# Gauss-Jordan help

#### tony77

##### New member
Hi all, I could use some help on setting up and solving (mainly setting up) two word problems involving Gauss-Jordan for linear Algebra.

1. "a coffee dealer has 3 kinds of beans costing $3.00,$3.75, and $4.00 per pound. If the dealer wants 100 ln of a blend that costs$3.60 per lb, what amounts of the 3 beans can they use?"
2. UC browser SHAREit Appvn
2) "A company has on hand 60 lb of chemical A and 50 lb of chemical B that it wants to use up. Each unit of product 1 uses 3 lb of A and 2 lb of B, which each unit of product 2 uses 1 lb of A and 1 lb of B. Can the company use up all of A and B in making the two products, if so how many units of each should it make? (# of each units of each product need not be an integer)"

Any help will be appreciated!

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#### MarkFL

##### La Villa Strangiato
Math Helper
1.) Let's let $$x$$ be the amount of the $3.00 per pound bean, $$y$$ be the amount of the$3.75 per pound bean and $$z$$ be the amount of the $4.00 per pound bean used to make the blend. The dealer wants to have 100 pounds of the blaned, so we have: $$\displaystyle x+y+z=100$$ Because the dealer wants the blend to cost$3.60/pound, we have:

$$\displaystyle 3.00x+3.75y+4.00z=100\cdot3.60$$

Or:

$$\displaystyle 12x+15y+16z=1440$$

Because we have 3 variables, and only two constraints, we can let $$z=t$$, and set up our augmented matrix as follows:

$$\displaystyle \left[\begin{array}{ccc|c}1 & 1 & 1 & 100 \\ 12 & 15 & 16 & 1440 \\ 0 & 0 & 1 & t \end{array}\right]$$

The first operation I would perform is $$\displaystyle R_2-12R_1$$ to get:

$$\displaystyle \left[\begin{array}{ccc|c}1 & 1 & 1 & 100 \\ 0 & 3 & 4 & 240 \\ 0 & 0 & 1 & t \end{array}\right]$$

Next $$\displaystyle \frac{1}{3}R_2$$

$$\displaystyle \left[\begin{array}{ccc|c}1 & 1 & 1 & 100 \\ 0 & 1 & \dfrac{4}{3} & 80 \\ 0 & 0 & 1 & t \end{array}\right]$$

$$\displaystyle R_1-R_2$$

$$\displaystyle \left[\begin{array}{ccc|c}1 & 0 & -\dfrac{1}{3} & 20 \\ 0 & 1 & \dfrac{4}{3} & 80 \\ 0 & 0 & 1 & t \end{array}\right]$$

Finally $$\displaystyle R_1+\frac{1}{3}R_3$$ and $$\displaystyle R_2-\frac{4}{3}R_3$$

$$\displaystyle \left[\begin{array}{ccc|c}1 & 0 & 0 & 20+\dfrac{1}{3}t \\ 0 & 1 & 0 & 80-\dfrac{4}{3}t \\ 0 & 0 & 1 & t \end{array}\right]$$

Since no amount can be negative, we require:

$$\displaystyle 0\le t$$

$$\displaystyle 0\le80-\frac{4}{3}t\implies t\le60$$

And so we have:

$$\displaystyle (x,y,z)=\left(\frac{60+t}{3},\frac{4(60-t)}{3},t\right)$$ where $$0\le t\le60$$

#### Jason

##### Well-known member
HA. And @MarkFL said he didn't take linear algebra?

#### MarkFL

##### La Villa Strangiato
Math Helper
Gaussian elimination (aka the Gauss-Jordan method) is something I learned in College Algebra, and then again in Precalculus. Learning to work with a parameter is something I picked up along the way on the forums.