
Systems of Linear Equations: Definitions (page 1 of 7) Sections: Definitions, Solving by graphing, Substitition, Elimination/addition, Gaussian elimination. A "system" of equations is a set or collection of equations that you deal with all together at once. Linear equations (ones that graph as straight lines) are simpler than nonlinear equations, and the simplest linear system is one with two equations and two variables. Think back to linear equations. For instance, consider the linear equation y = 3x – 5. A "solution" to this equation was any x, ypoint that "worked" in the equation. So (2, 1) was a solution because, plugging in 2 for x: 3x – 5 = 3(2) – 5 = 6 – 5 = 1 = y On the other hand, (1, 2) was not a solution, because, plugging in 1 for x:
3x – 5 = 3(1) – 5 = 3 – 5 = –2 ...which did not equal y (which was 2, for this point). Of course, in practical terms, you did not find solutions to an equation by picking random points, plugging them in, and checking to see if they "work" in the equation. Instead, you picked xvalues and then calculated the corresponding yvalues. And you used this same procedure to graph the equation. This points out an important fact: Every point on the graph was a solution to the equation, and any solution to the equation was a point on the graph. Now consider the following twovariable system of linear equations: y = 3x – 2
In particular, this purple point marks the intersection of the two lines. Since this point is on both lines, it thus solves both equations, so it solves the entire system of equation. And this relationship is always true: For systems of equations, "solutions" are "intersections". You can confirm the solution by plugging it into the system of equations, and confirming that the solution works in each equation.
y = 3x – 2 To check the given possible solutions, I just plug the x and ycoordinates into the equations, and check to see if they work. Copyright © Elizabeth Stapel 20032011 All Rights Reserved checking (–1, –5): (–5) ?=? 3(–1) – 2 (–5) ?=? –(–1) – 6 Since the given point works in each equation, it is a solution to the system. Now I'll check the other point (which we already know, from looking at the graph, is not a solution): checking (0, –2): (–2) ?=? 3(0) – 2 So the solution works in one of the equations. But to solve the system, it has to work in both equations. Continuing the check: (–2) ?=? –(0) – 6 But –2 does not equal –6, so this "solution" does not check. Then the answer is: only the point (–1, –5) is a solution to the system Top  1  2  3  4  5  6  7  Return to Index Next >>


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