The area of a rectangle is $42 \mathrm{~m}^{2}$, and the length of the rectangle is $5 \mathrm{~m}$ more than twice the width. Find the dimensions of the rectangle.
Length : Øॉ $\mathrm{m}$
\[
\times \quad 5
\]

Step 1 ：Let's denote the width as x and the length as y. Then we have two equations: 1) y = 2x + 5 and 2) x*y = 42

Step 2 ：We can substitute the first equation into the second one to solve for x, which gives us the equation x*(2x + 5) = 42

Step 3 ：Solving this equation gives us two solutions for x: -6 and 7/2

Step 4 ：However, the dimensions of a rectangle cannot be negative. Therefore, the negative solution is extraneous and can be discarded

Step 5 ：So, the width of the rectangle is 7/2 or 3.5 m

Step 6 ：Substituting x = 3.5 into the first equation gives us y = 2*3.5 + 5 = 7

Step 7 ：So, the length of the rectangle is 7 m

Step 8 ：Final Answer: The dimensions of the rectangle are \(\boxed{3.5 \mathrm{~m}}\) (width) and \(\boxed{7 \mathrm{~m}}\) (length)