SOLUTION: I need help with a couple of fraction. I actually figured the others by myself which means I'm getting better. Simplify the complex fraction by any method.
1/3y + 1/6y / 1/2y
Algebra.Com
Question 429770: I need help with a couple of fraction. I actually figured the others by myself which means I'm getting better. Simplify the complex fraction by any method.
1/3y + 1/6y / 1/2y + 3/4y
2/y+4 / 3/y-4 - 1/y^2 - 16
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
[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]
----
Invert and multiply:
= [(3y/6)] * [(4/5y)]
-------------
= [12y/30y]
---
= (2/5)
===============
[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)]
---
= [2/(y+4)] / [(3y+12-1)/(y^2-16)]
----
= [2/(y+4)]
---
Cancel (y+4) from numerator and denominator to get:
------------------
= [2(y-4)]/(3y+11)
=====================
cheers,
Stan H.
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