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Stitz-Zeager_College_Algebra_e-book

Use the technique in example 823 to t a quadratic

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Unformatted text preview: ly, we see that the lines 6x + 3y = 9 and 4x + 2y = 12 are distinct and parallel, and as such do not intersect. 6. We can begin to solve our last system by adding the ﬁrst two equations x−y = 0 + (x + y = 2) 2x = 2 which gives x = 1. Substituting this into the ﬁrst equation gives 1 − y = 0 so that y = 1. We seem to have determined a solution to our system, (1, 1). While this checks in the ﬁrst two equations, when we substitute x = 1 and y = 1 into the third equation, we get −2(1)+(1) = −2 which simpliﬁes to the contradiction −1 = −2. Graphing the lines x − y = 0, x + y = 2, and −2x + y = −2, we see that the ﬁrst two lines do, in fact, intersect at (1, 1), however, all three lines never intersect at the same point simultaneously, which is what is required if a solution to the system is to be found. y y 6 5 4 3 2 1 −1 −2 −3 1 1 2 6x + 3y = 9 4x + 2y = 12 x x −1 y−x=0 y+x=2 −2x + y = −2 A few remarks about Example 8.1.1 are in order. It is clear that some systems of equations have solutions, and some do no...
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