Allysa spent $35 to purchase 12 chickens. She bought two different types of chickens. Americana chickens cost $3.75 each and Delaware chickens cost $2.50 each. Determine algebraically how many of each type of chicken Allysa purchased.
+499 Hey guest! In this problem, we're going to get involved with some systems of equations. First, we have that Allysa bought two different types of chickens: Americana and Delaware, Let's name the quantity that she purchased of each of these chickens \(a \) and \(d\) respectively. We have that she purchased 12 chickens total, so we can write the equation:
\(a + d = 12\)
The problem then tells us that Americana chickens cost $3.75 each, and Delaware Chickens cost $2.50 each. It also gives us that the total amount of money that Allysa spent was 35$. We can then write:
\(3.75a + 2.50d = 35\)
The reasoning for this equation is because each Americana is 3.75, so 3.75 * the number of Americanas purchased represents the total amount of money she spent on Americanas, and likewise for Delaware chickens.
Now, let's multiply our first equation by 2.50, and then subtract it from our second equation to get the value of a.
We get:
\(2.50a + 2.50d = 12 * 2.5 = 30\)
Subtract this from our second equation, and we get:
\(1.25a = 5\)
Dividing by 1.25 on both sides, we get:
\(a = \frac5{1.25} = 4\)
Now that we have the number of Americans purchased, a, we can just plug this back into our first equation \(a+d = 12\)
\(4 + d = 12\)
Subtracting 4 on both sides, we get:
\(d = 8 \)
+37146 12 chickens
a = #Americana then 12-a = Delaware
3.75 a + 2.50 (12-a) = 35 solve for a the number of Americana then #Delaware = 12-a
