A coffee grower sells three coffee blends. One bag of the house mix contains 300 grams of Colombian grain and 200 grams of French roast grain. One bag of the special blend contains 200 grams of Colombian grain, 200 grams of Kenyan grain and 100 grams of French roasted grain. One bag of gourmet blend contains 100 grams of Colombian grain, 200 grams of Kenyan grain and 200 grams of French roasted grain. The merchant has on hand 30 kilograms of Colombian grain, 15 kilograms of Kenyan grain, and 25 kilograms of French roast grain if he wants to use all the grains. (How many bags of each type of mix can you make?

Solution: Let x, y, and z be the amount of bags of house mix, special blend, and gourmet blend can be made, respectively. For the Colombian grain, the equation is:$$ 0.3x+0.2y+0.1z=30 $$ For the Kenyan grain, the equation is:$$ 0.2y+0.2z=15 $$ For the French roasted grain, the equation is:$$ 0.2x+0.1y+0.2z=25 $$ The problem becomes a system of linear equation in three variables. By isolating z in equation 2:$$ 0.2z=15-0.2y $$$$ z=75-y $$Substituting z to both equation 1 and 3:$$ 0.3x+0.2y+0.1\left(75-y\right)=30 $$$$ 0.3x+0.2y+7.5-0.1y=30 $$$$ 0.3x+0.1y=22.5 $$ $$ 0.2x+0.1y+0.2\left(75-y\right)=25 $$$$ 0.2x+0.1y+15-0.2y=25 $$$$ 0.2x-0.1y=10 $$Adding the two resulting equations:$$ 0.3x+0.1y+0.2x-0.1y=22.5+10 $$$$ 0.5x=32.5 $$$$ x=65 $$Substituting the value to any of the resultant equation (in this solution, use resultant equation 1):$$ 0.3\left(65\right)+0.1y=22.5 $$$$ 19.5+0.1y=22.5 $$$$ 0.1y=3 $$$$ y=30 $$Substituting y to equation 2:$$ 0.2\left(30\right)+0.2z=15 $$$$ 6+0.2z=15 $$$$ 0.2z=9 $$$$ z=45 $$ Answers: 65 bags of house mix, 30 bags of special blend, and 45 bags of gourmet blend.