Sarah would like to make a 6 lb nut mixture that is 60% peanuts and 40% almonds. she has several pounds of a mixture that is 80% peanuts and 20% almonds and several pounds of a mixture that is 50% peanuts and 50% almonds. a. what is the system that models this situation? b. what is the solution to the system? how many pounds of the 80/20 mixture? how many pounds of the 50/50 mixture?

The system of equations is (8x + 5Y = 36) and (2X + 5Y = 24).

The solution is 2 pounds of 80/20 mixture and 4 pounds of 50/50 mixture.

Step-by-step explanation:

Let's assume they use "X" lbs of 80/20 mixture and "Y" lbs of 50/50 mixture to make 6 lbs of 60/40 mixture.

So peanuts would be 80% of X, 50% of Y, and 60% of 6 lbs.

Mathematically, (80% of X) + (50% of Y) = 60% of 6.

80X + 50Y = 60*6

8X + 5Y = 36

Similarly almonds would be 20% of X, 50% of Y, and 40% of 6 lbs.

Mathematically, (20% of X) + (50% of Y) = 40% of 6.

20X + 50Y = 40*6

2X + 5Y = 24

So, the system to model this situation is (8X + 5Y = 36) and (2X + 5Y = 24).

To solve this system, we can subtract second equation from first equation.

(8X + 5Y) - (2X + 5Y) = 36 - 24

6X = 12

X = 2

We can plug this x=2 into any equation to solve for y.

8(2) + 5Y = 36

16 + 5Y = 36

5Y = 36 - 16 = 20

Y = 4

So, they use 2 pounds of 80/20 mixture and 4 pounds of 50/50 mixture.

translate k to n and reflect across the line containing jk.