DefinitionA 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 ExampleMr. 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?

SolutionTo 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 SystemTo 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 = 60Mr 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. Matrices2. Graphing – Intercepts3. Graphing – Functions (y =)4. Linear Combination5. Substitution – One Equation into the Other6. Substitution – Equations Equal to Each Other

Solving the SystemMethod 1: MatricesUsing the system of equations:

5c + 4s = 60 c + 2s = 18

Re-write as a matrix equation:

cs⎡⎣⎢⎤⎦⎥

51⎡⎣⎢

42⎤⎦⎥ =

6018⎡⎣⎢

⎤⎦⎥

Solving the SystemMethod 1: Matrices (Continued)Solve the system using inverse matrices:

Therefore, cost of one cheesesteak is c = $8,and the cost of one large soda is s = $5.

cs⎡⎣⎢⎤⎦⎥

51⎡⎣⎢

42⎤⎦⎥

= 6018⎡⎣⎢

⎤⎦⎥

cs⎡⎣⎢⎤⎦⎥

51⎡⎣⎢

42⎤⎦⎥

= 6018⎡⎣⎢

⎤⎦⎥

51⎡⎣⎢

42⎤⎦⎥

51⎡⎣⎢

42⎤⎦⎥

−1 −1

cs⎡⎣⎢⎤⎦⎥ = 8

5⎡⎣⎢⎤⎦⎥

Solving the SystemMethod 2: Graphing - InterceptsFor each equation:

Find the x-intercept: Point (x,0)Find the y-intercept: Point (0,y)

Let x = the cost of a cheesesteakLet y = the cost of a large sodaThen our system becomes

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

Solving the SystemMethod 2: Graphing – Intercepts

5x + 4y = 60 (Equation 1)

X-intercept: 5x + 4(0) = 60 5x = 60 5 5 x = 12 Pt. (12,0)

Y-intercept:5(0) + 4y = 60 4y = 60 4 4y = 15 Pt. (0,15)

x + 2y = 18 (Equation 2)

X-intercept:x + 2(0) = 18x = 18 Pt. (18,0)

Y-intercept:(0) + 2y = 182y = 18 2 2 y = 9 Pt. (0, 9)

5x + 4y = 60 (Equation 1) x + 2y = 18 (Equation 2)

Solving the SystemMethod 2: Graphing – Intercepts (Cont.)

Step 1: Graph Equation 1 – Using the intercepts

Solving the SystemMethod 2: Graphing – Intercepts (Cont.)

Step 2: Graph Equation 2 – Using the intercepts

Solving the SystemMethod 2: Graphing – Intercepts (Cont.)

Step 2: Graph Equation 2 – Using the intercepts

Solving the SystemMethod 2: Graphing – Intercepts (Cont.)

Step 3: Find the intersection - the solution.

Solving the SystemMethod 2: Graphing – Intercepts (Cont.)

Step 3: Find the intersection - the solution.

Solving the SystemMethod 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 SystemMethod 3: Graphing - Functions (y =)Let x = the cost of a cheesesteakLet y = the cost of a large sodaThen our system becomes

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

Now, we must solve each equation for “y”

Solving the SystemMethod 3: Graphing - Functions (y =)

5x + 4y = 60 (Equation 1)

-5x -5x 4y = 60 – 5x

4 4

x + 2y = 18 (Equation 2)

-x -x 2y = 18 - x

2 2

y =604

−54

x

y =15 −1.25x

y =182

−x2

y =9 −0.5x

Solving the SystemMethod 3: Graphing - Functions (y =)Type each equation into “y =“ in calculator

Set an appropriate “WINDOW”For example: Xmin = -5, Xmax = 20, Xscl = 5

Ymin = -5, Ymax = 20, Yscl = 5, Xres = 1Press “GRAPH”

Find the intersection point

y1 =15 −1.25xy2 =9 −0.5x

Solving the SystemMethod 3: Graphing - Functions (y =)Using the Window above, graph should look like:

Solving the SystemMethod 3: Graphing - Functions (y =)To find the intersection point:Press “2nd” “TRACE” “5:intersect” “ENTER”Press “ENTER” – 3 more timesSolution is:

Intersection:X = 8 Y = 5The 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 SystemMethod 4: Linear CombinationFrom 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 SystemMethod 4: Linear CombinationIn 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 (Add Equations) 3c = 24(Solve for c) 3c = 24 3 3 c = 8 (Cost of a cheesesteak)

Now use this to solve for the other variable.

