# Solving a System of 2 Linear Equations by F6yLEoY

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• pg 1
Utilizing 6 Methods:
•Matrices
•Graphing – Intercepts
•Graphing – Functions (y = )
•Linear Combination
•Substitution – One Equation into the Other
•Substitution – Equations Equal to Each Other
Definition
 A system of 2 linear equations is: 2 linear equations,
both of which contain the same 2 variables.

 Example:
5x + 4y = 60     (Equation #1)
x + 2y = 18     (Equation #2)
Real-World Example
Mr. K goes to the concession stand at a baseball game.
He buys 5 cheesesteaks and 4 large sodas and pays a
total of 60 dollars. Later in the game, Mr. S goes to the
concession stand. Mr. S buys 1 cheesesteak and 2 large
sodas and pays a total of 18 dollars.
What is the cost of one cheesesteak, and the cost of one
large soda?
Solution
 To find a solution to a system of 2 linear equations
means to find the point(s) in common (if possible)
between the two equations.
 In our example, this means to find the cost of one
cheesesteak and the cost of one large soda.

 NOTE: Sometimes a solution does not exist, or there
are an infinite number of solutions.
Setting Up the System
To set-up a system, do the following:
1. Identify the variables (unknowns) in the problem.
   Example:         Let c = Cost of one Cheesesteak
Let s = Cost of one large Soda
2. Write 2 separate equations, using the variables above,
to model the situation.
    Example: Each trip to the concession stand is an
equation:
Mr. K’s trip:        5c + 4s = 60
Mr S’s trip:          c + 2s = 18
Solving the System
 This tutorial will use 6 different methods to solve a
system of 2 linear equations:
1.   Matrices
2.   Graphing – Intercepts
3.   Graphing – Functions (y =)
4.   Linear Combination
5.   Substitution – One Equation into the Other
6.   Substitution – Equations Equal to Each Other
Solving the System
Method 1: Matrices
 Using the system of equations:
5c + 4s = 60
c + 2s = 18
 Re-write as a matrix equation:

5 4  c         60 
1 2          18 
     s         
Solving the System
Method 1: Matrices (Continued)
 Solve the system using inverse matrices:
5
4       c          60 
1               
2
s 
 
18 
 
 5 4  1  5 4       c          5 4  1  60 
1         1 2                   1 2      18 
   2         
s 
                    
c         8 
s        5 
           
Therefore, cost of one cheesesteak is c = \$8,
and the cost of one large soda is s = \$5.
Solving the System
Method 2: Graphing - Intercepts
 For each equation:
 Find the x-intercept: Point (x,0)
 Find the y-intercept: Point (0,y)

 Let x = the cost of a cheesesteak
 Let y = the cost of a large soda
 Then our system becomes

5x + 4y = 60
x + 2y = 18
Solving the System
Method 2: Graphing – Intercepts
5x + 4y = 60 (Equation 1)
x + 2y = 18 (Equation 2)
5x + 4y = 60 (Equation           x + 2y = 18 (Equation 2)
1)
X-intercept:
X-intercept:                               x + 2(0) =     18
5x + 4(0) =      60                 x       =      18  Pt.
(18,0)
5x       =       60
5               5

x = 12  Pt.         Y-intercept:
(12,0)
(0) + 2y =     18
Y-intercept:                                       2y =   18
5(0) + 4y =   60                            2      2
4y = 60                            y=    9  Pt. (0,
9)
Solving the System
Method 2: Graphing – Intercepts (Cont.)
Step 1: Graph Equation 1 – Using the intercepts
Solving the System
Method 2: Graphing – Intercepts (Cont.)
Step 2: Graph Equation 2 – Using the intercepts
Solving the System
Method 2: Graphing – Intercepts (Cont.)
Step 2: Graph Equation 2 – Using the intercepts
Solving the System
Method 2: Graphing – Intercepts (Cont.)
Step 3: Find the intersection - the solution.
Solving the System
Method 2: Graphing – Intercepts (Cont.)
Step 3: Find the intersection - the solution.
Solving the System
Method 2: Graphing – Intercepts (Cont.)

The intersection point (8,5) is the solution to the system
of equations.

