Answer:
The simplified sum is [tex]\frac{11p^{2}-23p+24}{(p-8)4p}[/tex]
Step-by-step explanation:
we have
[tex]\frac{2p+1}{p-8}+\frac{3p-3}{4p}=\frac{4p(2p+1)+(p-8)(3p-3)}{(p-8)4p}\\ \\=\frac{8p^{2}+4p+3p^{2}-3p-24p+24}{(p-8)4p}\\ \\ =\frac{11p^{2}-23p+24}{(p-8)4p}[/tex]
Answer:
[tex]\frac{11p^2-23p+24}{4p(p-8)}[/tex]
Step-by-step explanation:
Given expression,
[tex]\frac{2p+1}{p-8}+\frac{3p-3}{4p}[/tex]
[tex]=\frac{4p(2p+1)+(p-8)(3p-3)}{4p(p-8)}[/tex]
[tex]=\frac{8p^2+4p+3p^2-24p-3p+24}{4p(p-8)}[/tex]
By combining like terms,
[tex]=\frac{11p^2-23p+24}{4p(p-8)}[/tex]
Since, further simplification is not possible because numerator is not the perfect square trinomial,
Hence, the required simplified sum is,
[tex]\frac{11p^2-23p+24}{4p(p-8)}[/tex]