Question 343457
{{{1/3<(3-2x)/6<3/8 }}}
Let's multiply the equation by {{{24}}}.
{{{8<4(3-2x)<9}}}
Break up the problem into two inequalities and solve them separately,
{{{8<4(3-2x)}}}
{{{2<3-2x}}}
{{{2x<1}}}
{{{highlight(x<1/2)}}}
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.
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{{{4(3-2x)<9}}}
{{{3-2x<9/4}}}
{{{-2x<9/4-12/4}}}
{{{-2x<-3/4}}}
{{{highlight(x>3/8)}}}
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.
.
Then put the intervals together.
{{{3/8<x<1/2}}}
or in interval notation,
({{{3/8}}},{{{1/2}}})