I saw steps in an equation that added a 1 and a -1. Why is this okay?

Question

$(x^2+x)-(x+1)$=
$(x \cdot x + x \cdot 1)+(-1)\cdot(x+1)$

Does this have to be a multiplication by +1 and -1? I'm not sure why the - in the middle changed to a + when the (-1) appeared.

(Sorry I don't know the notation for the dot)

Answer 1

I'm not sure why the - in the middle changed to a + when the (-1) appeared.

By definition, $$a - b := a + (-b) = a + (-1)\cdot b\;,$$ whereas $(-b) = (-1)\cdot b$ is the inverse element of $b$. You could also have written $$\begin{split}(x^2 + x) - (x+ 1) &= (x\cdot x + x\cdot 1) - 1\cdot (x + 1)\end{split} $$

Answer 2

It might be easier to keep track of everything if you use a number instead of the variable $x$. If we substitute $x = 3$, the equation becomes $$ (3^2+3)-(3+1) = (3 \cdot 3 + 3 \cdot 1)+(-1)\cdot(3+1). $$ It's easy to see that $(3^2 + 3)$ is the same as $(3 \cdot 3 + 3 \cdot 1)$. $3 \cdot 3$ and $3^2$ are both just $9$, and multiplying $3 \cdot 1$ just leaves us with $3$. So, $3^2 + 3$ and $3 \cdot 3 + 3\cdot 1$ are both just $9+3 = 12$.

Now, on the left side we're supposed to take $12$ and subtract $3 + 1$. On the right side, we're supposed to take $12$ and add $(-1)\cdot (3+1)$. In other words, on one side we subtract $4$, and on the other we add $-4$. Can you see why these operations lead to the same result?

The point now is that there's nothing special about the number $3$ here; this would work if we plugged anything else in. In other words, it is always true that $x^2 = x \cdot x$, that $x \cdot 1 = x$, and that adding $(-1)\cdot (x+1)$ is the same as subtracting $x+1$. So, it is indeed true that $$ (x^2 + x) - (x+1) = (x\cdot x + x\cdot 1) + (-1) \cdot (x+1). $$