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Stitz-Zeager_College_Algebra_e-book

if he wants to make a 10 pound mix with a budget of

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Unformatted text preview: 4 5 6 7 x 1 3, − 1 2 2 3 4 x −1 −3 −4 x − 4y 3 5 2x +y 9 3 2 =7 5 =1 2 2x − 4y = 6 3x − 6y = 9 (Same line.) See Section 1.2 for a review of this. Note that we could have just as easily chosen to solve 2x − 4y = 6 for x to obtain x = 2y + 3. Letting y be the parameter t, we have that for any value of t, x = 2t + 3, which gives {(2t + 3, t) : −∞ < t < ∞}. There is no one correct way to parameterize the solution set, which is why it is always best to check your answer. 3 8.1 Systems of Linear Equations: Gaussian Elimination 453 5. Multiplying both sides of the ﬁrst equation by 2 and the both sides of the second equation by −3, we set the stage to eliminate x 12x + 6y = 18 + (−12x − 6y = −36) 0 = −18 As in the previous example, both x and y dropped out of the equation, but we are left with an irrevocable contradiction, 0 = −18. This tells us that it is impossible to ﬁnd a pair (x, y ) which satisﬁes both equations; in other words, the system has no solution. Graphical...
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