Systems of Equations


Unit Overview

In this unit, students will be able to:

·        Check solutions for a system of equations

·        Solve a system of equations by graphing

·        Identify the number of solutions for a system of equations

·        Use system of equations with word problems

 

Key Concepts

·        Systems of equations

·        Parallel lines

 

Ohio’s Learning Standards

·        8.EE.8 Analyze and solve pairs of simultaneous linear equations graphically.

o   a. Understand that the solution to a pair of linear equations in two variables corresponds to the point(s) of intersection of their graphs, because the point(s) of intersection satisfy both equations simultaneously.

o   b. Use graphs to find or estimate the solution to a pair of two simultaneous linear equations in two variables. Equations should include all three solution types: one solution, no solution, and infinitely many solutions. Solve simple cases by inspection. For example, 3x + 2y = 5 and 3x + 2y = 6 have no solution because 3x + 2y cannot simultaneously be 5 and 6.

o   c. Solve real-world and mathematical problems leading to pairs of linear equations in two variables. For example, given coordinates for two pairs of points, determine whether the line through the first pair of points intersects the line through the second pair. (Limit solutions to those that can be addressed by graphing.)

 

Calculators

·        Here is a link to the Desmos scientific calculator that is provided for the Ohio State Test for 8th Grade Mathematics in the spring.

·        You are strongly encouraged to use the Texas Instruments TI-30XIIS handheld scientific calculator.  It is extremely user friendly.

 

System of Equations

·        A system of equations is a set of two or more linear equations.

 

o   The solution is any ordered pair that makes both equations true.

 

Example A:  Find the solution for this system of equations:

 

                  x + y = 6

                        x – y = 2

 

Each equation has an infinite number of solutions.

 

                                    x + y = 6                               x – y = 2

x

y

 

X

Y

0

6

 

-1

-1

1

5

 

2

0

2

4

 

3

1

3

3

 

4

2

4

2

 

5

3

5

1

 

6

4

6

0

 

7

5

7

-1

 

8

6

 

o   There is just one ordered pair that is a solution for both equations:  (4, 2).  That point is the solution for this system of equations.

 

o   Let’s graph the solutions (points) of both equations.


 

o   Notice that the lines cross at the solution, (4, 2).

           

·        The solution to any system of equations is the point where their graphed lines cross.

 

·        To check a solution, substitute (x, y) into both equations.

 

o   Both equations must produce true statements

o   Let’s check the solution for Example A, (4, 2).

 

                              x + y = 6 →   4 + 2 = 6

                              x – y = 2         4 – 2 = 2

 

 

Example B:  Solve this system of equations:

 

                        2x + 4y = 20

                        y = 3x – 2

 

§  Let’s graph each line by finding 2 points that are solutions, and then drawing a line through those points.

 

 

                     

 

 

§  We can also graph either using the slope-intercept form:  y = mx + b.

 

 


 

      Let’s check the solution, (2, 4) by substituting the coordinates into both equations.

 

                  2x + 4y = 20  →       2 • 2 + 4 • 4 = 20   → 4 + 16 = 20   → TRUE

                        y = 3x – 2                  4 = 3 • 2  – 2    → 4 = 6 – 2  → TRUE

 

 

For further explanation and practice on graphing system of equations:

 



Let’s Practice.

 

 

 

 

 

Parallel lines

·        Parallel lines have no solutions → they never cross, and thus have no points in common.

 

o   Parallel lines have the same slope (and a different y-intercept).  If they are in slope-intercept form, you can quickly identify parallel lines.

 

o   Example C:  Solve this system of equations:

 

                                    y = ½ x + 6

                                    y = ½ x – 2

 

§  Use the slope intercept form to graph each line.

 

y = ½ x + 6 → slope = ½ , y-intercept = 6

y = ½ x – 2 → slope = ½ , y-intercept = -2

 

o   Both equations have a slope of ½ (with different y-intercepts).

