Verified answer

Answer:

The perimeter is 31 meters.

Step-by-step explanation:

• Name all the given units into specific variables.
  [tex]l=length\\w=width\\A=area\\P=perimeter[/tex]
• Translate the statements into an algebraic equation.
  [tex]L=2w-1\\A=55m^{2}[/tex]
• Take note that the formula for finding a rectangle's area is A=length x width.
• Use the formula for finding the values of your length and width.
  [tex]A=l(w)\\A=(2w-1)(w)\\A=2w^2-w[/tex]
• Since our area is 55 sq. meters, we can equate this to our given expression as A=A
  [tex]A=55m^2\\A=2w^2-w\\A=A\\2w^2-w=55m^2[/tex] You may omit the unit m^2 to avoid confusion.
• The equation is quadratic as you subtract 55 on both sides. With this, you can find the value of w (width) by factoring.
  [tex]2w^2-w=55m^2\\2w^2-w-55m^2=0\\(2w-11)(w+5)=0\\w=\frac{22}{4}; w=-5[/tex]
• There are two values of w, but 22/4 is the true value of w since there is no negative width. Thus, w=-5 is an extraneous solution.
• Since we now have the value for width, we must now find the value of length to find the perimeter of our rectangle.
  [tex]l=2w-1\\l=2(\frac{22}{4} )-1\\l=11-1\\l=10[/tex]
• Take note that the perimeter for a rectangle is [tex]P=2l+2w[/tex]
  [tex]P=2l+2w\\P=2(10)+2(\frac{22}{4}) \\P=20+11\\P=31m[/tex]