We are asked to find the amount of three different foodstuffs to prepare in order to meet the requirements of particular vitamins/minerals. The amount of each for each foodstuff is given below:

`([Food I,Food II, Food III, Total],[400,1200,800,6800],[110,570,340,2590],[90,30,60,810])`

We can set up a system of linear equations. As suggested, let...

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We are asked to find the amount of three different foodstuffs to prepare in order to meet the requirements of particular vitamins/minerals. The amount of each for each foodstuff is given below:

`([Food I,Food II, Food III, Total],[400,1200,800,6800],[110,570,340,2590],[90,30,60,810])`

We can set up a system of linear equations. As suggested, let x be the number of ounces of Food I, y the oz of Food II, and z the oz of Food III.

The amount of vitamin A we get from Food I is 400x. (For each ounce of I we get 400 units of vitamin A.) Likewise the amount from II is 1200y and the amount from III is 800z. The total is the sum from each food type and must equal 6800 so:

400x+1200y+800z=6800

In a similar fashion we find that for vitamin C we can write:

110x+570y+340z=2590

and for calcium we get:

90x+30y+60z=810

Thus the system of three equations in three unknowns is:

400x+1200y+800z=6800

110x+570y+340z=2590

90x+30y+60z=810

There are three possibilities. If the system is consistent and independent then there is only one solution, i.e. only one combination of the the food types to satisfy the equations. If the system is inconsistent then there is no solution to the system and no combinations of food types satisfies the conditions. If the system is consistent and dependent, there will be many ways to solve the system.

We can solve the system using linear combinations, substitution, putting the augmented matrix in reduced row echelon form, using inverse matrices, or using Cramer's method among other methods.

Using Gauss-Jordan reduction we put the augmented matrix in reduced row echelon form:

`([400,1200,800,6800],[110,570,340,2590],[90,30,60,810])` ==> `([1,0,.5,8],[0,1,.5,3],[0,0,0,0])`

The last row being all zero indicates that the system is consistent and dependent. Rewriting as a set of equations we get:

`x+1/2z=8`

`y+1/2z=3`

**We can write the solution in terms of z as `(8-1/2z,3-1/2z,z)` .**

For example, if we do not include food III we use 8oz of I and 3oz of II. (Check that we get 8(400)+3(1200)=6800 units vit A, 8(110)+3(570)=2590 units of vit C, and 8(90)+3(30)=810 units of calcium.)

We could also use 5 units of z, etc...

z has an upper limit of 6 since the amount of food II should not be negative.

**Further Reading**