Respuesta :

Answer : option 2,4 and 5 are true statements

[tex]\frac{3p+1}{6p} - \frac{2p-3}{2p^2}[/tex]

We have 6p  and 2p^2 in the denominator

Least common denominator LCD = 6p^2

Now we need to make the denominators same

[tex]\frac{(3p+1)*p}{6p*p} - \frac{(2p-3)*3}{2p^2*3}[/tex]

[tex]\frac{(3p^2+p)}{6p^2} - \frac{(6p-9)}{6p^2}[/tex]

So first fraction becomes [tex]\frac{(3p^2+p)}{6p^2}[/tex]

We find the difference by subtracting the numerators

[tex]\frac{(3p^2+p-6p + 9)}{6p^2}[/tex]

[tex]\frac{(3p^2- 5p + 9)}{6p^2}[/tex]

The resulting is a rational expression

So option 2, 4   and 5 are true statements



Answer:

option 2,4 and 5 are true statements

\frac{3p+1}{6p} - \frac{2p-3}{2p^2}

Step-by-step explanation:

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