Answer
$\displaystyle \frac{y+3}{3y+8}$
Work Step by Step
To subtract rational expressions with the same denominator,
subtract numerators and place the difference over the common denominator.
If possible, factor and simplify the result.
---
Don't forget to place the second numerator in parentheses when subtracting.
$\displaystyle \frac{4y^{2}+5}{9y^{2}-64}-\frac{y^{2}-y+29}{9y^{2}-64}= \displaystyle \frac{4y^{2}+5-y^{2}-y+29)}{9y^{2}-64}$
$= \displaystyle \frac{4y^{2}+5-y^{2}+y-29}{9y^{2}-64}$
$= \displaystyle \frac{3y^{2}+y-24}{9y^{2}-64}\qquad$... factor what we can...
... Numerator:
$... $ two factors of $ac=-72$ whose sum is 1... are $-8$ and $+9$.
... Rewrite bx and factor in pairs:
$3y^{2}+y-24=3y^{2}+9y-8y-24$
$=3y(y+3)-8(y+3)=(y+3)(3y-8)$
... Denominator: a difference of squares, $(3y)^{2}-8^{2}$
$=\displaystyle \frac{(y+3)(3y-8)}{(3y+8)(3y-8)}$
... reduce (divide the numerator and denominator with common factors)
$=\displaystyle \frac{y+3}{3y+8}$
