Solving Absolute Value Equations Containing TWO Absolute Value Expressions - Example 1

[km-sckz]

in the second equation why didn't you put p+1 also in brackets and put a negative sign outside it?

[piya]

Do you not take into consideration p cannot be 3? Because if it is then the equation will be undefined. I'm trying to use the same concept for inequalities and I'm slightly confused

[michaelkingDJ]

What do I do when I have a the modulus of my function being the integrand? Do I just remove the bars or??

[naminejackson]

For the second equation why did you make -12 -p become positive do you always have to do that?

[yadira]

So would I do it the same way if my equation was |P+1|/4=2 ? exept obviously instead of having two digits at the bottom it's just the 4?

[eternallyever]

@GME777Gregz
The alternate answer will still be the same regardless of the expression the negative sign is at.

[cenorib]

I think in the second equation he actually meant:

p + 1 = (12 (3 - p)) . (-1) => the .(-1) multiplying the other side expresses the negative difference this equality has to zero
p + 1 = (36 - 12p) . (-1)
p + 1 = -36 + 12p
p -12p = -36 -1
-11p = -37
-p = -37 / -11
p = 37/11

The result is the same as in the video, he just made it in a way that it wasn't very clear.