Mathematical Expressions; School Homework $y + 3$ is always $5$ more than $y – 2$ 
so $y + 3 – (y – 2) = 5$

$(y + 4 ) – (y – 3) = ~ ?$
$(y - 2) – (y - 3) = ~ ?$
How would you work it out? I know that $y + 3$ is 5 more than $y - 2$, so should I do trial method or is there a method that's easier and faster? 
Oh and why type of mathematical equation is this? Please make it easy. Thanks!
 A: When be subtract a quantity $b$, we can think of it as "adding $-b$. So $a - b = a + -b$. 
When we have an expression of the form $a - (b + c) = a + -(b + c)$ we can always distribute $-1$ (i.e., multiply through by $-1$)  
$$
\begin{align} a + -1\times(b + c) 
& = a + -1\times b + -1\times c \\ \\
& = a + - b + - c \\ \\
& = a - b - c
\end{align}
$$
Using these facts, we can proceed:
$$
\begin{align} (y + 4 ) – (y – 3) & = y + 4 + -1\cdot (y + - 3)) \\ \\
& = y + 4 + -1\cdot y + (-1)(-3) \\ \\
& = y + 4 - y + 3 = 7 \\ \\ 
& = (y - y) + (4 + 3) \\ \\
& = 7
\end{align}
$$
So $\displaystyle\quad (y + 4) - (y - 3) = 7 \implies \quad (y + 4)$ is 7 more than $(y - 3)$
We can think of is also as just distributing the $-$ sign over the quanity:
$$
\begin{align} (y – 2) – (y – 3) & = (y - 2) + (- y - (-3))\\ \\
& = y - 2 - y +  3 \\ \\
& = -2 + 3 \\ \\
& = 1
\end{align}$$
So $\displaystyle \quad(y - 2) - (y - 3) = 1  \implies \quad (y - 2)$ is 1 more than $(y - 3)$
A: The distributive property says that $-\rm something = -1 \times \rm something$. Hence, $y + 3 - (y - 2)  = y + 3 - 1(y - 2)$. 
The original definition is that $a(b + c) = ab + ac$.
So $-1(y - 2) = (-1\times y) + (-1\times-2) = -y + 2 $.
You are left with 
$y+3 - y + 2$ 
$= y - y + 3 +2$
$= 0 + 3+2$ 
$= 5$.
For the second example, you have
$y + 4 - (y - 3)$
$= y + 4 - 1(y - 3)$
$= y + 4 - y + 3$
$= y -y + 4 + 3$
$= 7$
Try doing the third yourself. 
