<PRE><B>Question 429770</B>
[1/3y + 1/6y] / [1/2y + 3/4y]
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= [(2/6)y + (1/6)y] / [(2/4)y + (3/4)y]
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= [(3/6)y] / [(5/4)y]
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Invert and multiply:
= [(3y/6)] * [(4/5y)]
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= [12y/30y]
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= (2/5)
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[2/(y+4)] / [(3/(y-4)) - (1/(y^2 - 16))]
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[2/(y+4)] /[(3(y+4)/(y^2-16)) - 1/(y^2-16)]
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= [2/(y+4)] / [(3y+12-1)/(y^2-16)]
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= [2/(y+4)]*[(y^2-16)/(3y+11)]
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Cancel (y+4) from numerator and denominator to get:
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= [2(y-4)]/(3y+11)
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cheers,
Stan H.</PRE>