SOLUTION: how do i find the length and width of a rectangle if the perimeter is 46 and the area is 42?

Algebra ->  Rectangles -> SOLUTION: how do i find the length and width of a rectangle if the perimeter is 46 and the area is 42?      Log On


   



Question 199556: how do i find the length and width of a rectangle if the perimeter is 46 and the area is 42?
Found 2 solutions by ankor@dixie-net.com, edjones:
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
How do i find the length and width of a rectangle if the perimeter is 46
and the area is 42?
:
Write perimeter equation:
2L + 2W = 46
Simplify, divide by 2:
L + W = 23
Arrange for substitution
W = (23-L)
:
Write the area equation:
L * W = 42
;
Substitute (23-L) for W
L *(23-L) = 42
:
-L^2 + 23L - 42 = 0
Multiply by -1
L^2 - 23L + 42 = 0
Factors to
(L - 21)(L - 2) = 0
The greater solution will be the length
L = 21, then W = 2
;
Length = 21
Width = 2
:
Check solutions by finding the perimeter

Answer by edjones(8007) About Me  (Show Source):
You can put this solution on YOUR website!
L*w=42
2(L+w)=46
.
L+w=23
L=23-w
.
w(23-w)=42 substitute for L.
-w^2+23w=42
w^2-23w=-42 multiply each side by -1
w^2-23w+42=0
(w-21)(w-2)=0
w=2, L=21 or vice versa
.
Ed