Applications of Systems of Linear Equations

The relevance of systems of linear equations in various applications.

Systems of linear equations can be used in various subject areas and are used to solve for unknowns. With n equations, the highest possible number of unknowns that can be solved for is n. Complications occur when the number of unknowns is unequal to the number of equations in the system.

Linear systems of equations with more or less equations than unknowns occur quite often. When there are fewer equations than unknowns, it is impossible to solve for all the unknowns algebraically. Because of this, there are usually infinitely many solutions for each variable (it is hard to narrow it down). When there are more equations than unknowns (an overdetermined system), there are usually no solutions. This occurs because there may be solutions that satisfy two of the three equations, but not necessarily all three. The only way there would be solutions to an overdetermined system (for example, one with three equations and two unknowns) would be if all the equations are equivalent, if two of the three equations are equivalent and the other equation intersects both, or if all three equations intersect each other at a single point.


I. Linear Equations can be used to find out the best method of maximizing profits. Let’s say a farmer produces 2 goods, yogurt and ice cream. Each quart of yogurt requires 0.4 quarts of milk and 0.2 quarts of cream, while each quart of ice cream requires 0.2 quarts of milk and 0.4 quarts of cream. The profit gained from a quart of yogurt is 8 cents and the profit gained from a quart of ice cream is 10 cents. If the farmer has 10 quarts of milk and 14 quarts of cream, how much of each good should he make to maximize his profits?

Let y = number of quarts of yogurt and let i = number of quarts of ice-cream. We now have an equation for the total amount of milk required:

0.4y + 0.2i

The equation for cream would look like this:

0.2y + 0.4i

Since there are only 10 quarts of milk and 14 quarts of cream, we have to alter these equations slightly:

0.4y + 0.2i ≤ 10

0.2y + 0.4i ≤ 14

An equation for total profit can also be made, as we know how much profit is obtained from a quart of each good. Let p = profit.

p = 8y + 10i

We know that y and i cannot be negative, because there is no such thing as a negative quart value. Using the two inequalities we obtained, we can create a graph which will give us a region of points which are feasible. This region is a convex polygon, so we must test all the corner points (excluding the origin) to find the maximum profit to be obtained. These points are A(0,35), B(10,30) and C(25,0). The value of the objective function is largest at point B, so we will plug the values of point B into our equations.

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