You have one type of nut that sells for $2.80/lb and another type of nut that sells for $9.60/lb. you would like to have 20.4 lbs of a nut mixture that sells for $6.60/lb. how much of each nut will you need to obtain the desired mixture?

11.396Ibs of nuts that cost $9.60/Ib and 9.014Ibs that cost $2.80/Ib

Step-by-step explanation:

First we find the cost of the supposed mixture we are to get by selling it $6.60/Ib which weighs 20.41Ibs

Which is 6.6 x 20.41 = $134.64

Now we label the amount of mixture we want to get with x and y

x = amount of nuts that cost $2.8/Ib

y = amount of nuts that cost $9.6/Ib

Now we know the amount of mixture needed is 20.41Ibs

So x + y = 20.41Ibs

And then since the price of the mixture to be gotten overall is $134.64

We develop an equation with x and y for that same amount

We know the first type of nut is $2.8/Ib

So for x amount we have 2.8x

For the second type of nut that is $9.6/Ib

For y amount we have 9.6y

So adding these to equate to $134.64

2.8x + 9.6y = 134.64

So we have two simultaneous equations

x + y = 20.41 (1)

2.8x + 9.6y = 57.148 (2)

We can solve either using elimination or factorization method

I'm using elimination method

Multiplying the first equation by 2.8 so that the coefficient of x for both equations will be the same

2.8x + 2.8y = 57.148

2.8x + 9.6y = 134.64

Subtracting both equations

-6.8y = -77.492

Dividing both sides by -6.8

y = -77.492/-6.8 =11.396

y = 11.396Ibs which is the amount of nuts that cost $9.6/Ib

Putting y = 11.396 in (1)

x + y = 20.41 (1)

x +11.396 = 20.41

Subtract 11.396 from both sides

x +11.396-11.396 = 20.41-11.396

x = 9.014Ibs which is the amount of nuts that cost $2.8/Ibs

10 you would take 50/ .2

step-by-step explanation: