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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 =
+ (−20x − 30y = −45)
From this we get y = − 2 . We can temporarily avoid too much unpleasantness by choosing to
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 .
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
into both x − 45 = 7 and 29 + y = 2 . We leave it to the reader to verify that the solution
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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