Can anyone help with this algebra problem? A rectangle is 4 times as long as it is wide. A second rectangle is 5 centimeters longer and 2...

We assume  the first rectangle measures x cm wide.  Then by given details it is 4x cm long .Then its algebraic area = 4x^2 cm^2.                               (1)

By the given facts the second rectangle measures 4x+5cm long and x+2cm wide. Therefore its  area algebraically = (4x+5)(x+2) cm^2                       (2)

By the given facts,the ralation  between the areas of rectangles is that  the area of second  rectangle  is greater than that of the first by 270 cm ^2.

Therefore,(2)-(1)  gives us: (4x+5)(x+2)-4x^2=270 (3).

So,we have set up the equation of this word probem. Now we solve for x to get the width, x and then the  length, 4x of the original rectangle  from  the equation ,(3):

4x^2+(4x)(2)+5x+10-4x^2=270

13x+10=270

13x=270-10

13x=260

13x/13=260/13

x=20cm, width of the original rectangle

4x=80cm, length of the original rectangle.

Tally:

Area of the first(original) rectangle =80*20=1600cm^2

Area of the 2nd rectangle:(80+5)(20+2)=85*22=1870.

1870-1600=270. So, the fact that the 2nd rectangle is greater by 270cm^2 in area  is verified.