Question #78b61

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Answer 1 · 2017-04-26T13:04:19

#x<3#

#color(blue)((x+3)/x<2#

This is an inequality problem. Solve it by balancing both sides by the same operation

Simplify the left hand side

#rarrx/x+3/x<2#

#rarr1+3/x<2#

Subtract #1# from both sides

#rarr3/x<1#

Divide both sides by #3#

#rarr1/x<1/3#

Multiply both sides by #3x# (Reverse the sides)

#color(green)(rArrx<3#

Answer 2 · 2017-04-26T13:32:44

The solution is #x in (-oo,0) uu (3,+oo)#

We cannot do crossing over.

So,

#(x+3)/x<2#

#(x+3)/x-2<0#

#(x+3-2x)/x<0#

#(3-x)/x<0#

Let,

#f(x)=(3-x)/x#

We build a sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaaaaa)##0##color(white)(aaaaaaaa)##3##color(white)(aaaa)##+oo#

#color(white)(aaaa)##x##color(white)(aaaaaaaa)##-##color(white)(aaaa)##||##color(white)(aaaa)##+##color(white)(aaaa)##+#

#color(white)(aaaa)##3-x##color(white)(aaaaa)##+##color(white)(aaaa)##||##color(white)(aaaa)##+##color(white)(aaaa)##-#

#color(white)(aaaa)##f(x)##color(white)(aaaaaa)##-##color(white)(aaaa)##||##color(white)(aaaa)##+##color(white)(aaaa)##-#

Therefore,

#f(x)<0# when #x in (-oo,0) uu (3,+oo)#
graph{(3-x)/x [-16.02, 16.02, -8.01, 8.01]}