Solving -2A - 2B = 2

Question

I should know this, but when simplifying $$-2A - 2B = 2$$ When I divide the LHS by 2, do I divide -2A AND -2B or just one of them? I always thought you did one of them, and then the next. So order of operations would be $$-2A/-2$$ then RHS $$2/-2$$ then LHS $$-2B/-2$$ and lastly RHS $$-1/-2$$

Answer 1

$-2A-2B=2$

Now divide each side by $-2$. (We divide both $-2A$ and $-2B$ together, not what you said).

$A+B=-1$

That is all.

Answer 2

You divide the whole side of LHS and whole side of RHS. In this case you divide both -2A and -2B.

$$-2A-2B=2$$

$$\frac{1}{2}(-2A-2B)=\frac{1}{2} (2)$$

Answer 3

Dividing the both sides of $$-2A-2B=2$$ by $2$ gives you $$\frac{-2A+(-2B)}{2}=\frac 22\iff \frac{-2A}{2}+\frac{-2B}{2}=\frac 22\iff -A-B=1.$$ Here, note that $$\frac{b+c}{a}=\frac{b}{a}+\frac ca.$$

Answer 4

Multiply both sides of the original equation by $\frac12$. The RHS obviously becomes $1$, and this is how you simplify the LHS: $$\tfrac12\cdot(-2A-2B)$$ $$\tfrac12\cdot[(-2)A+ (-2)B]$$ $$\tfrac12\cdot[(-2)A]+ \tfrac12\cdot[(-2)B]$$ $$[\tfrac12\cdot(-2)]A+ [\tfrac12\cdot(-2)]B$$ $$(-1)A + (-1)B$$ $$-A-B$$

Answer 5

When you are wondering about something like this, a good strategy is to replace $A$ and $B$ by some actual numbers and check if what you are claiming is true; here you could take $A=1$ and $B=-2$ (so we indeed have $(-2)1-2(-2)=2$, as we're supposed to). Your suggestion is that we should have $A+B=1/2$, but $1+(-2)=-1$, so this is not correct. If you do the calculations in the other answers, you get $A+B=-1$ (or maybe $-A-B=1$), which checks out. Of course, you could just get lucky with your choices, so if you're really unsure, try a few actual numbers!

The other thing that's helpful is to remember why things are true, rather than trying to remember sequences of "allowable" manipulations. In this case, you have two numbers that are equal to each other; one of them is written in a strange way, but that doesn't matter. So if you perform any operation, you'll get the same answer independent of which expression you use - in your example, you have $-2A-2B=2$, so if you want to calculate half of $-2A-2B$, you can just calculate half of $2$ (because it's the same!) and get $\frac{1}{2}(-2A-2B)=1$. Now you need to remember why $\frac{1}{2}(-2A-2B)=-A-B$...

I appreciate this may have been more detail than you really needed, but this kind of thing comes up a lot, so I've tried to write an answer that might be useful to people with similar problems!