$y + 3$ 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!
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 $(y - 3)$
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.