I'm helping my little sister with her HW and its been years since i've seen this stuff...
How would I solve
z-6 - 5z
2z+3 2z+3
$\displaystyle \frac {z-6}{2z+3} - \frac {5z}{2z+3}$
They both have the same denominator, so you can combine them:
$\displaystyle =\frac {z-6-5z}{2z+3}$
z and -5z have the same variable so you could look at it like this: z -z -z -z -z -z
it should be clear then that they combine to equal -4z
$\displaystyle =\frac {-4z-6}{2z+3}$
both terms in the numerator are multiples of -2, so factor out a -2
$\displaystyle =\frac {-2(2z+3)}{2z+3}$
And since -2 is multiplied and divided by 2z+3, the 2z+3 cancels out to equal one
$\displaystyle =-2*1$
Anything times 1 is itself
$\displaystyle =-2$
The minus sign applies to both terms look at it like this:
$\displaystyle \frac {y-3}{y+5} - \frac {2y-7}{y+5}$
combine
$\displaystyle \frac {y-3-(2y-7)}{y+5}$
distribute the minus sign
$\displaystyle \frac {y-3-2y+7}{y+5}$
combine like terms
$\displaystyle \frac {-y+4}{y+5}$
There are other ways to simplify from here, but "simplify" is kind of ambiguous, this answer is probably what her book is looking for.
Anyway, with that negative sign, it applies to the whole term. You could think of it like this also:
$\displaystyle \frac {y-3}{y+5} +(-1)* \frac {2y-7}{y+5}$
Or it may help to work backwards:
$\displaystyle \frac {-2y+7}{y+5}$
= $\displaystyle \frac {(-1)2y+(-1)(-1)7}{y+5}$
= $\displaystyle \frac {(-1)(2y+(-1)7)}{y+5}$
= $\displaystyle \frac {(-1)(2y-7)}{y+5}$
= $\displaystyle (-1)\frac {(2y-7)}{y+5}$
= $\displaystyle -\frac {2y-7}{y+5}$
However you want to look at it, just understand the sign out front needs to get distributed.
with this one, you cannot combine $\displaystyle z^2$ and z. Another way of looking at it which might help is to factor out the Z.
so z+19z = z(1+19) = 20z
but $\displaystyle 3z^2 + 19z = z(3z+19)$
3z and 19 are not like terms. So $\displaystyle z^2$ and z are not like terms.
You would do it like this:
$\displaystyle \frac {z+6+3z^2+19z+19}{z+5}$
$\displaystyle = \frac {3z^2+20z+25}{z+5}$
Split up the numerator
$\displaystyle = \frac {3z^2+15z+5z+25}{z+5}$
group them (you don't have to do all these steps I'm doing, but I want to hit every step along the way to make it easier to see)
$\displaystyle = \frac {(3z^2+15z)+(5z+25)}{z+5}$
Factor out a 3z
$\displaystyle = \frac {3z(z+5)+(5z+25)}{z+5}$
Factor out a 5
$\displaystyle = \frac {3z(z+5)+5(z+5)}{z+5}$
Factor out a (z+5)
$\displaystyle = \frac {(z+5)(3z+5)}{z+5}$
z+5 is in the numerator and the denominator, so it cancels out
$\displaystyle = 3z+5$
The only difference here is that you don't have a common denominator, so you will have to make one before you can combine the fractions.
In simple arithmetic, you combine fractions without a common denominator by first finding the least common multiple of both denominators, and then multiplying the numerator and denominator of both fractions by the necessary amount to change the denominator to that multiple. For example:
$\displaystyle \frac12 + \frac23$
The least common multiple of 2 and 3 is 6, so we do
$\displaystyle \frac12 + \frac23$
$\displaystyle =\frac{1\color{red}\cdot3}{2\color{red}\cdot3} + \frac{2\color{red}\cdot2}{3\color{red}\cdot2}$
$\displaystyle =\frac36 + \frac46 = \frac{3 + 4}6 = \frac76$
This basic process doesn't change with the introduction of variable expressions. We have:
$\displaystyle \frac{4y^2+zy-3}{5y+1} - \frac{3y^2-2y-4}{z+5}$
The least common multiple of $\displaystyle 5y + 1$ and $\displaystyle z + 5$ is $\displaystyle (5y + 1)(z + 5)$, so we simplify thus:
$\displaystyle \frac{4y^2+zy-3}{5y+1} - \frac{3y^2-2y-4}{z+5}$
$\displaystyle =\frac{\left(4y^2+zy-3\right){\color{red}(z + 5)}}{(5y+1){\color{red}(z + 5)}} - \frac{\left(3y^2-2y-4\right){\color{red}(5y + 1)}}{{\color{red}(5y + 1)}(z+5)}$
$\displaystyle =\frac{\left(4y^2+zy-3\right)(z + 5) - \left(3y^2-2y-4\right)(5y + 1)}{(5y + 1)(z+5)}$
You should be able to take it from here.