Simplify $3^2 + (-8\div2)$.
I know its not possible to divide $-8$ into $2$, so is there a way to make the 8 positive before working the problem?
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Hint: $(-8)\div2=-(8\div2)$. In fact, you should do division first before considering the minus sign.
On the contrary, it is possible to divide $-8$ by $2$. The definition of division is that $x \div y = z$ if and only if $x = z \cdot y$ and $y \neq 0$. We know that $-8 = -4 \cdot 2$; therefore, $-8 \div 2 = -4$.
From http://en.wikipedia.org/wiki/Order_of_operations, the order of operations are as:
So, if you come across a problem like this, you should follow the steps above in order, from 1 to 3.
Now to address your question, for $3^2 + (-8÷2)$, we do the exponents first:
$$3^2 + (-8\div2)=3\times3+(-8\div2),$$ Then we can take the minus sign out of $8$ and multiply $3\times3$, so
Now, we can operate $8 \div 2$,
$$9+((-1)\times 8\div2)=9+((-1)\times 4).$$
Next, any number $\alpha$ multiplied by $(-1)$ will become $-\alpha$. So, if we have $\alpha$, then $\alpha\times(-1)=-\alpha$. For example if we have $\alpha=5$, then $5\times(-1)=-5$
So, we have $$9+((-1)\times 4)=9+(-4),$$
since here $\alpha=4$, so $4\times(-1)=(-1)\times 4=-4$.
where we subtracted.
Hope this helps.