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I'm a bit lost here....

Equation 1: $(5p − 6) + (1 − p)$.

Shouldn't I apply distributive property here? By distributing the '$+$' sign into $(1 - p)$ to give $(1 + p)$? If that is the case, then the new formula reworded is: $$(5p - 6) + 1 + p = 6p - 5,$$ right?

But the book has a different answer and it is, $4p - 5$ instead.... deductively examining where I went wrong, it seems the '$+$' sign isn't distributed and thus the $p$ in $(1 - p)$ didn't change into a positive

If the book has the right answer, then this procs my title question, when do we use distributive property?

Consider the following equation: $$−10 − 4(n − 5),$$ the $-4$ is distributed into $n$ and $-5$.... If I'm seeing how the formula is worded, whats the difference between this and the case above? Don't they both prompt distributive property cycle? The above case just has an invisible $+1$ right?

I got all the wrong answers in my math test on this part lol but i'm determined to know why.

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    $\begingroup$ Why do you think that $(1-p)$ will become $(1+p)$ ? $(5p − 6) + (1 − p)= 5p-6+1-p$. $\endgroup$ – Mauro ALLEGRANZA Oct 19 '18 at 9:26
  • $\begingroup$ i thought the + has an invisible 1 that it can distribute... is this not true? but then again whenever a positive times a negative it'll still be a negative anyways, so p won't turn positive i believe, just figured this out now lol $\endgroup$ – Moorease Oct 19 '18 at 9:32
  • $\begingroup$ $1(1 - p) = 1 \cdot 1 + 1 \cdot (-p) = 1 - p$. $\endgroup$ – N. F. Taussig Oct 19 '18 at 9:45
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If you multiply +1 by -p, you will get -p not +p

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