Question 196726
{{{a/(4a^2-1)+4/(2a-1)+2}}} Start with the given expression.



{{{a/((2a-1)(2a+1))+4/(2a-1)+2}}} Factor the first denominator.



Take note that the LCD is {{{(2a-1)(2a+1)}}}



{{{a/((2a-1)(2a+1))+(4(2a+1))/((2a-1)(2a+1))+2}}} Multiply the second fraction by {{{(2a+1)/(2a+1)}}} to get the denominator equal to the LCD.



{{{a/((2a-1)(2a+1))+(8a+4)/((2a-1)(2a+1))+2}}} Distribute



{{{a/((2a-1)(2a+1))+(8a+4)/((2a-1)(2a+1))+2(((2a-1)(2a+1))/((2a-1)(2a+1)))}}} Multiply the third term "2" by {{{((2a-1)(2a+1))/((2a-1)(2a+1))}}} to get the denominator equal to the LCD.



{{{a/((2a-1)(2a+1))+(8a+4)/((2a-1)(2a+1))+(2((2a-1)(2a+1))/((2a-1)(2a+1)))}}} Multiply



{{{a/((2a-1)(2a+1))+(8a+4)/((2a-1)(2a+1))+(2(4a^2-1))/((2a-1)(2a+1))}}} FOIL



{{{a/((2a-1)(2a+1))+(8a+4)/((2a-1)(2a+1))+(8a^2-2)/((2a-1)(2a+1))}}} Distribute



{{{(a+8a+4+8a^2-2)/((2a-1)(2a+1))}}} Now that all the denominators are equal, we can combine the fractions.



{{{(8a^2+9a+2)/((2a-1)(2a+1))}}} Combine like terms.



{{{(8a^2+9a+2)/(4a^2-1)}}} FOIL the denominator.



So {{{a/(4a^2-1)+4/(2a-1)+2=(8a^2+9a+2)/(4a^2-1)}}} where {{{a<>-1/2}}}, or {{{a<>1/2}}}