Solve and write in lowest terms?

Question

(7/(p^2+p))+(14/(p^2-2p))

Answer 1

[7/(p²+p)] + [14/(p²-2p)] =   ...factor the denominators of both fractions

[7/(p(p+1))] + [14/(p(p-2))] =

[7/(p(p+1))] + [14/(p(p-2))] =  

You need a common denominator to add these together.  Multiply the first fraction by (p-2)/(p-2) and the second fraction by (p+1)/(p+1)

[(p-2)/(p-2)][7/(p(p+1))] + [(p+1)/(p+1)][14/(p(p-2))] =  

[(7(p-2))/(p(p-2)(p+1))] + [14(p+1)/(p(p-2)(p+1))] =  

[7(p-2) + 14(p+1)] / [(p(p-2)(p+1))] =  

[7p - 14 + 14p + 14] / [(p(p-2)(p+1))] =  

[7p + 14p] / [(p(p-2)(p+1))] =  

[21p] / [(p(p-2)(p+1))] =  

The p in the numerator cancels with the p in the denominator since p/p = 1

21 / [((p-2)(p+1))] =  

21 / ((p-2)(p+1))

Hope this helps you!