SOLUTION: Simplify the complex fractions by any method -2/(ab^2 )over( 5)/a 3y+2/7y^3 over x^3/x-1 (1/3y)+(1/6y) over (1/2y)+(3/4y) 2/y+4 over (3/y-4)-(1/y^2-16)

Click here to see ALL problems on Polynomials-and-rational-expressions

Question 484538: Simplify the complex fractions by any method
-2/(ab^2 )over( 5)/a
3y+2/7y^3 over x^3/x-1
(1/3y)+(1/6y) over (1/2y)+(3/4y)
2/y+4 over (3/y-4)-(1/y^2-16)

Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
Simplify the complex fractions by any method
-2/(ab^2 )over( 5)/a
---
= (-2/(ab^2)) / (5/a)
---
Invert the denominator and multiply:
= (-2/(ab^2)) * (a/5)
---
= (-2/(5b^2))
==============================

3y+2/7y^3 over x^3/x-1
Invert the denominator and multiply:
3y+2/7y^3 times (x-1)/x^3
---
= [(3y+2)(x-1)] / [7x^3y^3]
==============================

(1/3y)+(1/6y) over (1/2y)+(3/4y)
----
= [ (2+1)/(6y)] / [(2+3)/(4y)]
----
= (1/(2y)) * (4y/5)
---
= (4y)/(10y)
---
= 2/5
================================

[2/(y+4)] / (3/y-4)]-(1/y^2-16)
-------
Invert the denominator and multiply:
= [2(y-4)/3(y+4)] - 1/(y^2-16)
----
= [2(y-4)/3(y+4)] - 1/(y^2-16)
----
lcd = 3(y^2-16) which equals 3(y-4)(y+4)
----
= [2(y-4)^2/lcd] - [3/lcd]
---
= [2(y-4)^2-3]/[3(y^2-16)]
============================
Cheers,
Stan H.