# Removing brackets and negative values within equations

So I wondered if anyone here can clarify something for me. I'm reading a maths book and one of the exercises has me a little confused:

Part of the -3(q-2) is simplified to -3 x -2.

It confuses me that the negative is taken into account for the two terms. Shouldn't 4(q+1) and 3(q-2) be independent units due to dealing with the brackets first and then the later subtracted from the former?

Making it:

4(q+1) = 4q + 4

3(q-2) = 3q - 6

= 4q + 4 - 3q - 6

Taking the same logic and a simple example 10 - 2(3-1) Why would it be -2 x -1?

I always imagined I could have negative and positive values outside the brackets for example I could have:

10 - -2(3-1) = 10 - -2 x 3 - -2 x 1

= 10 - -6 - -2 = 14

Or I could have:

10 - 2(3-1) = 10 - 2 x 3 - 2 x 1

= 10 - 6 - 2 = 2

If I use their logic it doesn't seem to allow this choice and is very confusing. Any help on why it should be done like in the book would be greatly appreciated.

The gist of the matter is that $$-(q-2)=-q+2$$ That is, it's the same as $$(-1)\times(q-2)$$ and by the same token, $$-3(q-2)=(-3)(q-2)=(-3)(q)+(-3)(-2)=-3q+6$$
In your evaluation of $$10-2(3-1),$$ you insert a second minus sign, making it $$10--2(3-1).$$ There is no justification for this. The two expressions are not equal.
I'll start by presenting a simpler example and then I'll go to the computations: $$5 - 4 \times(3-2) = 5 - 4 \times 1 = 5 - 4 = 1$$ Or, alternatively, using the distributive property first, we will get a more complicated way of getting the same result but that (hopefully) explains the care we should take when there is a subtraction: $$5 - (4 \times 3 - 4 \times 2) = 5 - 4 \times 3 + 4 \times 2 = 5 - 12 + 8 = -7 + 8 = 1$$
So, going back to the equation, I'll try to breakdown in shorter steps, trying to make it easier: $$4(q+1) - 3(q-2) = 7$$ Recall that the product has priority over the subtraction, therefore I will place some superfluous parenthesis that do not change the result: $$(4(q+1)) - (3(q-2)) = 7$$ Now, I will use the distributive property of the product with the respect to the sums: $$(4q+4) - (3q-6) = 7$$ Now I can remove parenthesis, remembering that the negative is applied to all terms in the second parenthesis: $$4q+4 - 3q+6 = 7$$ From this point onwards I think you don't have any doubts.
Alternatively, going back to the initial equation: $$4(q+1) - 3(q-2) = 7$$ We can think that it is $$-3$$ that is being multiplied by the second parenthesis: $$4(q+1) + (-3)(q-2) = 7$$ Hence we get $$4q+1 + (-3)q-(-3)\times2 = 7$$ $$4q+1 -3q-(-6) = 7$$ $$4q+1 -3q+6 = 7$$ So, the same result.