Can I get help with some Algebra problem solving?

11. The length is L, the width is W

2L+2W=44

LW=40

W=40/L

2L+2(40/L)=44

2L+80/L=44

2L^2+80=44L

L^2-22L+40=0

(L-2)(L-20)=0

L=2 or L=20

W=2 or W=20

The rectangle is 2 by 20

12.The length is L, the width is W

2L+2W=38

LW=84

W=84/L

2L+2(84/L)=38

L+84/L=19

L^2+84=19L

L^2-19L+84=0

(L-12)(L-7)=0

The rectangle is 12 by 7

13. (x+2)(x+5)=130

x^2+7x+10=130

x^2+7x-120=0

(x+15)(x-7)=0

x=-15 or x=7

The side of the square is 7

14. L=2W

(L-3)(w+2)=30

(2w-3)(w+2)=30

2w^2+w-6=30

2w^2+w-36=0

(2w+9)(w-4)=0

W=-9/2 or W=4

W=4, L=8

The rectangle is 8 by 4

15. L=3W

(L+2)(W-1)=68

(3W+2)(W-1)=68

3w^2-W-2=68

3W^2-W-70=0

(3W+14)(W-5)=0

w=-14/3 or w=5

The rectangle is 5 by 15

B.11. Rectangle has length L and width W. L+W = (1/2)perimeter =(1/2)44 = 22. Therefore W = 22 - L. Now area

= 40. Therefore LW = 40 so L(22- L) = 40, ie., L^2 - 22L = - 40, ie., (L-11)^2 = 121-40 = 81 = 9^2, L = 11(+/-)9.

Clearly, L = 20, W = 2. [linear units = cm, area units = cm^2].

B.12. L+W =(1/2)38 =19. Therefore W =19 - L. Now area = 84, ie.,LW = 84, ie., L(19-L) =84, ie.,L^2 -19L+84=0,

ie., (L-7)(L-12) = 0, ie., L = 12, W = 7 [units: linear m, area m^2, m = meter.]

B.13. Dimensions of resulting rectangle are L by W, where L = x+5 and W = x+2. Then rectangle area, LW, =

(x+5)(x+2) =130, ie., x^2 + 7x - 120 =0, ie.,(x-8)(x+15) =0, ie., x =8 {since x > 0} Square side length = 8 cm.