8, the x-coordinate, is the cost of a cheesesteak = \$8
5, the y-coordinate, is the cost of a large soda = \$5
Solving the System
Method 3: Graphing - Functions (y =)
 Let x = the cost of a cheesesteak
 Let y = the cost of a large soda
 Then our system becomes

5x + 4y = 60
x + 2y = 18

 Now, we must solve each equation for “y”
Solving the System
Method 3: Graphing - Functions (y =)
5x + 4y = 60 (Equation 1)   x + 2y = 18 (Equation 2)
-5x       -5x               -x          -x
4y = 60 – 5x                2y = 18 - x
4      4                        2        2

60 5                       18 x
y    x                   y   
4 4                         2 2

y  15  1.25x             y  9  0.5x
Solving the System
Method 3: Graphing - Functions (y =)
Type each equation into “y =“ in calculator
y1  15  1.25x
y2  9  0.5x
Set an appropriate “WINDOW”
For example: Xmin = -5, Xmax = 20, Xscl = 5
Ymin = -5, Ymax = 20, Yscl = 5, Xres =
1
Press “GRAPH”

Find the intersection point
Solving the System
Method 3: Graphing - Functions (y =)
Using the Window above, graph should look like:
Solving the System
Method 3: Graphing - Functions (y =)
To find the intersection point:
 Press “2nd” “TRACE”  “5:intersect” “ENTER”
 Press “ENTER” – 3 more times
 Solution is:
 Intersection:
 X=8                      Y=5
 The ordered pair (8,5)
8, the x-coordinate, is the cost of a cheesesteak = \$8
5, the y-coordinate, is the cost of a large soda = \$5
Solving the System
Method 4: Linear Combination
 From the original system of equations:
5c + 4s = 60 (equation 1)
c + 2s = 18 (equation 2)
 Linear Combination is the process of eliminating one
of the variables. This is done by multiplying one (or
both) of the equations by a number, so that when
adding the equations together, one variable is canceled
out.
 Example: 2x + (-2x) = 0
Solving the System
Method 4: Linear Combination
 In our equations: we can cancel the +4s in equation 1,
by multiplying equation 2 by (-2).
5c + 4s = 60        5c + 4s = 60
-2 (c + 2s = 18)      -2c +-4s = -36
(Solve for c) 3c        = 24
3         3
c = 8
(Cost of a
cheesesteak)

Now use this to solve for the other variable.
Solving the System
Method 4: Linear Combination
You can substitute c = 8 into either of the original equations, and
solve for the other variable.
Equation 1                             Equation 2
5c + 4s = 60                              c + 2s = 18
5(8) + 4s = 60                            8 + 2s = 18
40 + 4s = 60                      -8         -8
-40          -40                      2s = 10
4s = 20                                  2 2
4     4                                 s=5
s=5
So, s, the cost of a soda = \$5. And, the solution of:
c = 8 and s = 5, satisfies both original equations.
Solving the System
Method 5: Substitution – One Equation into
the Other
 From the original system of equations:
5c + 4s = 60 (equation 1)
c + 2s = 18 (equation 2)
 By solving one equation for one of the variables, and
substituting into the other equation, the system
becomes one equation with one variable, since there is
a one common point (c,s) that satisfies both equations.
 So, we need to solve one of our equations for c (or s)
and then substitute the result into the other equation.
Solving the System
Method 5: Substitution – One Equation into
the Other
 Solving equation 2 for “c”:
c + 2s = 18
- 2s    - 2s
c       = 18 – 2s   (NOTE: This is NOT “16s”)

