can you solve this for me?please?
2008-10-29 2:34 pm
A certain rectangle has an area of 80 square units. Its length is one more than three times its width. What are the dimensions of the rectangle? Draw a diagram, solve the problem, and write an equation
回答 (6)
2008-10-29 2:40 pm
✔ 最佳答案
Length = LWidth = 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
2008-10-30 5:49 pm
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
â´ (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.)
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
â´ (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.)
2008-10-29 10:02 pm
Length = 16
Width = 5
Width = 5
2008-10-29 9:45 pm
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. :)
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. :)
參考: simple algebra
2008-10-29 9:42 pm
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
(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
2008-10-29 9:38 pm
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.
3x^2 + x - 80 = 0.
Now factorise and solve and discard - ve root.
(x-5)(3x+16)=0 so x =5 and length 16.
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