Jeff went fishing yesterday and caught a large fish. Upon measuring it he found that the head was 6 inches long. He also noted that the tail was as long as the head plus one-third of the back and that the back was as long as the tail plus twice the head. How long was this fish?
To find the length of the fish, let's break down the information given into equations:
Let H be the length of the fish's head.
The tail is as long as the head plus one-third of the back: T = H + 1/3B.
The back is as long as the tail plus twice the head: B = T + 2H.
We can substitute the value of T from the first equation into the second equation:
B = (H + 1/3B) + 2H.
To simplify this equation, let's get rid of the fraction by multiplying both sides by 3:
3B = 3H + B + 6H.
Combine like terms on the right side of the equation:
3B - B = 9H.
2B = 9H.
Divide both sides of the equation by 9 to solve for B:
B = (9H) / 2.
Now we can substitute the value of B back into the first equation to find T:
T = H + 1/3B.
T = H + (1/3)((9H) / 2).
T = H + (3H / 2).
T = (2H + 3H) / 2.
T = 5H / 2.
Finally, the length of the fish is the sum of the head, tail, and back:
Length = H + T + B.
Length = H + (5H / 2) + (9H / 2).
Length = (2H + 5H + 9H) / 2.
Length = 16H / 2.
Length = 8H.
Therefore, the length of the fish is 8 times the length of its head. To find the length of the fish, we would need to know the length of the head, as it is not given in the question.