Answer :
By setting up a system of linear equations, the solution to the number of big bucks, little bucks, and does bagged by the Johnson party is found to be 3 big bucks, 6 little bucks, and 17 does.
To find out how many big bucks, little bucks, and does the Johnson party got, we'll assign variables and create equations based on the information provided. Let's denote:
Little bucks as x,
Does as y,
Big bucks as z.
The problem states that:
The number of does (y) is 5 more than twice the amount of little bucks (x), so we have y = 2x + 5.
The number of big bucks (z) was half as many as little bucks (x), so we have z = x/2.
The total number of deer is 26, so x + y + z = 26.
Now we can substitute the first two equations into the third to solve for x:
x + (2x + 5) + (x/2) = 26
3.5x + 5 = 26
x = (26 - 5) / 3.5
x = 6
There were 6 little bucks. Applying this to the other equations:
y = 2 × 6 + 5
y = 17
There were 17 does.
z = 6 / 2
z = 3
There were 3 big bucks.
Therefore, the Johnson party got 3 big bucks, 6 little bucks, and 17 does.
9514 1404 393
Answer:
- 3 big bucks
- 6 little bucks
- 17 does
Step-by-step explanation:
Let B, L, D represent the numbers of big bucks, little bucks, and does, respectively.
B + L + D = 26 . . . . . . total bagged
D = 5 + 2L . . . . . . . . . 5 more does than twice the number of little bucks
B = L/2 . . . . . . . . . . . . half as many big bucks as little bucks
__
We can substitute for B and D in the first equation:
L/2 + L + (5 +2L) = 26
7/2L = 21 . . . . . . . . . . . . . subtract 5, collect terms
L = 6 . . . . . . . . . . . . . . . . . multiply by 2/7
D = 5 + 2·6 = 17
B = 6/2 = 3
The party bagged 3 big bucks, 6 little bucks and 17 does.
