can you solve this for me?please?

Length = L
Width = W

L*W = 80
L=3*W+1

Plug the second into the first:

(3*W+1)*W = 3*W^2+W = 80

Resulting quadratic equation:
3W^2+W-80=0

The solutions are W=5 or W=-5 1/3

Since W must be positive, we're left with W=5

L=80/W=80/5=16

The rectangle is 5*16

x = length
y = width

xy = 80 (solve by using substitution)
x = 1 + 3y

xy = 80
(1 + 3y)y = 80
y + 3y^2 = 80
3y^2 + y - 80 = 0
3y^2 + 16y - 15y - 80 = 0
(3y^2 + 16y) - (15y + 80) = 0
y(3y + 16) - 5(3y + 16) = 0
(3y + 16)(y - 5) = 0

3y + 16 = 0
3y = -16
y = -16/3

y - 5 = 0
y = 5

xy = 80
x(-16/3) = 80
x = 80/(-16/3)
x = -80(3/16)
x = -15

xy = 80
x(5) = 80
x = 80/5
x = 16

∴ (x = -15 , y = -16/3) , (x = 16 , y = 5) (<== the answer should be positive, so the length is 16 units and the width is 5 units.)

Length = 16
Width = 5

5 and 16

one side can be x. the other is 3x+1.

area is side x side. so (x)(3x+1)=80

then by quadratic equation u get x = 5. :)

you draw the diagram, I solve it
(3w + 1)w = 80 whence 3w^2 + w -- 80 = 0 Or (3w + 16)(w -- 5) = 0
giving w = 5 and length = 3*5+1 = 16

If the breadth is x (>0)the length is 3x+1 and so x(3x+1) = 80
3x^2 + x - 80 = 0.
Now factorise and solve and discard - ve root.
(x-5)(3x+16)=0 so x =5 and length 16.