 

o   When graphed, you see the parallel lines.  Thus, the answer for this system is no solutions.

 

o   Some parallel lines may not be in slope-intercept form, so you will not be able to identify the equal slopes from the equations.

 

o   Example D:  Solve this system of equations:

 

 

 

o   When graphed, these parallel lines both have a slope of -2.  Thus, the answer for this system is no solutions.

 

Identical lines

·       Identical lines have infinite solutions:  they have every point in common.

 

o  They usually are written in different forms, so the equations do not appear to be identical.

 

o   Example E:  Solve this system of equations:

 

y = 3x – 6       → slope = 3, y-intercept = -6

6x – 2y = 12

      -6x            -6x

       -2y = 12 – 6x

                                           ÷-2     ÷-2      ÷-2     

                                                y = -6 + 3x     → slope = 3, y-intercept = -6

 

 

o  The equations produce identical lines.  Thus, there are infinite solutions.

 

For a further explanation on number of solutions for systems:

 

 

 

Let’s practice.

 

 


 

 

System of Equations Word Problems

·        If a word problem gives you a system of equations, simply graph them and find the point of intersection.  Be sure to answer what the question is asking for.

 

·        If you need to write a system of equations from a word problem:

o   Identify your variables

o   Identify your rate(s) of change,

§  If there is just one rate of change, it is often the slope (m).

§   It is usually multiplied by one or both variables.

o   If you are given a starting value, that is usually your y-intercept (b).

o   If given a slope and y-intercept into the equation, y = mx + b

o   Identify the total, which is usually a number or a variable.

 

Example F:

At “Cory’s Health Club”, you must pay an initial $80 fee, as well as $30 per month.  George’s Gym costs $35 per month with no membership fee. 

 

Part A:     Write a system of equations showing the relationship between the number of months as a member (M) and the total membership cost (T) for both places.  Then find the number of months when the costs are equal.

·        Create your equations using the variables T and  M:

 

o   Corey’s Health Club:    T = 30M + 80 → $30 per month, plus a one-time $80 fee

o   George’s Gym:              T = 35M          → $35 per month, no fee

 

Part B:  At how many months are the costs of the two gyms equal?

            A.  12 months          B.  16 months          C.  20 months          D.  24 months

·        You could graph these, but it is hard to be precise with such large numbers.  If you plug in the answers for both equations, you will find that 16 months produces the same cost:

 

o   Corey’s Health Club:    T = 30M + 80 → T = 30 • 16 + 80 = $560

o   George’s Gym:              T = 35M          → T = 35 • 16 = $560

Example G:

120 tickets were sold for the school play.  Adult tickets (A) cost $8 each.  Student tickets (S) cost $3 each. Total ticket sales were $720.

Part A:  Write a system for this scenario

·        Create your equations using the variables A and S:

o   120 total tickets sold (adults + students):  A + S = 120

o   $720 of tickets sold(adult sales  + students sales):  8A + 3S = 720

 

Part B:  What are the number of tickets sold?

A. 72 adult tickets and 48 student tickets

      B.  80 adult tickets and 20 student tickets

      C.  45 adult tickets and 100 student tickets

      D.  63 adult tickets and 57 student tickets

·        Check the solutions with the equations. 

o   You can quickly eliminate B and C, since they don’t add up to 120 tickets sold

o   Answer A (72 adult tickets and 48 student tickets) is the correct answer

 

§  A + S = 120         →72 + 48 =120            → true

§  8A + 3S = 720    → 8 • 72 + 3 • 48 = 720 → true

 

o   Answer D:  the total cost is wrong → 8 • 63 + 3 • 57 = 675

 

·        You could graph these, but it is hard to be precise with such large numbers.

o   Here is what the graph looks like using the Desmos graphing calculator that you will use in the Algebra I Ohio State Test:

 

 

o   They cross at 72, 48 à 72 adult tickets and 48 student tickets.

 

For a further explanation and practice on system word problems:

 

 

Let’s practice.