The length of a rectangle is $3 \mathrm{~m}$ less than twice the width, and the area of the rectangle is $27 \mathrm{~m}^{2}$. Find the dimensions of the rectangle.
Length : Џ $\mathrm{m}$
Width :
$\mathrm{m}$

Step 1 ：We are given that the length of a rectangle is 3m less than twice the width, and the area of the rectangle is 27m². We need to find the dimensions of the rectangle.

Step 2 ：We know that the area of a rectangle is given by the formula \(Area = Length \times Width\). We also know that the length of the rectangle is 3m less than twice the width. We can set up a system of equations to solve for the length and width of the rectangle.

Step 3 ：Let's denote the width as \(w\) and the length as \(l\). Then we have: \(l = 2w - 3\) and \(l \times w = 27\).

Step 4 ：We can substitute the first equation into the second to solve for \(w\), and then substitute \(w\) back into the first equation to solve for \(l\).

Step 5 ：The solutions for \(w\) and \(l\) are \(w = \frac{9}{2}\) and \(l = -9\). However, the length of a rectangle cannot be negative. Therefore, the solution \(l = -9\) is not valid.

Step 6 ：The valid solution is \(w = \frac{9}{2}\) and substituting \(w\) into the first equation, we can find the corresponding \(l\).

Step 7 ：Final Answer: The dimensions of the rectangle are \(\boxed{w = \frac{9}{2} \mathrm{m}}\) and \(\boxed{l = 6 \mathrm{m}}\).