Answer
$\displaystyle \frac{2y+1}{y+1}$
Work Step by Step
Factoring trinomials $ax^{2}+bx+c:$
search for integer factors of $a\times c$, whose sum is $b.$
If we find these factors, say m and n, rewrite the trinomial as
$ax^{2}+mx+nx+c$ and factor the expression "in pairs".
Factor the numerator:
we search for integer factors of 6$\times$4=24, whose sum is 11
We find +8 and +3,
$6y^{2}+11y+4$=$6y^{2}+3y+8y+4$
$=3y(2y+1)+4(2y+1)=(3y+4)(2y+1)$
Factor the denominator:
we search for integer factors of 3$\times$4=12, whose sum is 7
We find +3 and +4,
$3y^{2}+7y+4=3y^{2}+3y+4y+4$
$=3y(y+1)+4(y+1)=(3y+4)(y+1)$
$\displaystyle \frac{6y^{2}+11y+4}{3y^{2}+7y+4}=\frac{(3y+4)(2y+1)}{(3y+4)(y+1)}$
... reduce
$=\displaystyle \frac{2y+1}{y+1}$
