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Ответ:

talahtoo6659

31.01.2021

$\frac{4p-3}{3p+1}+\frac{2p+5}{3p+1}=2$

Step-by-step explanation:

we are given

$\frac{4p-3}{3p+1}+\frac{2p+5}{3p+1}$

we can see that

denominators of both terms are 3p+1

So, both has same denominators

so, we can combine numerators

and we get

$\frac{4p-3}{3p+1}+\frac{2p+5}{3p+1}=\frac{4p-3+2p+5}{3p+1}$

now, we can combine like terms

$\frac{4p-3}{3p+1}+\frac{2p+5}{3p+1}=\frac{4p+2p-3+5}{3p+1}$

$\frac{4p-3}{3p+1}+\frac{2p+5}{3p+1}=\frac{6p+2}{3p+1}$

now, we can factor out common term

$\frac{4p-3}{3p+1}+\frac{2p+5}{3p+1}=\frac{2(3p+1)}{3p+1}$

we can see that 3p+1 is on both top and bottom

so, we can cancel it

and we get

$\frac{4p-3}{3p+1}+\frac{2p+5}{3p+1}=2$

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