Working Out Easy Equations does anyone know how to do this equation? I know it's easy but I can't work out what the question means.

When I expanded the first equation:
$(y+4)-(y-3)$
$y^2 -3y +4y - 12$
$y^2-1y-12$
Not sure what I should do after.
Can someone explain how would you work it out in easy terms?
Thank you
 A: $$(y + 4) - (y - 3) = (y + 4) - (y) - (-3) = (y+4) + (3 - y) = (y - y) + (3 + 4) = 7$$
$$(y-2) - (y - 3) = (y - 2) - y - (-3) = (y-2) + (3 - y) = (y - y) + (3 - 2) = 1$$
In general,
$$(y-a) - (y - b) = b - a$$
In the same way that $x + (y + z) = x + y + z$, $x - (y + z) = x - y - z$. The minus sign distributes to all terms and reverses their sign. Mainly because 
$$z - (x + y) = z + (-1)(x + y) = z + (-1)x + (-1)y = z - x - y$$
A: Notice that, for any numbers $a$ and $b$, we have that 
$$
-(a-b)=(-1)(a-b)=(-1)(a)-(-1)(b)=-a-(-b)=-a+b=b-a.
$$
Use this observation for your problems below.

Problem 1: What is $(y+4)-(y-3)$?
Answer. We have that
$$
(y+4)-(y-3)=y+4-y+3=7.
$$
Problem 2: What is $(y-2)-(y-3)$?
Answer. We have that
$$
(y-2)-(y-3)=y-2-y+3=1.
$$

If you make sure to use the observation given at the beginning of the answer, the above calculations should make sense. 
A: Perhaps, by this you can understnad : 
$(y+4)-(y-3)=(y+4)+[-(y-3)]=y+4+[-y+3]=y+4-y+3=7$.
A: When you made the jump from $(y+4)-(y-3)$ to $y^2-3y+4y-12$, you were applying the expansion rules for quantities multiplied together, not subtracted.
Multiplication: $(y+4)(y-3)=y^2-3y+4y-12$
Subtraction: $(y+4)-(y-3)=y+4-y+3  = 7$
Therefore, your answer for the first question above should really be seven.
A: It might help to look at it like this:
$$\begin{align}
(y+4) & \\
-\quad(y-3) & \\
 & \\
\hline & \\
7
\end{align}$$
A: Keep it simple.


*

*Distribute the central minus sign:
$$
(y+3)-(y-2)=5 \hspace{5mm} \Rightarrow \hspace{5mm} (y+3)+(-y+2)
$$

*Drop the parentheses:
$$
(y+3)+(-y+2) \hspace{5mm} \Rightarrow \hspace{5mm} y+3-y+2
$$

*Cancel the variables:
$$
y+3-y+2 \hspace{5mm} \Rightarrow \hspace{5mm} 3+2
$$

*I think you've got it from here. Rinse and repeat.

A: What you did incorrectly is that you mistook the expression to say $(y+4)(y+3)$ because you expanded to turn it into a quadratic.
The expression is actually a difference so what you should get is:
$(y+4)-(y-3)=y+4-y+3=7$
and
$(y-2)-(y-3)=y-2-y+3=1$
A: This is an identity, not an equation. It is true for any real number y. Hence you can't solve the equation to get a single number.
A: First do (y + 4) - (y - 3)
= y + 4 - y + 3
= 7
Then do (y - 2) - (y - 3)
= y - 2 - y + 3
= 1
