How do you add or subtract #(5y)/(1-4y^2 ) - ( 2y)/(2y+1) + (5y)/(4y^2-1)#?

Question

How do you add or subtract #(5y)/(1-4y^2 ) - ( 2y)/(2y+1)  +  (5y)/(4y^2-1)#?

Savannah Anderson · Accepted Answer

First, notice that #4y^2-1# = #-(1-4y^2#), then rewrite #(5y)/(4y^2-1)# as #-(5y)/(1-4y^2)#, which allows you to get rid of the first #(5y)/(1-4y^2)# term
and you are left with only #-(2y)/(2y+1)#.

Jameson Allen · Answer

undefined To add or subtract the given expressions, we need to find a common denominator. The common denominator for the three expressions is (1-4y^2)(2y+1)(4y^2-1).

To add or subtract fractions with different denominators, we need to rewrite each fraction with the common denominator.

The first fraction, (5y)/(1-4y^2), can be rewritten as (5y)(2y+1)(4y^2-1)/[(1-4y^2)(2y+1)(4y^2-1)].

The second fraction, (2y)/(2y+1), can be rewritten as (2y)(1-4y^2)(4y^2-1)/[(1-4y^2)(2y+1)(4y^2-1)].

The third fraction, (5y)/(4y^2-1), can be rewritten as (5y)(2y+1)(1-4y^2)/[(1-4y^2)(2y+1)(4y^2-1)].

Now, we can combine the numerators and keep the common denominator:

[(5y)(2y+1)(4y^2-1) - (2y)(1-4y^2)(4y^2-1) + (5y)(2y+1)(1-4y^2)] / [(1-4y^2)(2y+1)(4y^2-1)].

Simplifying the numerator:

[10y^4 - 5y^2 - 8y^4 + 2y^2 + 10y^4 - 5y^2] / [(1-4y^2)(2y+1)(4y^2-1)].

Combining like terms:

[-y^2] / [(1-4y^2)(2y+1)(4y^2-1)].

Therefore, the simplified expression is -y^2 / [(1-4y^2)(2y+1)(4y^2-1)].