Question 161797
simplify a complex fraction
:
{{{(1/9)+(1/(3x))}}}
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{{{(x/9)-(1/x)}}}
:
Convert both to a single fraction over a common denominator
{{{((x+3))/(9x)}}}
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{{{(x(x)-9(1))/(9x)}}}
:
which is:
{{{((x+3))/(9x)}}}
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{{{((x^2-9))/(9x)}}}
:
Invert the dividing fraction and multiply, cancel out 9x
{{{((x+3))/(9x)}}} * {{{(9x)/((x^2-9))}}} = {{{((x+3))/((x^2-9))}}}
:
Note that x^2-9 is the "difference of squares" so factor that. Cancel out x+3
{{{((x+3))/((x+3)(x-3))}}} = {{{1/((x-3))}}}