Question

Anyone want to help me with this? Could i have it sometime tonight.

Answer 1

so we know the area if 44 cm², and we also know the "L" is 3 more than twice "W", hmmm what's twice W anyway? well, just 2W, and 3 more than that? well 2W + 3, hmmm so

`A=LW ~~ \begin{cases} L=length\\ W=width\\[-0.5em] \hrulefill\\ A=44 \end{cases}\implies 44=LW\hspace{5em}\stackrel{\textit{3 more than twice W}}{L=2W+3} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{substituting on the 1st equation}}{44=(\underset{L}{2W+3})W}\implies 44=2W^2+3W \\\\\\ 0=2W^2+3W-44 \implies 0=(2W+11)(W-4) \\\\[-0.35em] ~\dotfill\\\\ 0=2W+11\implies -11=2W\implies \cfrac{-11}{2}=W ~~ \bigotimes \\\\[-0.35em] ~\dotfill\\\\ 0=W-4\implies 4=W ~~ \textit{\LARGE \checkmark}`

now, the negative value is a valid value for W, however for this case, the width cannot be a negative number, it must be positive, thus we do away with it, and since L = 2W+3, that means L = 2(4)+3 = 11.