Solving two equations with two unknowns is a process that is not complicated but it does require some thought and a little bit of algebra.  If you look around you will find that there are many equations in real life situations.  Once you discover these equations, you just need to apply the numeric values for known constants and assign terms for unknown values.  You would need to have at least two equations which relate your two unknowns to establish constraints, otherwise you may have a situation where any number of values could be solutions to your unknowns.  These numerous solutions will also not necessarily reflect optimal values.  It’s best to use as much information as you can to find the best solutions, that’s basically what this article is about.

Let’s take a look at an example to see how this works.

Say you are a farmer and you grow pumpkins and cantaloupes.  You have a contract to sell a combined 300,000 lbs of pumpkins and cantaloupes to a large retailer.  You can harvest 15,000 lbs of pumpkins per acre and 12,000 lbs of cantaloupes per acre.  Your fixed cost, including labor, fertilizer, pesticides, etc. per acre of pumpkins is \$1,300 and \$900 per acre of cantaloupes. You have determined that if your fixed costs exceed \$24,500 your profits begin to decrease and you lose money. How many acres of pumpkins and how many acres of cantaloupes should you grow to maximize profits?

So looking at our situation, we have several pieces of information to help us determine our most optimal solution, the number of acres of pumpkins and cantaloupes to grow.  We first write down our situation in the form of two mathematical equations.  Let’s define our unknowns, acreage of pumpkins to grow, and acreage of cantaloupes to grow, as P and C respectively.

With a total combined harvest of 300,000 lbs, we can write, (15,000 lb/acre)(P) + (12,000 lb/acre)(C) = 300,000 lb, this is our first equation.

Now we also have the constraint or upper limit of \$24,500 for our fixed costs, so we can write, (\$1,300/acre)(P) + (\$900/acre)(C) = \$24,500.  This is our second equation.

We have two equations and two unknowns, so we will now use substitution to solve.  To do this we will use algebra to write one unknown in terms of the other, then substitute in to the other equation and solve.

(15,000 lb/acre)(P) + (12,000 lb/acre)(C) = 300,000 lb

(15,000 lb/acre)(P) = 300,000 lb – (12,000 lb/acre)(C)

P = [300,000 lb – (12,000 lb/acre)(C)] / (15,000 lb/acre)

Now substitute this equation in to the second.

(\$1,300/acre)(P) + (\$900/acre)(C) = \$24,500

(\$1,300/acre)([300,000 lb - (12,000 lb/acre)(C)] / (15,000 lb/acre)) + (\$900/acre)(C) = \$24,500

We now use algebra and solve.

(1,300/15,000)[300,000 - (12,000)(C)] + (900)(C) = 24,500

26,000 – 1,040(C) + 900(C) = 24,500

26,000 – 140(C) = 24,500

-140(C) = -1,500

C = 10.7 acres, so lets round down to planting 10 acres of cantaloupes

Now we plug this value in to one of our original two equations.

(14,000 lb/acre)(P) + (12,000 lb/acre)(C) = 300,000 lb

(14,000 lb/acre)(P) + (12,000 lb/acre)(10 acres) = 300,000 lb

(14,000 lb/acre)(P) = 300,000 lb – 120,000 lb = 180,000 lb

P = (180,000 lb) / (14,000 lb/acre)

P = 12.9 acres, so we will plant 12 acres of pumpkins

Let’s go back to check that our solutions meet our constraints:

Equation 1:

(15,000 lb/acre)(P) + (12,000 lb/acre)(C) = 300,000 lb

(15,000 lb/acre)(12 acres) + (12,000 lb/acre)(10 acres) = 300,000 lb, Good

Equation 2:

(\$1,300/acre)(P) + (\$900/acre)(C) =\$24,500

(\$1,300/acre)(12 acres) + (\$900/acre)(10 acres) = \$24,600, which is close enough

So our optimal solution is to plant 12 acres of pumpkins and 10 acres of cantaloupes.

Examples like this can be found in all kinds of situations in life.  The ability to recognize these equations with unknowns and translate them in to mathematical equations to then solve and find the optimal solution is an incredibly valuable skill which can save you time, money, and energy.