 Now, substitute this new expression for c into equation
1 for c:
5c + 4s = 60
5(18 – 2s) + 4s = 60
Solving the System
Method 5: Substitution – One Equation into
the Other
 Now, solve this equation for s:
5(18 – 2s) + 4s = 60
90 – 10s + 4s = 60
90 – 6s = 60
-90       - 90
- 6s = - 30
-6    -6
s=5
Solving the System
Method 5: Substitution – One Equation into
the Other
 Now, substitute s = 5 into either of the original equations and solve for
c.
5c + 4s = 60      (equation 1)
5c + 4(5) = 60
5c + 20 = 60
- 20 - 20
5c      = 40
5             5
c=8
So, s, the cost of a soda = \$5. And, c, the cost of a
cheesesteak = \$8.
Solving the System
Method 6: Substitution – Equations Equal to
Each Other
 From the original system of equations:
5c + 4s = 60 (equation 1)
c + 2s = 18 (equation 2)
 By solving each equation for the same variable, and
setting the expressions equal to each other, the system
becomes one equation with one variable, since there is
a one common point (c,s) that satisfies both equations.
 So, we need to solve both of our equations for c (or s),
and then set them equal to each other.
Solving the System
Method 6: Substitution – Equations Equal to
Each Other
5c + 4s = 60 (equation 1)      c + 2s = 18 (equation 2)
Solving for c:                Solving for c:
5c + 4s = 60                  c + 2s = 18
- 4s     - 4s                - 2s     - 2s
5c      = 60 – 4s             c      = 18 – 2s
5             5
c = 60 - 4s
5     5
OR c = 12 – 0.8s
Solving the System
Method 6: Substitution – Equations Equal to
Each Other
Now, set the two expression for c equal to each other and
solve for s:
12 – 0.8s = 18 – 2s
+ 0.8s        + 0.8s
12         = 18 – 1.2s
-18           -18
-6          = -1.2s
-1.2           -1.2
5=s
Solving the System
Method 6: Substitution – Equations Equal to
Each Other
Now substitute s = 5 into either of the expressions found
previously for c.
c = 18 – 2s
c = 18 – 2(5)
c = 18 – 10
c=8

So, s, the cost of a soda = \$5. And, c, the cost of a
cheesesteak = \$8.
Verifying the Solution to the System
The “Check-Step”
All of the methods of solving the system resulted in the
same solution: s = 5, c = 8.
To verify this solution satisfies both original equations,
substitute the solution values into the original
equations.
Equation 1:                 Equation 2:
5c + 4s = 60                c + 2s = 18
5(8) + 4(5) = 60                   8 + 2(5) = 18
40 + 20 = 60                8 + 10 = 18
60 = 60 ✓                          18 = 18 ✓
IMPORTANT
 In the example given in this tutorial, one solution
exists. However, not all systems of 2 linear equations
have exactly one solution.
 It is possible to have either:
1.   An infinite number of solutions, OR
2.   No solution
 If a system has an infinite number of solutions, this
means that the equations are actually the same
equation (the same line if graphing), just written in
different forms.
 If a system has no solutions, this means that the
equations have the same slopes, but different y-
intercepts (parallel lines if graphing).
IMPORTANT
 To determine if one solution exists, write each
equation in Standard Form: Ax + By = C.
 For a system of 2 linear equations, this would look like:
Ax + By = C (equation 1)
Dx + Ey = F (equation 2)
 If the system has infinite solutions, then:
A B C
 
D E F
 If the system has no solution, then:
A B C
 
D E F
 Otherwise, the system has ONE solution.
Practice Problems
Solve each of the following systems of linear equations
by the method indicated. First, make sure the system
does, in fact, have one solution. When finished with
each problem, be sure to verify the solution found.

1.) Method: Matrices
2x + 3y = 2
3x – 4y = -14
Practice Problems
2.) Method: Graphing - Intercepts
2x + 9y = 36
2x – y = 16

3.) Method: Graphing – Functions (y=)
5x – 6y = 48
2x + 5y = -3
Practice Problems
4.) Method: Linear Combination
4x – 3y = 17
5x + 4y = 60

5.) Method: Substitution – One Equation into the Other
8x – 9y = 19
4x + y = -7
Practice Problems
6.) Method: Substitution – Equations Equal to Each
Other
4x – y = 6
3x + 2y = 21

7.) Method: You Choose
8x – 4y = 23
4x – 2y = -17
Practice Problems
8.) Method: You Choose
-2x + 5y = 9
y = 13 - x

9.) Method: You Choose
5x + 3y = 12
15x + 9y = 36
Practice Problems
10.) Method: You Choose
Suppose that the promotions manager of a minor
league baseball team decided to have a giveaway of
tote bags and t-shirts to the first 150 fans present. The
team owner agrees to a budget of \$1350 for the
products to be given away. One bag costs \$10 and one
t-shirt costs \$7. How many bags and how many t-
shirts should be given away?

Set up the system of 2 linear equations and solve.

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