Answer:
[tex]\frac{4p-3}{3p+1}+\frac{2p+5}{3p+1}=2[/tex]
Step-by-step explanation:
we are given
[tex]\frac{4p-3}{3p+1}+\frac{2p+5}{3p+1}[/tex]
we can see that
denominators of both terms are 3p+1
So, both has same denominators
so, we can combine numerators
and we get
[tex]\frac{4p-3}{3p+1}+\frac{2p+5}{3p+1}=\frac{4p-3+2p+5}{3p+1}[/tex]
now, we can combine like terms
[tex]\frac{4p-3}{3p+1}+\frac{2p+5}{3p+1}=\frac{4p+2p-3+5}{3p+1}[/tex]
[tex]\frac{4p-3}{3p+1}+\frac{2p+5}{3p+1}=\frac{6p+2}{3p+1}[/tex]
now, we can factor out common term
[tex]\frac{4p-3}{3p+1}+\frac{2p+5}{3p+1}=\frac{2(3p+1)}{3p+1}[/tex]
we can see that 3p+1 is on both top and bottom
so, we can cancel it
and we get
[tex]\frac{4p-3}{3p+1}+\frac{2p+5}{3p+1}=2[/tex]
