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SYSTEMS OF EQUATIONS AND LINEAR EQUALITIES


Unit Overview
A system of equations is a collection of equations in the same variables.  In this unit you will solve systems of equations using 3 techniques; graphing, substitution, and elimination.  At the conclusion of the unit, you will be graphing linear inequalities.


Graphing Systems of Equations

The solution of a system of two linear equations in x and y is any ordered pair, (x, y) that satisfies both equations.  The solution (x, y) is also the point of intersection of the graphs.


There are 3 possible solutions.

To graph a line, remember to solve for y and use the slope intercept form of  y = mx + b.  Plot the y-intercept, use the slope ratio of    to plot more points, and then connect the points using a straight edge. 

Quicktime_Video_Icon  Graphing Systems of Equations -- Freediving (02:59) 

Let's review consistent and inconsistent systems. 

question1What is the classification of this system shown below?

The system is inconsistent.

"Click here" to check the answer.


question1How many solutions does this system have?

There are no solutions.

"Click here" to check the answer.



question1What is the classification of this system shown below?

The system is consistent and independent.

"Click here" to check the answer.


question1How many solutions are there?

There is one solution.

"Click here" to check the answer.



question1What is the classification of the system shown below?  
Note:  The graph shows two lines, one coinciding with the other; thus, one lies atop the other?

The system is consistent and dependent.

"Click here" to check the answer.


question1How many solutions are there?

There are many solutions.

"Click here" to check the answer.


Stop!  Go to Questions #1-5 about this section, then return to continue on to the next section.

Substitution Method for Solving Systems of Equations

To solve a system by substitution:
          1.)  solve one of the equations for a variable (hint: solve for a variable that has a coefficient of 1
          2.)  substitute this value into the other equation to find the value of one of the variables
          3.)  substitute this value back into either of the equations to find the second variable

Let's take a look at this problem again, but this time, solve the first equation for y.

Quicktime_Video_Icon  Solving Systems of Equations with Substitution (04:24) 

Stop!  Go to Questions #6-8 about this section, then return to continue on to the next section.

Elimination Method for Solving Systems of Equations

To solve a system by elimination:

1.)  the coefficients of the same variable must be the same
2.)  if the coefficients are the same, either subtract the equations or add the equations to eliminate that variable, depending on the signs (same sign – subtract, different signs – add)
3.)  substitute this value back into one of the equations to find the other variable

Quicktime_Video_Icon  Solving Systems of Equations:  Two Approaches (02:59) 

Quicktime_Video_Icon  Elimination and the Addition Property (04:33) 

Quicktime_Video_Icon  Elimination and the Multiplication Property of Equality (04:06) 

Sometimes it is necessary to multiply both equations in a system to eliminate one of the variables.  Multiply each equation by values so that one of the variables has the same or opposite coefficient in both equations.

Let's see how this works.

Stop!  Go to Questions #9-13 about this section, then return to continue on to the next section.


Applications of Linear Systems

Quicktime_Video_Icon  Systems of Equations and Algebra -- Scubadiving (02:56)

Quicktime_Video_Icon  Word Problem:  Rate, Time, and Distance (04:12)

Systems of equations are used in business to determine the break-even point.  The break-even point is the point at which income equals the cost.

Example #4:  The Jones family plans to rent a car for a one-day trip.  The car rental company offers two rental plans.  The first plan allows them to rent a car for $42 per day plus 25 cents per mile.  The second plan allows them to rent a car for $90 a day with unlimited mileage.  Write an equation that represents the cost of renting each car.  Find the break-even point of the rental cars.  If the Jones family plans to drive 150 miles, which plan should they choose?

                       Let       m = the number of miles the car is driven

                                    c = the total cost to rent the car for one day

Since the first plan costs $42 plus 25 cents per mile, the first equation is:

            c = 42 + 0.25m

Since the second plan costs $90 a day with unlimited mileage, the second equation is:

             c = 90

Write and solve the system of equations.

             c = 42 + 0.25m
             c = 90

Since both equations represent c, use substitution to solve the system.

                        c = 42 + 0.25m
                        90 = 42 + 0.25m          Substitute c = 90.
                        48 = 0.25m                  Subtract 42 from both sides.
                        192 = m                       Divide both sides by 0.25           

Thus, the cost for the rental plans is the same if they plan to drive 192 miles.  This is the break-even point.

If the Jones family plans to drive less than 192 miles the first plan is cheaper.  If the Jones family plans to drive more than 192 miles the second plan is cheaper.

question1Which plan is cheaper for the Jones family outing?

Rental Plan 1 is cheaper since they are only traveling 150 miles.

"Click here" to check the answer.



Stop!  Go to Questions #14-17 about this section, then return to continue on to the next section.

Linear Inequalities

A solution to a linear inequality in two variables, x and y, is an ordered pair (x, y) that satisfies the inequality.  The solution to a linear inequality is a region of the coordinate plain and is called a half-plane bounded by a boundary line.

          *If a linear inequality is a  <  or  > , the boundary line will be a dashed line. ----------   
          *If a linear inequality is a  ≤  or  ≥ , the boundary line will be a solid line. _______

To graph a linear inequality:
1.)  solve the inequality for y
2.)  plot the y-intercept
3.)  use the slope   to plot more points
4.)  determine whether to connect with a dashed line or a solid line
5.)  shade a region of the coordinate plane (this is determined by either  < ,  > ,  ≤ ,  or  ≥ ).  If  <  or  ≤ , shade below the boundary line;  if  >  or  ≥ , shade above the boundary line. 

You can also choose a point to test, such as (0, 0), as it is easy to test if it does not lie on the line.  If you plug 0 in for x and y and get a true statement, shade the portion of the coordinate plane that contains that point.  If you plug 0 in for x and y and get a false statement, shade the portion of the coordinate plane that does not contain that point.  If the point (0, 0) is on the line, choose another point as the test point.

Example #1:   y  ≤  –x – 3

Since this is a  ≤  inequality, the line is solid.
Shading will occur below the boundary line.

To test the point (0, 0), replace x and y with 0.  Since we chose to shade the region that does not include (0, 0), we should get a false
statement.

      0  ≤  –(0) – 3
      0  ≤  –3

This is a false statement; therefore, we chose
to shade the correct half-plane.



Example #2:   y > 2x – 2

Since this is  > , the boundary line will be dashed.

Shading will occur above the line or up and left.

To test the point (0, 0), replace x and y with 0.
Since we chose to shade the region that
does include (0, 0), we should get a true
statement.

      0  >  2(0) – 2
      0  >  –2

This is a true statement;  therefore, we chose
to shade the correct region, the one that
includes the point (0, 0).

Stop!  Go to Questions #18-25 to complete this unit.




Below are additional educational resources and activities for this unit.
 
Click on the icon to find and practice topics for this unit.
 
Solving Systems of Equations
 
Graphing Systems of Equations