Question 684210
{{{((s^2+3s-18)/(s+4))/((s^2+3s-18)/(s+6))}}}
Division is the inverse of multiplication, therefore,
{{{((s^2+3s-18)/(s+4))((s+6)/(s^2+3s-18))}}}
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Cancel out terms
{{{highlight_green((s+6)/(s+4))}}}
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