Respuesta :
Answer:
a. The system of equations that models the situation is....
[tex]0.80x+0.50y=3.60\\\\ 0.20x+0.50y=2.40[/tex]
b. The solution to the system: x = 2 and y = 4
The amount of 80/20 mixture is 2 pounds and the amount of 50/50 mixture is 4 pounds.
Step-by-step explanation:
Suppose, the amount of 80/20 mixture is [tex]x[/tex] pounds and the amount of 50/50 mixture is [tex]y[/tex] pounds.
So, the amount of peanuts in 80/20 mixture [tex]= 0.80x[/tex] pound and the amount of almonds in 80/20 mixture [tex]=0.20x[/tex] pound.
And the amount of peanuts in 50/50 mixture [tex]=0.50y[/tex] pound and the amount of almonds in 50/50 mixture [tex]=0.50y[/tex] 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 [tex]=(6\times 0.60)=3.60[/tex] pounds
and the amount of almonds in that mixture [tex]=(6 \times 0.40)= 2.40[/tex] pounds.
So, the system of equations will be.........
[tex]0.80x+0.50y=3.60 ...................(1)\\\\ 0.20x+0.50y=2.40...................(2)[/tex]
Subtracting equation (2) from equation (1), we will get.....
[tex](0.80x+0.50y)-(0.20x+0.50y)=3.60-2.40\\ \\ 0.60x=1.20\\ \\ x= \frac{1.20}{0.60}=2[/tex]
Now, plugging this [tex]x=2[/tex] into equation (1), we will get......
[tex]0.80(2)+0.50y=3.60\\ \\ 1.60+0.50y=3.60\\ \\ 0.50y=3.60-1.60=2\\ \\ y=\frac{2}{0.50}=4[/tex]
So, the amount of 80/20 mixture is 2 pounds and the amount of 50/50 mixture is 4 pounds.
