Question 95746
{{{(3/5)x-2<3/10-x}}} <==== OK
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{{{(3/5)x-2<3/10-x}}}
+x +x <==== adding x to both sides OK
+2 +2 <=== and adding 2 to both sides OK
{{{(3/5)x+x<(3/10)+2}}} <==== OK
{{{(3/5)x+x<23/10 }}} <==== OK
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then divided 3/5x by 3/5 <=== Nope.  You forgot the +x on the left side. When you add 
{{{(3/5)x + x = (3/5)x + (5/5)x}}} you get {{{(8/5)x}}} on the left side. At this point
the equation becomes: {{{(8/5)x <23/10}}}
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and divided 23/10 by 3/5 <==== divide both sides by {{{8/5}}} to solve for x
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flipped the 3/5 (should flip the 8/5) and multiplied the two fractions on the right like this...
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x+x<23/10*5/3 <=== should be {{{(5/8)*(8/5)x < (5/8)* (23/10)}}}
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2x<115/30 <=== disregard this. In the preceding step do the multiplication of both sides
by {{{5/8}}}. The left side becomes just x because in the {{{(5/8)*(8/5)}}} the
numerators 
cancel the denominators.  On the other side (the right side) the 5 in the numerator
reduces the 10 in the denominator to 2 ... so the multiplication becomes just:
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{{{(1*23)/(8*2)}}} and this simplifies to {{{23/16}}}. So the inequality is reduced to:
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{{{x < 23/16}}} 
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the interpretation of this is that x can take any value less than {{{23/16}}} ... all the
way down to -infinity. So the answer is (-infinity,23/16)
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Hope this helps you to understand the minor error you made.  When you had {{{(3/5)x + x}}} 
you couldn't just divide the one term containing x by {{{3/5}}}. You either had to add the
two terms containing x or you also had to divide the x by {{{3/5}}}.