3 \\ w = > 7 \: feet[/tex]
