Stitz-Zeager_College_Algebra_e-book

# 472 systems of equations and matrices like all good

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Unformatted text preview: We multiply both sides of the ﬁrst equation by 15 and both sides of the second equation by 18 to obtain the kinder, gentler system 5x − 12y = 21 4x + 6y = 9 Adding these two equations directly fails to eliminate either of the variables, but we note that if we multiply the ﬁrst equation by 4 and the second by −5, we will be in a position to eliminate the x term 20x − 48y = 84 + (−20x − 30y = −45) −78y = 39 1 From this we get y = − 2 . We can temporarily avoid too much unpleasantness by choosing to 1 substitute y = − 2 into one of the equivalent equations we found by clearing denominators, say into 5x − 12y = 21. We get 5x + 6 = 21 which gives x = 3. Our answer is 3, − 1 . 2 At this point, we have no choice − in order to check an answer algebraically, we must see if the answer satisﬁes both of the original equations, so we substitute x = 3 and y = − 1 2 y x 1 into both x − 45 = 7 and 29 + y = 2 . We leave it to the reader to verify that the solution 3 5 3 is correct. Graphing both of the lines involved with considerable care yields an intersection point of 3, − 1 . 2 4. An eerie calm settles over us as we cautiously approach our fourth sys...
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