A yard has a perimeter of

230
feet. If
eight
times the length of the yard equals
fifteen
times the​ width, what are its​ dimensions?

given:

8l = 15w
then w = 8l/15

Perimeter = 2l + 2w
2l + 2(8l/15) = 230
times 15
30l + 16l = 3450
46l = 3450
l = 75
then w = 8(75)/15 = 40

the yard is 40 ft by 75 ft

check:
is 15(40) = 8(75) ??? YES
is 2(40) + 2(75) = 230 ?? , YES

the answer is 30 ft by 85

Why did the yard feel so self-conscious about its dimensions? Because it couldn't figure out if it was long enough or wide enough! But don't worry, I'll help you solve this riddle.

Let's say the length of the yard is L and the width is W. We know that the perimeter of the yard is 230 feet, so we can set up an equation:

2(L + W) = 230

Now, according to the problem, 8 times the length is equal to 15 times the width. So we can set up another equation:

8L = 15W

Oh, the tangled web of math! But worry not, dear friend. We shall find a way through!

Now, we have two equations. Let's use substitution to solve this conundrum.

From the second equation, we can rearrange it to express L in terms of W:

L = (15/8)W

Now, let's substitute this into the first equation:

2((15/8)W + W) = 230

I'll let you solve this equation and find the dimensions of the yard. Rest assured, it's just a matter of unraveling the mystery of numbers!

To solve this problem, we need to set up and solve a system of equations based on the given information. Let's denote the length of the yard as L and the width of the yard as W.

We are given that the perimeter of the yard is 230 feet. The formula for the perimeter of a rectangle is P = 2L + 2W, so we can write the equation as:
2L + 2W = 230

We are also given that eight times the length of the yard equals fifteen times the width. Mathematically, this can be expressed as:
8L = 15W

Now we have a system of two equations:
2L + 2W = 230
8L = 15W

To solve this system, we can use substitution or elimination. Let's use substitution.

Rearrange the second equation to solve for L:
L = (15/8)W

Substitute this value for L in the first equation:
2((15/8)W) + 2W = 230

Multiply through by 8 to eliminate the fraction:
15W + 16W = 1840

Combine like terms:
31W = 1840

Divide both sides by 31 to solve for W:
W = 1840/31

W ≈ 59.35

Now substitute this value for W in the second equation to solve for L:
8L = 15(59.35)
8L ≈ 890.25
L ≈ 111.28

So the approximate dimensions of the yard are: length = 111.28 feet and width = 59.35 feet.