Question 213072
They trick you a little: They are asking for the area, but first you have to solve for the actual width. Fortunately this works out pretty easily. Let width above be w and rectangles A and B.

Area of rectangle A: 
Width = w
Length = w + 5
Area rectangle A: (Length * Widgth) AreaA = (w)*(w+5)

Area of rectangle B:
Width = w + 1
Length = (w + 5) - 2
Area rectangel B: (Length * Width) AreaB = {{{(w+1)*(w+5-2) = (w+1)*(w+3)}}}

Now the problem says the two areas are equal. Set each area computation equal to the other, and solve for width w. 

AreaA = AreaB
{{{(w)*(w+5) = (w+1)*(w+3)}}}
Simplify with distributive property:
{{{(w*w) + (5*w) = (w*w) + (1*3) + (3*w) + (1*w)}}}
Combine like terms.
{{{w^2 + 5w = w^2 + 3 + (3w + w)}}}
{{{w^2 + 5w = w^2 + 3 + 4w}}}
Subtrace the w^2 from each side to get rid of it. The square term cancels itself out, which is really convenient so we don't have to take a root of anything (!).
{{{w^2 + 5w (-w^2) = w^2 + 3 + 4w (-w^2)}}}
{{{5w = 3 + 4w}}}
Subtract 4w from both sides to isolate the number 3.
{{{5w - 4w = 3 + 4w (- 4w)}}}
{{{5w - 4w = 3}}}
{{{w = 3}}}

Then plug in w to one of the rectangles to get the area. 
AreaA = Length * Width = {{{w * (w+5) = 3 * (3+5) = 3*8 = 24}}}

Check your answer by subbing w in the other rectangle:
AreaB = Length * Width = {{{(w+5-2) * (w+1) = (3+5-2) * (3+1) = 6*4 = 24}}}

AreaA = AreaB !