algebra question involving perimeter.?

Question

the lenght of a rectangular field is two yards more than three times its width , If the perimeter of the field is 444 yards, what are the dimensions of the field?

okay i know. this problem seem seasy and all. i know what im looking for but i just dont know how to set up this problem. please help me dont just give me the answer. thanks. :D

Answer 1

x=width
3x+2=length

x+x+3x+2+3x+2=444
8x+4=444
8x=440
x=55

so width is 55
length is 3x+2
3(55)+2
165+2=167

check it:
167+167+55+55
334+110=444
so it works

Answer 2

Help is on the way.

First, draw a picture of the problem: a rectangle.
Label the four sides:
the left and right sides each have 'x' for the width

the top and bottom sides each have '3x + 2'
(*That is 3 times the width plus an extra 2 *)

The perimeter is found by adding the four sides.

You get the following EXPRESSION

x + (3x + 2) + x + (3x + 2)

This is supposed to equal 444. So we get the EQUATION

x + (3x + 2) + x + (3x + 2) = 444

Simplfy the left-hand side (LHS) and you get:

8x +4 = 444

Solve the equation using inverse operations

8x + 4 - 4 = 444 = 4 (* Subtract 4 from each side *)
8x = 440 (* Simplify LHS and RHS *)
8x/8 = 440/8 (* Divide both sides by the coefficieint of x*)
x = 55 (* Simplify LHS and RHS *)

That means that the width (x) is 55 yards
The length is 3 * 55 [ 3 times the width ] + 2 =167 yards

Final check:

Find w + l + w + l

55 + 167 + 55 + 167 = 444

Yep, your problem is solved.

Have a Good day

Answer 3

ok, your length is 2 more yards then 3 times its width. You need to find the width & length. Use x for width. The problem states the length is 2 more than 3 times the width so that is saying the length is 3x+2. to solve for perimeter its 2 times the length + 2 times the width. To set the problem up it is:

2(3x+2)+2(x)=444(perimeter of the field). Now use the distributive property & eliminate the parenthises.
Combine like terms & solve for x(which was the width). Subsitute x into the length term & you will find the length. Hope this helped.

Answer 4

let the width is x .then the length is 3x+2
the perimeter =2(length+width)
then 444 =2(3x+2 +x) then 222=4x +2 then 4x=220
then the width x = 220/4=55 then the length= 3(55)+2=165+2=167
then the dimentions are : length= 167yard ,width= 55yard

Answer 5

Let x = width.

Equation:
2(3x + 2 + x) = 444
2(4x + 2) = 444
8x + 4 = 444
8x = 440
x = 55
3x + 2 = 167

Answer: width = 55, length = 167

Proof:
2(167 + 55) = 444
2 * 222 = 444
444 = 444

Answer 6

L - lenght
W - width

"three times its width":-
3W

"two yards more than three times its width":
3W + 2

"the lenght of a rectangular field is two yards more than three times its width":

L = 3W + 2

"perimeter of the field is 444":

2L + 2W = 444

Now find W and L (=solve the equations)
-

Answer 7

l = 3w + 2
P = 444

P = 2(l + w)

444 = 2((3w + 2) + w)
222 = 4w + 2
4w = 220
w = 55

l = 3(55) + 2
l = 167

ANS :

Width = 55 yards
Length = 167 yards

Answer 8

Perimeter = 2 (L + W)
L = 3W + 2
W = W

2(3W + 2) + 2W = 444
8W + 4 = 444
8W = 440
W = 55

Answer:

L = 167
W = 55


L + W = 222

Perimeter = 2 (L + W ) = 2 (222) = 444

Answer 9

System of equations:

Length(L)=3*width(W)+ 2
2w+2L=444

2w+[2(3w+2)]=444
2w+6w+4=444
8w+4=444
8w=440
w=55 yards

2w+2L=444
2(55)+2L=444
110+2L=444
2L=334
L=167 yards

w=55 yards, L=167 yard

Answer 10

Since b is our basis for everything, we'll use that. a= 2b b= b c= a+1 and since a=2b, c=2b+1 All these added together are 76 so 2b + b + 2b + 1 = 76. Simplified, 5b + 1 = 76. Subtract 1 and get 5b = 75 b = 15. Now, using that, we can find our sides. a is 2(15) or 30 cm, b is 15 cm and c is 2(15)+1 or 31 cm.