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Stitz-Zeager_College_Algebra_e-book

# Solve the following systems of linear equations using

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Unformatted text preview: d ‘parameter’, and get y = 2 t − 3 . Our set of solutions can 2 1 3 3 For speciﬁc values of t, we can generate then be described as t, 2 t − 2 : −∞ < t < ∞ . solutions. For example, t = 0 gives us the solution 0, − 3 ; t = 117 gives us (117, 57), 2 and while we can readily check each of these particular solutions satisfy both equations, the question is how do we check our general answer algebraically? Same as always. We claim that for any real number t, the pair t, 1 t − 3 satisﬁes both equations. Substituting x = t and 2 2 3 1 y = 1 t − 2 into 2x − 4y = 6 gives 2t − 4 2 t − 3 = 6. Simplifying, we get 2t − 2t +6 = 6, which 2 2 is always true. Similarly, when we make these substitutions in the equation 3x − 6y = 9, we get 3t − 6 1 t − 3 = 9 which reduces to 3t − 3t + 9 = 9, so it checks out, too. Geometrically, 2 2 2x − 4y = 6 and 3x − 6y = 9 are the same line, which means that they intersect at every point on their graphs. The reader is encouraged to think about how our parametric solution says exactly that. y y 2 1 −1 −1 −2 1 1 2...
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