Solving Systems Algebraically

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Solving Systems of Equations Algebraically

Alternative Methods to Solving Linear Systems

In real-life applications, systems of equations will oftentimes have solutions that are not exact (integers).  Consider the rental car industry, for example.

Example 1:

Which is the better value when renting a vehicle?

Rent-A-Hunk o' Junk charges $29.95 per day and 43˘ per mile.
Tom's Total Wrecks charges $45 per day plus 32˘ per mile.

If you knew when the costs were the same, you could easily determine the better value for your individual needs.

The Solution Process:

First, let's choose some meaningful variables to describe our situation.

Let m= the total miles to be driven.
Let c = the total rental cost for each company.

Then write a system of two equations using the two unknown variables.

c = 29.95 + 0.43m

Cost for Rent-A-Hunk o' Junk
c = 45 + 0.32m

Cost for Tom's Total Wrecks

By using the Substitution Method the system can be solved algebraically.

Since both equations are names for cost, c, we can set them equal to each other.  This will let us solve for miles, m. 

 

Solving for m will tell us where the cost will be the same:


Therefore, the cost is the same for both rental companies if you plan to drive somewhere around 137 miles.

For distances under 137 miles Rent A Hunk o' Junk is cheaper.
For distances over 137 miles Tom's Total Wrecks is cheaper.

The Substitution Method as it applies to the Learning Process:

Before students are ready to take on practical applications, they must first practice the skill. Systems that would could be solved by the substitution method are of the type shown below:

This system is easily solved by substitution because one equation has the y-term isolated ready to be "substituted into the other equation."



This system could be solved by substitution by first solving the second equation for "a", then "substituting" that expression back into the first equation for "a."

 



This system is similar to the real-world application at the beginning.  Both equations represent a name for "m."  Therefore, simply set the equations equal to each other and solve for "n."
Example 2:   

    We will use the second example above to show the procedure.

  
  
Find the solution set to the system given.

 
 
Solve the  second equation for "a."

   
 
"Substitute" the new name for variable "a" into the first equation.
Solve for "b."

   

"Substitute" the value for "b" into the second equation, to determine the value for 'a."
    (4, -2)
Therefore, the system's solution is at the coordinates (4, -2).  This would graphically mean the two linear equations intersect at this point in the coordinate plane.

 

In summary, 

bulletif a variable in one equation of a system of equations is alone on one side of the equation, you can substitute for that variable in the other equation.
bulletSometimes, neither equation has a variable alone on one side.  We can solve one equation for one of the variables and proceed as above.

 

The Elimination Method as it applies to the Learning Process:

Another method for solving systems of equations is called the elimination or addition method.  It is most useful when both equations are written in standard form Ax + By = C. (Chapter 2, Section 2).

 

Example 3:
Solve the system.

Draw a line under the equations, and add.
The result is an equation with only one variable, which we can solve. Thus, x = 5 gives the x-value of the solution to the given system.


To find the y-value, substitute 5 for x in either of the two equations of the system.

The solution to the system is the ordered pair shown at right.  This solution would be difficult to see visually on a graph.

 

Using the Addition Property and Multiplication:

Example 4:
Solve the system.

If we add the equations, no variables will be eliminated.  However, if the -y in the first equation were -2y, the y-terms would be additive inverses and would be eliminated when added.  
We can use multiplication to multiply every member of the first equation  by 2 in order to "create" inverses.


Rewrite the system by replacing the first equation with our new equation from the step above.

Now add the equations and proceed as before.

To find the y-value, substitute 3 for x in either of the two equations of the system.


The solution to the system is the ordered pair shown at right.  
(3,1)

 

In summary,

bulletif a system of equations is written in standard form, you can eliminate one of the variables by using the multiplication property of equality to create additive inverses for one variable.
bulletRewrite the system of equations to show at least one variable with additive inverses.  Then add the system together, you will then have one equation with one variable only, known as a linear combination of the original equations.
bulletSolve for the remaining variable, then substitute this value back into one of the original equations to find the missing coordinate for the second variable.

Some Special Cases:

Example 5:
Solve the system.
Multiply the second equation in the system by -2 to create additive inverses on the x-values.

Rewrite the system using equation 1 and your new equation above.

Add the system together to create a linear combination of the system.
Notice that all variables are inverses of each other and results in the left side of the equation becoming zero while the right side's sum is 16.


System has no solution.
Lines are parallel.

 

Example 6:
Solve the system.

Multiply the second equation in the system by -2 to create additive inverses on the x-values.

Rewrite the system using equation 1 and your new equation above.
Add the system together to create a linear combination of the system.
Notice that all variables are inverses of each other and results in the left side of the equation becoming zero while the right side's sum is 16.

System has infinitely many solutions.
Lines coincide.