Solving the SystemMethod 4: Linear CombinationYou 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 = 185(8) + 4s = 60 8 + 2s = 1840 + 4s = 60 -8 -8-40 -40 2s = 10

4s = 20 2 2 4 4 s = 5

s = 5So, s, the cost of a soda = $5. And, the solution of: c = 8 and s = 5, satisfies both original equations.

Solving the SystemMethod 5: Substitution – One Equation into the OtherFrom 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 SystemMethod 5: Substitution – One Equation into the OtherSolving equation 2 for “c”:

c + 2s = 18 - 2s - 2sc = 18 – 2s (NOTE: This is NOT “16s”)

Now, substitute this new expression for c into equation 1 for c:

5c + 4s = 605(18 – 2s) + 4s = 60

Solving the SystemMethod 5: Substitution – One Equation into the OtherNow, solve this equation for s:

5(18 – 2s) + 4s = 6090 – 10s + 4s = 6090 – 6s = 60

-90 - 90 - 6s = - 30

-6 -6 s = 5

Solving the SystemMethod 5: Substitution – One Equation into the OtherNow, substitute s = 5 into either of the original equations and solve for c.

5c + 4s = 60 (equation 1)5c + 4(5) = 605c + 20 = 60 - 20 - 205c = 405 5c = 8

So, s, the cost of a soda = $5. And, c, the cost of a cheesesteak = $8.

Solving the SystemMethod 6: Substitution – Equations Equal to Each OtherFrom 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 SystemMethod 6: Substitution – Equations Equal to Each Other5c + 4s = 60 (equation 1) c + 2s = 18 (equation 2)Solving for c:5c + 4s = 60 - 4s - 4s5c = 60 – 4s5 5 c = 60 - 4s 5 5 OR c = 12 – 0.8s

Solving for c:c + 2s = 18 - 2s - 2sc = 18 – 2s

Solving the SystemMethod 6: Substitution – Equations Equal to Each OtherNow, set the two expression for c equal to each

other and solve for s:12 – 0.8s = 18 – 2s + 0.8s + 0.8s12 = 18 – 1.2s-18 -18-6 = -1.2s-1.2 -1.2 5 = s

Solving the SystemMethod 6: Substitution – Equations Equal to Each OtherNow substitute s = 5 into either of the

expressions found previously for c.c = 18 – 2sc = 18 – 2(5)c = 18 – 10c = 8

So, s, the cost of a soda = $5. And, c, the cost of a cheesesteak = $8.

Verifying the Solution to the SystemThe “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 = 185(8) + 4(5) = 60 8 + 2(5) = 1840 + 20 = 60 8 + 10 = 18 60 = 60 ✓ 18 = 18 ✓

IMPORTANTIn 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, OR2. 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).

IMPORTANTTo 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:

If the system has no solution, then:

Otherwise, the system has ONE solution.

AD

=BE

=CF

AD

=BE

≠CF

Practice ProblemsSolve 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: Matrices2x + 3y = 23x – 4y = -14

Practice Problems2.) Method: Graphing - Intercepts

2x + 9y = 362x – y = 16

3.) Method: Graphing – Functions (y=)5x – 6y = 482x + 5y = -3

Practice Problems4.) Method: Linear Combination

4x – 3y = 175x + 4y = 60

5.) Method: Substitution – One Equation into the Other

8x – 9y = 194x + y = -7

Practice Problems6.) Method: Substitution – Equations Equal to

Each Other4x – y = 63x + 2y = 21

7.) Method: You Choose8x – 4y = 234x – 2y = -17

Practice Problems8.) Method: You Choose

-2x + 5y = 9y = 13 - x

9.) Method: You Choose5x + 3y = 12

15x + 9y = 36

Practice Problems10.) 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.