Together these two turkeys weigh twenty pounds," said the butcher. The little fellow sells for two cents a pound more the big one." Mrs. Smith bought the little one for 82 cents and Mrs. Brown paid $2.96 cents for the big turkey. How much did each turkey weigh?
I need the work also
Let the little fellow weigh x lbs
then the big fellow weighs 20-x lbs
let the cost of the big fellow be c cents per pound
then the cost of the little fellow is c+2 cents
x(c+2) = 82
cx + 2x = 82 #1
c(20-x) = 296
20c - cx = 296 #2
add #1 and #2
2x + 20c = 378
x + 10c = 189
x = 189-10c #3
I will let you finish it,
sub #3 into #1 to find c
then back into #3 to find x
cheap turkeys, must be old question
I don't get the last part:
I will let you finish it,
sub #3 into #1 to find c
then back into #3 to find x
Please help me through this!
Is it
(cx+2x) - (189-10c)?
x = 189-10c
then in #1
c(189-10c) + 2(189-10c) = 82
expanding this and lining it up as a quadratic
10c^2 - 169c - 296 = 0
I used the quadratic formula to get
c = 18.5
now back into #3
x = 189 - 10(18.5)
x = 4
so the little fellow weighed 4 lbs
and the big one weighed 16 lbs
the big one cost 18.5 cents a pound
the little one cost 20.5 cents a pound
check: 20.5x4 = 82
16x18.5 = 296
To find the weight of each turkey, let's set up a system of equations:
Let's assume the weight of the big turkey is x pounds, and the weight of the little turkey is y pounds.
According to the information given, the total weight of both turkeys is 20 pounds:
x + y = 20 Equation 1
It is also mentioned that the little turkey, which weighs y pounds, is sold for two cents per pound more than the big turkey. So the price of the little turkey per pound is (x + 2) cents.
Mrs. Smith bought the little turkey for 82 cents. Since the price per pound is (x + 2) cents, we can set up the following equation:
y(x + 2) = 82 Equation 2
Furthermore, it is stated that Mrs. Brown paid $2.96 for the big turkey. So we can set up another equation:
x(y) = 296 Equation 3
Now, we have a system of equations consisting of Equations 1, 2, and 3. We can solve them simultaneously using substitution or elimination.
Start by solving Equation 3 for x:
x = 296/y Equation 4
Substitute Equation 4 into Equation 1:
(296/y) + y = 20
Multiply through by y to get rid of the fraction:
296 + y^2 = 20y
Rearrange the equation:
y^2 - 20y + 296 = 0
Now, we can factor or use the quadratic formula to solve this equation. In this case, the equation can be factored as:
(y - 8)(y - 12) = 0
So we have two possible solutions for y:
y = 8 or y = 12
Substitute these values back into Equation 4 to find the corresponding weights of each turkey:
For y = 8:
x = 296/8 = 37
For y = 12:
x = 296/12 = 24.67 (approximately)
Therefore, there are two possible combinations of turkey weights:
1. The big turkey weighs 37 pounds and the little turkey weighs 8 pounds.
2. The big turkey weighs approximately 24.67 pounds and the little turkey weighs 12 pounds.