A fish has a head 9" (inches) long. The tail is equal to the size of the head plus one-half the size of the body. The body is the size of the head plus the tail. How long is the fish?
Please, those of you who know how to do it, help me to understand answer this question through basic elaboration.
So, I going to denote head to h, body to b, and tail to t.
The tail is equal to the head (9 inches) plus half the body.
\(t=9+\frac{1}{2}b\)
The body is the size of the head plus the tail.
\(b=9+t\)
Now, plug the first equation into the second for t.
\(b=9+(9+\frac{1}{2}b)\)
Simplify.
\(b=18+{1\over2}b\)
Multiply everything by 2 to get rid of fractions.
\(2b=36+b\)
Subtract b from both sides.
\(b=36\)
So, the body is 36 inches long
Now, plug 36 into the first equation.
\(t=9+{36\over2}\)
Simplify.
\(t=9+18\)
\(t=27\)
So, the head is 9 inches long, the body is 36 inches long, and the tail is 27 inches long.
For the full length it is h+b+t, or \(9+36+27\).
\(9+36+27=72\)
So your fish is 72 inches long (now that's a catch!)
I hope I get this right...
h = head length = 9
b = body length = 9 + t
t = tail length = 9 + 1/2b
We need to solve for the body or tail length: solve(b=9+t,t=9+1/2b) = {b=36, t=27}
I don't know what's wrong with 72 as an answer, CPhill. Or were you just making another pun?
A fish has a head 9 inches long. | → | h = 9 |
The tail equals the size of the head plus one-half the size of the body. | → | t = h + b/2 |
The body is the size of the head plus the tail. | → | b = h + t |
b = h + t Since h = 9 , we can substitute 9 in for h .
b = 9 + t
t = h + b/2 Substitute 9 + t in for b and 9 in for h .
t = 9 + (9 + t)/2 Multiply through by 2 .
2t = 18 + 9 + t
2t = 27 + t Subtract t from both sides.
t = 27
b = h + t Substitute 9 in for h and 27 in for t .
b = 9 + 27
b = 36
And the total length of the fish = h + b + t
h + b + t = 9 + 36 + 27 = 72 inches
I also get 72 inches, like AT and helperid.