Question 196914
I'm assuming that you want to perform a partial fraction decomposition...



{{{(2y+1)/(y^2-y-6)}}} Start with the given expression.



{{{(2y+1)/((y+2)(y-3))}}} Factor the denominator




Since both factors are linear and are of multiplicity 1, this means that the fraction decomposes into:


{{{(2y+1)/((y+2)(y-3))=A/(y+2)+B/(y-3)}}}



{{{cross((y+2)(y-3))((2y+1)/(cross((y+2)(y-3))))=cross((y+2))(y-3)(A/cross((y+2)))+(y+2)cross((y-3))(B/cross((y-3)))}}} Multiply EVERY term by the LCD {{{(y+2)(y-3)}}} to clear out the fractions.



{{{2y+1=A(y-3)+B(y+2)}}} Simplify



{{{2y+1=Ay-3A+By+2B}}} Distribute



{{{2y+1=Ay+By-3A+2B}}} Rearrange the terms



{{{2y+1=(A+B)y-3A+2B}}} Combine like terms.




Since the "y" coefficient on the right side is {{{A+B}}}, this means that {{{A+B=2}}}. Also, the constant term {{{-3A+2B=1}}}




So this means that we have the system:


{{{system(A+B=2,-3A+2B=1)}}}



Solve this system of equations (using any method) to get the solutions:


{{{A=3/5}}} and {{{B=7/5}}}



Now plug these values back into the original equation to get:



{{{(2y+1)/((y+2)(y-3))=(3/5)/(y+2)+(7/5)/(y-3)}}} 




And simplify



{{{(2y+1)/((y+2)(y-3))=3/(5(y+2))+7/(5(y-3))}}}


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Answer:


So {{{(2y+1)/(y^2-y-6)=3/(5(y+2))+7/(5(y-3))}}}