a gardener wants to use 62 feet of fencing to enclose a rectangular-shaped garden. find the length and width of the garden if its length is to be 4ft longer than twice its width.
i wrote 2 equation and wonder which one is correct
4 + x + 2x = 62 and
4+ 2x = 62
are either those correct? if not please suggest. solving not necessary just the equation
2006-11-20 06:43:17 · 11 answers · asked by blinded_by_you01 1 in Science & Mathematics ➔ Mathematics
lets say the width is x
then the lenghth is 4+2x
so the final equation would be 2(4+2x+x)=62
multiply by 2 becaues its 2 lenth and 2 widths.
4+3x=31
3x=27
x=9 Width
length is 4+2x = 4+18 =22
now try to see if it works
22+22+9+9=62
it worked!!!!
2006-11-20 06:45:55 · answer #1 · answered by Anonymous · 0⤊ 0⤋
Neither are correct. You need two variables, like L(ength) and W(idth), not just x and we need to setup a simultaneous equation.
Here's how to set a simultaneous equation using W and L as the variables. We need to have a little knowledge about rectagles and perimeters as well. The perimeter of a rectangle is 2 time the length, plus 2 times the width. We know the perimeter of the rectangle from the information given. So, our first equation is:
2L + 2W = 62
Now, we setup the second equation, based on the other information given in the problem. The Length is 4 feet longer than twice the Width:
L -4 = 2W
Solve for L:
L = 2W +4
Now, we have the two equations:
2L + 2W = 62
L = 2W +4
Now, substitute (2W + 4) in the orgininal equation like this:
2(2w +4) + 2W = 62
Simplifies to:
4W + 8 + 2W = 62
Now that the equation is all in terms of W, you should be able to solve for W, then solve for L.
2006-11-20 07:03:44 · answer #2 · answered by bkeypurr 2 · 0⤊ 0⤋
Say width is 'x' and length is 'y':
Length is twice the width, +4;
y = 2x + 4
and 'x' times 'y' must equal 62;
xy = 62 (or y = 62/x).
So if y = 62/x, you can substitute it in the first equ:
62/x = 2x +4
See if you can do the rest from there ;)
2006-11-20 06:47:55 · answer #3 · answered by Aidan J 2 · 0⤊ 0⤋
I came up with the following:
width = w
length = 4+2w
total = 62
2w + 2(4+2w) = 62
2w + 8 + 4w = 62
6w = 54
w=9
2006-11-20 06:49:11 · answer #4 · answered by Jason C 2 · 0⤊ 0⤋
The sides are X and 2X+4
so if you multiply both side by 2 you get
2X and 4X+8
Therefore 2X+4X+8=62
6X=56
X=9
then plug back in. . .
2(9)+4=22
9+9+22+22=62!
2006-11-20 06:55:04 · answer #5 · answered by J w 1 · 0⤊ 0⤋
niether,
W = width
L = length
P = perimeter = 62
You'll need these two equations to solve the problem
2W + 2L = P (general property of a rectangle)
L = 4+2W (from the question: 4 ft longer then twice the width).
Cheers!
2006-11-20 06:48:06 · answer #6 · answered by agentman18 1 · 0⤊ 0⤋
length = 4 + 2 * width
2*length + 2*width = 62
from these two equations you can solve the length and the width.
and it is not probelm but porelbm ;)
2006-11-20 06:47:57 · answer #7 · answered by gjmb1960 7 · 0⤊ 0⤋
I think you need to post some examples of the kind of word problems you need to learn to solve. Then somebody can explain those problems and describe the technique. I don't see how it is possible to type a meaningful description that would clearly tell how to solve all possible word problems that could appear in an Algebra 1 course.
2016-05-22 00:12:01 · answer #8 · answered by Anonymous · 0⤊ 0⤋
62=2(x)+2(2x+4)
62=6x+8
2006-11-20 06:50:59 · answer #9 · answered by Alex W 2 · 0⤊ 0⤋
2(x+ 2x+4)=62
x is width, 2x+4 is length, and there two of each, so youhave a total of 4 sides in a rectangle.
Aidan below is computing area, not perimeter :)
2006-11-20 06:47:01 · answer #10 · answered by Anonymous · 0⤊ 0⤋