Question 210669
{{{-3x<=3x+7<=1/2}}}
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Note we can get two inequalities out of this:

{{{highlight(-3x<=3x+7)<=1/2}}}  AND {{{-3x<=highlight(3x+7<=1/2)}}}

So we write:

{{{-3x<=3x+7}}} AND {{{3x+7<=1/2}}}

Add -3x to both sides of the first inequality.
Multiply the second inequality through by 2 to
clear of fractions:

{{{-6x<=7}}} AND {{{6x+14<=1}}}

Divide the first inequality through by {{{-6}}}
which reverses the sign of inequality:
Add {{{-14}}} to both sides of the second

{{{x>=7/(-6)}}} AND {{{6x<=-13}}}

Simplify the first and divide the second through
by 6 (which BTW <i>does not</i> reverse the sign of
inequality):

{{{x>=-7/6}}} AND {{{x<=-13/6}}}

It's easier to see if we change the
improper fractions to mixed fractions

{{{x>=-1&1/6}}} AND {{{x<=-2&1/6}}}

As you can see this is a contradiction because
the same value of x cannot be to the right of
{{{-1&1/6}}}  AND also to the left of {{{-2&1/6}}}.

Thus the solution set is the empty, or null, set, Ø

Edwin</pre>