Co-Factors and Cramers rule

[Madi1316]

I'm having trouble, I keep getting zero for some reason?

Solve using co-factors and cramers rule:
A gift shop is making a mixture of almonds, pecans, and peanuts which cost $3.50 per pound, $4.00 per pound, and $2.00 per pound respectively. The store keeper wants to make 20 pounds of the mix to sell at $2.70 per pound. The number of pounds of peanuts is to be three times the number of pounds of pecans. Find the number of pounds of each to be used in the mixture.

 

[DrPhil]

I'm having trouble, I keep getting zero for some reason?

Solve using co-factors and cramers rule:
A gift shop is making a mixture of almonds, pecans, and peanuts which cost $3.50 per pound, $4.00 per pound, and $2.00 per pound respectively. The store keeper wants to make 20 pounds of the mix to sell at $2.70 per pound. The number of pounds of peanuts is to be three times the number of pounds of pecans. Find the number of pounds of each to be used in the mixture.

We can't tell where you are introducing a spurious zero unless we see your work.

You should be able to express all of the information in the problem as a system of three equations in three unknowns, and then you should write that in matrix form so you can use Cramer's Rule. Please repost and show us what you have done.

 

[soroban]

Hello, Madi1316!

Solve using co-factors and Cramers rule:

A shop is making a mixture of almonds, pecans, and peanuts which cost $3.50/lb, $4/lb., and $2/lb., resp.
The store keeper wants to make 20 pounds of the mix to sell at $2.70 per pound.
The number of pounds of peanuts is to be three times the number of pounds of pecans.
Find the number of pounds of each to be used in the mixture.

The total value of the mixture is: \(\displaystyle 20\times \$2.70 \:=\:\$54\)

Let \(\displaystyle x\) = lbs. of almonds.
Let \(\displaystyle y\) = lbs. of pecans.
Let \(\displaystyle z\) = lbs.of peanuts.

The mixture will be 20 pounds: .\(\displaystyle x + y + z \:=\:20\)

Its value is $54: .\(\displaystyle 3.5x + 4y + 2z \:=\:54 \quad\Rightarrow\quad 7x +8y+ 4z \:=\:108\)

Peanuts = 3 x Pecans: .\(\displaystyle z \:=\:3y \quad\Rightarrow\quad 3y - x \:=\:0\)


We have this system: .\(\displaystyle \left|\begin{array}{ccc|c} 1&1&1&20 \\ 7&8&4&108 \\ 0&3&1&0 \end{array}\right|\)

Is this what you had?