Question

Sarah would like to make 5 lb. nut mixture that is 60% walnuts and 40% cashews. She has several pounds of walnuts and several pounds of a mixture that is 20% walnuts and 80% cashews. Let w represent the number of pounds of walnuts needed to make the new mixture, and let m represent the number of pounds of the 80% cashew - 20% walnut mixture.

a) Write a system of linear equations that models this situation?
b) Which of the following is a solution to the system: 2 lb walnuts and 3 lb mixture OR 2.5 lb walnuts and 2.5 lb mixture? Show your work.

Answer 1

a) w+c = 5 ____equation(1) and w = c ____equation (2)

b) 2.5 lb walnuts and 2.5 lb cashews

Explanation:

Let w pounds be the weight of walnuts required and c pounds be the weight of cashews required to make the new mixture.

Total weight of the new mixture = 5 lb

So,

Weight of walnuts + Weight of cashews = Total weight of the new mixture

w+c = 5 ____equation (1)

Now,

60% of walnuts +40% of cashews = 20% of walnuts + 80% of cashews

0.60w+0.40c = 0.20w+0.80c

Subtracting 0.20w from both the sides of the equation, we get

0.60w+0.40c-0.20w = 0.20w+0.80c-0.20w

Cancelling out the 0.20w and -0.20w from the right side, we have

0.60w-0.20w+0.40c = 0.80c

=> 0.40w+0.40c=0.80c

Subtracting 0.40c from both sides, we get

0.40w+0.40c-0.40c=0.80c-0.40c

Cancelling out 0.40c and -0.40 c form the left side, we get

0.40w = 0.40c

Dividing both sides by 0.40, we have

`(0.40w)/(0.40) = (0.40c)/(0.40)`

Cancelling out the 0.40's from the top and bottom, we get

w = c ____equation (2)

Plugging in w=c into the equation 1, we get

w+c = 5

=> c+c =5

=> 2c = 5

Dividing both sides by 2, we get

`(2c)/(2) = (5)/(2)`

Cancelling out the 2's from the left, we get

c = 2.5

Plugging in c=2.5 into the equation 1, we get

w+c = 5

=> w + 2.5 = 5

Subtracting 2.5 from both sides, we get

w+ 2.5 -2.5 = 5 - 2.5

Cancelling out the +2.5 and -2.5 from the left side, we get

w = 2.5

So, we need 2.5 lb walnuts and 2.5 lb cashews to make the new mixture.