Answer:
<em>a. The system of equations that models the situation is....</em>
<em>
</em>
<em>b. The solution to the system: x = 2 and y = 4 </em>
<em>The amount of 80/20 mixture is 2 pounds and the amount of 50/50 mixture is 4 pounds.</em>
Step-by-step explanation:
Suppose, the amount of 80/20 mixture is
pounds and the amount of 50/50 mixture is
pounds.
So, the amount of peanuts in 80/20 mixture
pound and the amount of almonds in 80/20 mixture
pound.
And the amount of peanuts in 50/50 mixture
pound and the amount of almonds in 50/50 mixture
pound.
Now, Sarah would like to make a 6 pounds nut mixture that is 60% peanuts and 40% almonds.
So, the amount of peanuts in that mixture
pounds
and the amount of almonds in that mixture
pounds.
So, the system of equations will be.........

Subtracting equation (2) from equation (1), we will get.....

Now, plugging this
into equation (1), we will get......

So, the amount of 80/20 mixture is 2 pounds and the amount of 50/50 mixture is 4 pounds.