How do you solve #xy=12# and #5x-3y=-8#?

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Answer 1 · 2016-06-08T16:01:14

#y = 12/x#

#5x - 3(12/x) = -8#

#5x - 36/x = -8#

#(5x^2 - 36)/x = -8#

#5x^2 - 36 = -8x#

#5x^2 + 8x - 36 = 0#

#5x^2 - 10x + 18x - 36 = 0#

#5x(x - 2) + 18(x - 2) = 0#

#(5x + 18)(x - 2) = 0#

#x = -18/5 and 2#

#:. -18/5 xx y = 12 or 2 xx y = 12#

#y = 12/(-18/5) or y = 6#

#y = -10/3 or y = 6#

Our solution set is therefore #{-18/5, -10/3} and {2, 6}#.

Hopefully this helps!