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.

Economics

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.

P = 8(10) + 10(30)

P = $3.80

The farmer’s optimal profit would be $3.80 a day, using his materials to create 10 quarts of yogurt and 30 quarts of ice-cream.

**Profit maximization is very important for an economist. This is an invaluable source of revenue for an economist as a good plan will make much more money than a bad one. It is important to use linear equations in order to find out the best way to make money.

II. Linear equations can be used to find the equilibrium price and quantity to be supplied in a given market. Let’s say there are two products, orange juice and water, which are interrelated. Let p1 and q1 represent the price and quantity demanded respectively for product 1 (orange juice) and p2 and q2 represent the same for product 2 (water).

Demand Supply

Product 1: p1 = 2000 – 3q1 – 2q2 q1 = 100 + 2q1 + q2

Product 2: p2 = 2800 – q1 – 4q2 q2 = 200 + 3q1 + 2q2

For equilibrium to be achieved, both price expressions must be equal, so the following equations are obtained:

Product 1: 2000 – 3q1 – 2q2 = 100 + 2q1 + q2

Product 2: 2800 – q1 – 4q2 = 200 + 3q1 + 2q2

Simplifying these expressions gives us the equations

5q1 + 3q2 = 1900

4q1 + 6q2 = 2600

Elimination can now be utilized.

4(5q1 + 3q2 = 1900)

-

5(4q1 + 6q2 = 2600)

=

-18q2 = -5400, which is equal to q2 = 300. Plugging this value into the equation 4q1 + 6q2 = 2600 gives us 4q1 + 1800 = 2600, which can be simplified to q1 = 200.

By plugging these values (of the quantities) into the original demand equations, we find that the equilibrium price for product 1 is $800 and the equilibrium price for product 2 is $1400.

p1 = $800

q1 = 200 units

p2 = $1400

q2 = 300 units

**Finding the equilibrium price and quantity in a market is very important to an economist since beyond these values, nothing will be sold at a profit. Linear systems are used to find these crucial numbers.

Biology

Systems of linear equations can also be used in biology. Here is one example in which systems of linear equations can be used in this subject.

I. Let’s say there are 200 insects that an experiment is to be performed on. 18 adult insects were removed so that the ratio of juvenile insects to adult insects would be 3 : 4. The number of juveniles and adults must be found.

Let j = number of juvenile insects and a = number of adult insects.

We know that the total number of insects is equal to 200. This gives us the following equation:

j + a = 200

We also know that after 18 adult insects were taken out, there were three juvenile insects to every four adult insects. This can be represented by the following equation:

4j = 3(a – 18)

Rewriting the first equation, we can solve for j in terms of a. Subtract a from both sides of the equation to get this new one:

j = 200 – a

By plugging this in to the other equation, we obtain this:

4(200 – a) = 3(a – 18)

This can be simplified to 800 – 4a = 3a – 54, which can be further simplified to 800 = 7a – 54, 854 = 7a, a = 122. By plugging this value for “a” into the first equation (j + a = 200), we obtain this new equation: j + 122 = 200. We can now solve for j, which is equal to 78.

We can check this answer by subtracting 18 from 122 and obtaining 104. 78/104 simplifies to 3/4, so we know our answer is correct.

**Biologists perform many experiments on specimens while leaving the control group unaffected. Linear systems of equations come in handy in such simple tasks as finding out how many specimens must be experimented on.

Chemistry

Systems of linear equations can also be used in chemistry to balance chemical equations. Here is a simple example in which they are used for this purpose.

I. H2 + O2 → H2O

A variable has to be assigned to each molecule/compound in the equation.

aH2 + bO2 → cH2O

We have to create an equation for each different element in the equation. For Hydrogen, there are 2 atoms on both sides. When written with the assigned variables, this looks like 1a + 0b = 1c, or a = c.

Another equation must be written for the Oxygen in the compound. The oxygen is not balanced as written in the equation, so we must find the greatest common factor, which happens to be two. When we write the equation for Oxygen, it looks like this:

0a + 2b = c, or 2b = c.

We now have two equations, a = c and 2b = c. We can write this into one equation: a = 2b = c. We now know that the coefficient for the molecule O2 is equal to half of the coefficients of the other two molecules.

Every coefficient value in a chemical equation must be a positive integer. We know that 2b must be positive, and the lowest value that b can be is 1. 2 times 1 is equal to 2, so we know that “a” and “c” are equal to 2.

When we substitute the values for the variables, we get the balanced equation:

2H2 + O2 → 2H2O

**Chemists must balance every chemical reaction they perform in order to be able to obtain information about the reaction (such as percent yield, mole ratios, and amounts of reactant formed). Many chemical equations are difficult to solve, and this method makes the task much easier.

Physics

Systems of linear equations and matrices can be used to find missing currents with given voltages.

The currents I1, I2, and I3 (in amperes) are given by the solution of this system of linear equations.

2I1 + 4I3 = E1

I2 + 4I3 = E2

I1 + I2 – I3 = 0

Where E1 and E2 are voltages. Assuming E1 and E2 are equal to 14 and 28 volts respectively, the unknown currents, I1, I2, and I3, can be found.

This system of equations can be represented by the matrices shown:

A * X = B

2 0 4 I1 14

0 1 4 I2 = 28

1 1 -1 I3 0

Using the inverse of the coefficient matrix (matrix A), the solutions can be found as shown:

I1 5/14 -4/14 -4/14 14

I2 = -4/14 3/7 -4/7 28

I3 1/14 1/7 1/7 0

Solving for unknowns, this is:

I1 -3

I2 = 8

I3 5

**Many a times, the field of Physics is concerned with unknown currents. These currents must be identified so that one can determine the net charges they carry as well as the corresponding resistance therein.

• Twice as many Hydrogen molecules as
• Oxygen molecules combine to form water.
• Three systems with infinite solutions