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**Solving a System of 2 Linear Equations**

Utilizing 6 Methods: Matrices Graphing – Intercepts Graphing – Functions (y = ) Linear Combination Substitution – One Equation into the Other Substitution – Equations Equal to Each Other

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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)

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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?

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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.

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**Setting Up the System To set-up a system, do the following:**

Identify the variables (unknowns) in the problem. Example: Let c = Cost of one Cheesesteak Let s = Cost of one large Soda 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

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Solving the System This tutorial will use 6 different methods to solve a system of 2 linear equations: Matrices Graphing – Intercepts Graphing – Functions (y =) Linear Combination Substitution – One Equation into the Other Substitution – Equations Equal to Each Other

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**Solving the System Method 1: Matrices**

Using the system of equations: 5c + 4s = 60 c + 2s = 18 Re-write as a matrix equation:

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**Solving the System Method 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.

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**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

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**Solving the System Method 2: Graphing – Intercepts**

5x + 4y = 60 (Equation 1) x + 2y = 18 (Equation 2) 5x + 4y = 60 (Equation 1) X-intercept: 5x + 4(0) = 60 5x = x = 12 Pt. (12,0) Y-intercept: 5(0) + 4y = 60 4y = y = 15 Pt. (0,15) x + 2y = 18 (Equation 2) X-intercept: x + 2(0) = 18 x = 18 Pt. (18,0) Y-intercept: (0) + 2y = 18 2y = y = 9 Pt. (0, 9)

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**Solving the System Method 2: Graphing – Intercepts (Cont.)**

Step 1: Graph Equation 1 – Using the intercepts

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**Solving the System Method 2: Graphing – Intercepts (Cont.)**

Step 2: Graph Equation 2 – Using the intercepts

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**Solving the System Method 2: Graphing – Intercepts (Cont.)**

Step 2: Graph Equation 2 – Using the intercepts

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**Solving the System Method 2: Graphing – Intercepts (Cont.)**

Step 3: Find the intersection - the solution.

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**Solving the System Method 2: Graphing – Intercepts (Cont.)**

Step 3: Find the intersection - the solution.

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**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

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**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”

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**Solving the System Method 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

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**Solving the System Method 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 = 1 Press “GRAPH” Find the intersection point

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**Solving the System Method 3: Graphing - Functions (y =)**

Using the Window above, graph should look like:

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**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

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**Solving the System Method 4: Linear Combination**

From the original system of equations: 5c + 4s = (equation 1) c + 2s = (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

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**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) c +-4s = -36 (Add Equations) 3c = 24 (Solve for c) 3c = 24 c = 8 (Cost of a cheesesteak) Now use this to solve for the other variable.

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**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 = c + 2s = 18 5(8) + 4s = s = 18 s = s = 10 4s = 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.

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**Solving the System Method 5: Substitution – One Equation into the Other**

From the original system of equations: 5c + 4s = (equation 1) c + 2s = (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.

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**Solving the System Method 5: Substitution – One Equation into the Other**

Solving equation 2 for “c”: c + 2s = 18 - 2s s 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

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**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 - 6s = - 30 -6 -6 s = 5

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**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 5c = 40 c = 8 So, s, the cost of a soda = $5. And, c, the cost of a cheesesteak = $8.

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**Solving the System Method 6: Substitution – Equations Equal to Each Other**

From the original system of equations: 5c + 4s = (equation 1) c + 2s = (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.

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**Solving the System Method 6: Substitution – Equations Equal to Each Other**

5c + 4s = 60 (equation 1) c + 2s = 18 (equation 2) Solving for c: 5c + 4s = 60 - 4s s 5c = 60 – 4s c = s OR c = 12 – 0.8s Solving for c: c + 2s = s - 2s c = 18 – 2s

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**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 = -1.2s = s

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**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.

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**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) = (5) = = = = 60 ✓ 18 = 18 ✓

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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: An infinite number of solutions, OR 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).

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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: If the system has no solution, then: Otherwise, the system has ONE solution.

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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

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Practice Problems 2.) Method: Graphing - Intercepts 2x + 9y = 36 2x – y = 16 3.) Method: Graphing – Functions (y=) 5x – 6y = 48 2x + 5y = -3

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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

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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

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Practice Problems 8.) Method: You Choose -2x + 5y = 9 y = 13 - x 9.) Method: You Choose 5x + 3y = 12 15x + 9y = 36

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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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