How do you solve #(x+12)/x<7#?

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Answer 1 · 2015-05-09T23:13:16

Solve #(x+12)/x<7# .

#(cancelx/cancelx)+12/x<7#

#1+12/x<7#

Subtract #1# from both sides.

#12/x<7-1#

#12/x<6#

Multiply both sides by #x#.

#12/cancelx*cancelx<6*x# =

#12<6x#

Divide both sides by #6#.

#12/6< cancel(6)(x)/cancel6#

#2< x#

Flip to get #x# on the left side.(optional)

#x>2#.