Geomtry Rectangle word problem

Question

Can anyone tell me how solve this problem thnx The length of a rectangle is 6 ft. greater than its width. Were the width 2 ft. less and its length 3 ft. greater, the area would be the same as before. Find the dimensions of the rectangle.

Answer 1

l - length

w - width.

Relation 1:

length is 6 ft greater than width.

l = w+6.

Area is

A = lw

A = (w+6)w.

 

Relation 2:

Same area with "Were the width 2 ft. less and its length 3 ft. greater

A = (w-2)(l+3)

A = (w-2)(w+6+3).

 

We know that Area formulas in Relation 1 and Relation 2 have the same value, so we use the last expressions and solve the equation.

(w+6)w = (w-2)(w+6+3),

 

w² + 6w = (w-2)(w+9),

 

w² +6w = w² -2w+9w-18,

 

Check the math again and solve the equation above. You should have only one answer.