Why leave this negative sign?

[Guest]

Following this example online:

Evaluate the expression [tex]-x-2(-m+n)[/tex]

Where:

x = -1
m = -2
n = -3

[tex]= -(-1)-2(-(-2)+(-3))[/tex]

[tex]= 1-2(2-3)[/tex]

[tex]= 1-(-2)[/tex]

^^^ OKAY! This bit we just did, why did we distribute a positive 2 and leave the negative sign? I want to do [tex]-2(2-3) = 2[/tex]

[tex]= 1+2[/tex]

[tex]= 3[/tex]

Is there some specific reason it is performed this way?

[nathi123]

[tex]A= -x-2(-m + n) = -x+2m-2n; x=-1;m=-2;n=-3\Rightarrow A=-(-1) +2(-2)-2(-3)=1-4+6=7-4=3[/tex].

[Guest]

No, there is no particular reason.

Evaluate the expression \displaystyle -x-2(-m+n)−x−2(−m+n)

Where:

x = -1
m = -2
n = -3

= -(-1)-2(-(-2)+(-3))=−(−1)−2(−(−2)+(−3))
= 1-2(2-3)=1−2(2−3)

At this point you can continue as
1- (4- 6)= 1- (-2)= 1+ 2= 3
or
1+ (-4+ 6)= 1+ 2= 3.

Since they give the same answer you can use either way.

[Guest]

Thanks all of you

[GeradHum]

Yes! The key is the negative sign before the 2. You have −2(2−3)-2(2 - 3)−2(2−3), which means −2×(2−3)-2 \times (2 - 3)−2×(2−3). You must keep the negative because it multiplies the whole parenthesis. So:
−2(2−3)=−2×(−1)=+2-2(2 - 3) = -2 \times (-1) = +2−2(2−3)=−2×(−1)=+2
If you drop the negative, the value changes. That's why it’s kept!