SOLUTION: Can someone help me please? The width of a rectangular is 4ft. less than the length. The area is 5ft.^2. Find the length and the width. Thank you.

Algebra ->  Equations -> SOLUTION: Can someone help me please? The width of a rectangular is 4ft. less than the length. The area is 5ft.^2. Find the length and the width. Thank you.      Log On


   

Question 132293: Can someone help me please?
The width of a rectangular is 4ft. less than the length. The area is 5ft.^2. Find the length and the width.
Thank you.
Found 2 solutions by jim_thompson5910, edjones:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Let L=length, W=width


Since the "width of a rectangular is 4ft. less than the length", this means we have the equation:

W=L-4


Also since the "area is 5ft.^2", we have the second equation


A=L%2AW ----> 5=L%2AW Notice how I plugged in A=5


5=L%2A%28L-4%29 Now plug in W=L-4


5=L%5E2-4L Distribute


0=L%5E2-4L-5 Subtract 5 from both sides




0=%28L-5%29%28L%2B1%29 Factor the right side (note: if you need help with factoring, check out this solver)



Now set each factor equal to zero:
L-5=0 or L%2B1=0

L=5 or L=-1 Now solve for L in each case


So our answer is
L=5 or L=-1



However, since a negative length doesn't make any sense, the only possible solution is

L=5


So the length is 5 ft


Now go back to the equation W=L-4


W=5-4 Plug in L=5


W=1 Subtract


So the width is 1 ft

Answer by edjones(8007) About Me  (Show Source):
You can put this solution on YOUR website!
w=L-4
A=L*w
L(L-4)=5
L^2-4L-5=0
(L-5)(L+1)=0
L=5
w=5-4=1
.
Ed