MAT 150 Module 10 – Systems of Equations Lesson 1 – Systems of Linear Equations

Systems of Linear Equations A system of linear equations is a set of two linear equations in two variables, x and y. A solution to the system is the point (x, y) that satisfies both equations in the system. Graphically, the point (x, y) is the point where the lines formed by the two equations intersect.

Systems of Linear Equations

How do we decide whether a system has a unique solution, no solution, or infinitely many solutions? To decide, we want to put both equations in slope-intercept form (y = mx + b). Then it is easy to tell if the equations are different lines, parallel lines, or the same line.

Example 1 Without solving, decide whether the system have a unique solution, no solution, or many solutions. A. 6x - y = 10 x + 2y = 6 B. 2x + 3y = 6 4x + 6y = 12 C. 2x - y = 5 - 4x + 2y = -18

Example 1 - Solution Without solving, decide whether the system have a unique solution, no solution, or many solutions. a. 6x - y = 10 x + 2y = 6 6x – y = 10 -y = -6x + 10 y = 6x - 10 x + 2y = 6 2y = -x + 6 y = -½x + 3

Example 1 - Solution Without solving, decide whether the system have a unique solution, no solution, or many solutions. b. 2x + 3y = 6 4x + 6y = 12 2x + 3y = 6 3y = -2x + 6 y = -⅔x + 2 4x + 6y = 12 6y = -4x + 12 y = -⅔x + 2

Example 1 - Solution Without solving, decide whether the system have a unique solution, no solution, or many solutions. C. 2x - y = 5 - 4x + 2y = -18 2x - y = 5 -y = -2x + 5 y = 2x -5 -4x + 2y = -18 2y = 4x -18 y = 2x -9

Solving systems of linear equations There are three methods used to solve systems of linear equations: Any of these three methods will work on any problem. However, your goal is to use the method that makes the most sense for that problem. GraphingSubstitutionElimination

Solving systems by graphing To solve a system by graphing, graph both lines and find the point of intersection, if there is one.

Example 2 Solve the system of equations by graphing: -x + 3y = 6 2x + 3y = 15

Example 2 - Solution Solve the system of equations by graphing: -x + 3y = 6 2x + 3y = 15 The solution is the point (3, 3).

Solving systems by substitution To solve a system by substitution: Solve one equation for either x or y. Then substitute for either or y into the other equation. Solve for the variable you are left with. Plug back in to find the other variable.

Example 3 Solve the system by substitution: 2x – y = 14 x = y + 8

Example 3 - Solution Solve the system by substitution: 2x – y = 14 x = y + 8 Since we have x = y+8, Substitute y+8 in for x into the first equation. 2x –y = 14 becomes 2(y+8) – y = 14 2y+16 – y = 14Distribute Combine y terms y+ 16 = 14 Subtract 16 y = -2 Plug into x = y+8 to find x X = = 6 The solution is (6, -2)

Solving systems by elimination Elimination is a good method when the equations are written in general form, Ax + By = C. In this method, our goal is to cancel out either x or y, leaving an equation with only one variable. Multiply one or both equations by a constant Then add the two equations to cancel out x or y This will result in an equation you can solve for x or y. So that either the x terms or the y terms in both equations have the same coefficient with opposite signs. Plug that solution back in to solve for the other variable. Like 2x and -2x or -4y and 4y

Example 4 Solve the systems by elimination: a. 5x + 3y = 6 -3x – 2y = 2 b. 4x – 6y = 12 2x – 3y = 6 c. 7x = 8 – 14y x + 2y = 5

Example 4 - Solution Solve the systems by elimination: a. 5x + 3y = 6 -3x – 2y = 2 Want 6y and -6y 2(5x +3y = 6) 3(-3x -2y = 2) 10x + 6y = x - 6y = 6 x = 18 5(18)+3y = y = 63y = -84 y = -84/3 =-28

Example 4 - Part a Solution Solve the systems by elimination: a. 5x + 3y = 6 -3x – 2y = 2 To check the answer, plug in x = 19 and y = -28 for x and y into BOTH equations. 5(18) +3(-28) = 6? -3(18) -2(-28) = 2?

Example 4 – Part b solution Solve the systems by elimination: b. 4x – 6y = 12 2x – 3y = 6 4x - 6y = 12 -2(2x - 3y = 6) 4x - 6y = x + 6y = = 0 Want 4x and -4x Same line – infinitely many solutions

Example 4 – Part b solution

Example 4 – Part C Solution Solve the systems by elimination: c. 7x = 8 – 14y x + 2y = 5 7x + 14y = 8 x + 2y = 5 7x + 14y = 8 -7(x + 2y = 5) 7x + 14y = 8 -7x - 14y = = -27 No Solution

Example 4 – Part C solution

Applications of systems of linear equations

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