# Math Help - Absolute values 2

1. ## Absolute values 2

Ok our equation is now : |x+3|+|x-2|= 5

And examining the following 4 cases we have:

1) $x+3\geq 0\Longrightarrow x=2$ which is accepted

2) $x+3\geq 0\wedge x-2<0\Longrightarrow 5=5$

3) $x+3<0\wedge x-2\geq 0\Longrightarrow -5=5$

4) $x+3<0\wedge x-2<0\Longrightarrow x = -3$ 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 $-5\neq 5,\neg -3<-3$ we have :

x=2 or 5=5.

And the question is how do we get rid of 5=5

2. ## Re: Absolute values 2

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 $x$ lying in the interval $[-3,2[$

Now, summarise all your obtained solutions in one solution.

3. ## Re: Absolute values 2

Originally Posted by Siron
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 $x$ lying in the interval $[-3,2[$

Now, summarise all your obtained solutions in one solution.
Can you be so kind ,sir ,as to show me how on earth 5=5 implies that the equation holds for all x's in between -3 and 2 ???

4. ## Re: Absolute values 2

Consider case 2) $x+3\geq 0 \Leftrightarrow x\geq -3$ and $x-2<0 \Leftrightarrow x<2$, if we summarise these two conditions then we get: $-3\leq x<2$. By solving the equation we obtain: $x+3-x+2=5 \Leftrightarrow 5=5$ that means the equation holds for every $x$ subjected to the given condition $-3\leq x<2$. Therefore $-3\leq x <2$ is the solution of the equation.

5. ## Re: Absolute values 2

Originally Posted by Siron
Consider case 2) $x+3\geq 0 \Leftrightarrow x\geq -3$ and $x-2<0 \Leftrightarrow x<2$, if we summarise these two conditions then we get: $-3\leq x<2$. By solving the equation we obtain: $x+3-x+2=5 \Leftrightarrow 5=5$ that means the equation holds for every $x$ subjected to the given condition $-3\leq x<2$. Therefore $-3\leq x <2$ is the solution of the equation.
Is there ax axiom ,theorem ,or definition where you can base your argument?

6. ## Re: Absolute values 2

Originally Posted by psolaki
Is there ax axiom ,theorem ,or definition where you can base your argument?
Take any number $-3\le b\le 2$.
Is the distance from $b\text{ to }-3$ plus the distance from $b\text{ to }2$ equal to 5?

7. ## Re: Absolute values 2

Beside what Plato said, for example take the equation: $x+2-x+2=4$. What are the solutions (and why)?

8. ## Re: Absolute values 2

Originally Posted by Siron
Beside what Plato said, for example take the equation: $x+2-x+2=4$. What are the solutions (and why)?
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

9. ## Re: Absolute values 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").

10. ## Re: Absolute values 2

Originally Posted by HallsofIvy
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").
How do you know that the equation: x+2-x+2=4 holds for all real Nos??.

That the equation ,x+3 = 4 ,is true for x=1 can be shown in the following way: x+3=4 => (x+3)+(-3) = 4 -3 =>x+(3+(-3)) = 1 => x+0=1 => x=1

11. ## Re: Absolute values 2

Originally Posted by psolaki
How do you know that the equation: x+2-x+2=4 holds for all real Nos??.
Can you give me a value of $x$ for what the equation is not satisfied?

12. ## Re: Absolute values 2

Originally Posted by Siron
Can you give me a value of $x$ for what the equation is not satisfied?
Can i try all the x's to find out since they are infinite in No??

13. ## Re: Absolute values 2

It's clear why the equation holds for every $x\in [-3,2[$ because x is substracted by x in the equation (so x 'dissapears'). This is the only way I can explain it.

14. ## Re: Absolute values 2

Originally Posted by psolaki
How do you know that the equation: x+2-x+2=4 holds for all real Nos??

That the equation ,x+3 = 4 ,is true for x=1 can be shown in the following way: x+3=4 => (x+3)+(-3) = 4 -3 =>x+(3+(-3)) = 1 => x+0=1 => x=1
Do you understand what "=" means? x+ 2- x+ 2= (x- x)+ (2+ 2)= 0+ 4= 4 no matter what x is.

15. ## Re: Absolute values 2

Originally Posted by HallsofIvy
Do you understand what "=" means? x+ 2- x+ 2= (x- x)+ (2+ 2)= 0+ 4= 4 no matter what x is.
Is there a definition for "="?? I do not know,please go ahead and explain,but i think this is irrelavent to our argument

Page 1 of 2 12 Last