Question 192413
divide this rational expression
{{{(x^2+6x+9)/(2x^2y-18y)}}}
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{{{(6xy+18y)/(3x^2y-27y)}}}
:
Factor
{{{((x+3)(x+3))/(2y(x^2-9))}}}
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{{{(6y(x+3))/(3y(x^2-9))}}}
:
Cancel 3y into 6y, (the denominator fraction), factor (x^2-9)'s
{{{((x+3)(x+3))/(2y(x-3)(x+3))}}}
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{{{(2(x+3))/((x+3)(x-3))}}}
:
Cancel the x+3's, (both fractions)
{{{((x+3))/(2y(x-3))}}}
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{{{2/((x-3))}}}
:
Invert the dividing fraction and multiply, cancel (x-3)'s
{{{((x+3))/(2y(x-3))}}} * {{{(x-3)/2}}} = {{{(x+3)/(4y)}}} is the solution