case 3) is impossible like you notice (because of the conditions).
Indeed for case 2) you obtain 5=5 that means the equation holds for every lying in the interval
Now, summarise all your obtained solutions in one solution.
Ok our equation is now : |x+3|+|x-2|= 5
And examining the following 4 cases we have:
1) which is accepted
2)
3)
4) and since x+3<0 => x<-3 ,for x=-3 we have -3<-3
Hence the solutions of the equation are:
x=2 or 5=5 or -5=5 or -3<-3 .And since we have :
x=2 or 5=5.
And the question is how do we get rid of 5=5
case 3) is impossible like you notice (because of the conditions).
Indeed for case 2) you obtain 5=5 that means the equation holds for every lying in the interval
Now, summarise all your obtained solutions in one solution.
Yes, but what is the general expression ,or theorem,supporting the fact that the above equation has as a solution set the whole set of real Nos
For example ,when we have that: 2x+3 =5 implies 2x =5 ,that fact is supported by the axiom or theorem that ,when we add the same No to both sides of the equation the equation will not change.
Also by the axiom : x+(-x)=0 and by the theorem 5-3=2
It's not a matter of theorem but just what is meant by a "solution" to an equation. The set of solutions is simply those values of whatever variable you have that make the equation true.
The equation x+ 3= 4 is true only for x= 1. That is its "solution". The equation x+ 2- x+ 2= 4, which is the same as 2+ 2= 4 is true for all values of x. The set of solutions of that equation is "all real numbers" (assuming that the set of real numbers is our "universal set").