This is an expression that contains a negative in the denominator. If I were to take the negative and place it in the numerator, would this change all positive terms to negative and vice-versa?
Negative in denominator: $\frac{a+3-y^2}{-2}$
After flipping negative to the numerator: $\frac{-a-3+y^2}{2}$ (Is this correct?)
On
yes it's like if you put parentheses in the numerator and u replace whats inside by a letter or a number. So u would have to ditribute the negative sign to every term inside the parentheses. $\frac{-(a+3-y^3)}{2}$
On
This is not an equation. Rather, this is an expression. For the change, multiply both the numerator and the denominator by -1.
On
Yes, but your language is a little vague and misleading. Things to notice:
$1/1 = 1$ (Duh)
$1/(-1) = -1$ (Not so duh; but pretty easy)
$1*1 = 1; (-1)*1= 1*(-1) = -1; (-1)(-1) = 1$ ("A negative times a negative is a positive". To some this is obvious; to some this makes no sense. But most know this as a rule [$*$])
$a(b + c) = ac + ac$ (this is the distributive law)
So if we know all those thing we can conclude:
$\frac {a + 3 - y^2}{-2} = $
$\frac{a+3 - y^2}{(-1)2}=$
$\frac 1{-1}\frac{a+3 -y^2}{2} =$
$- \frac{a+3 -y^2}{2} =$
$\frac{(-1)(a+3-y^2)}2 =$
$\frac{(-1)a + (-1)3 -(-1)y^2}2=$
$\frac{-a -3 +y^2}2$
which can be simplified as "you flip the sign everywhere".
Post-script: Or, d'oh, we can do what everyone else suggests:
$\frac {a + 3 - y^2}{-2} = 1*\frac {a + 3 - y^2}{-2}=\frac{-1}{-1}\frac {a + 3 - y^2}{-2}= \frac{-(a+3-y^2}{-(-2)} = \frac {-a - 3 + y^2}{2}$
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[$*$] why is "a negative times a negative a positive"
Welp.... if $x = -5$ means $5 + x = 0$ and $3*5$ means $5+5+5$. So $(-3)*5= x$ means $x + 5+5+5 = 0$ so $(-3)*5 = -5 -5 -5$. So $(-3)*(-5) = -(5) -(-5) - (-5) = 5+5+5 = 3*5$.
multiplying numerator and denominator with $(-1)$ we get $$\frac{-a-3+y^2}{2}$$