Page 1 of 2 12 LastLast
Results 1 to 15 of 20

Math Help - Absolute values 2

  1. #1
    Member
    Joined
    Jul 2011
    Posts
    121
    Thanks
    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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Siron's Avatar
    Joined
    Jul 2011
    From
    Norway
    Posts
    1,250
    Thanks
    20

    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.
    Last edited by Siron; December 27th 2011 at 10:11 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jul 2011
    Posts
    121
    Thanks
    1

    Re: Absolute values 2

    Quote Originally Posted by Siron View Post
    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 ???
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Siron's Avatar
    Joined
    Jul 2011
    From
    Norway
    Posts
    1,250
    Thanks
    20

    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.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Jul 2011
    Posts
    121
    Thanks
    1

    Re: Absolute values 2

    Quote Originally Posted by Siron View Post
    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?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,615
    Thanks
    1578
    Awards
    1

    Re: Absolute values 2

    Quote Originally Posted by psolaki View Post
    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?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor Siron's Avatar
    Joined
    Jul 2011
    From
    Norway
    Posts
    1,250
    Thanks
    20

    Re: Absolute values 2

    Beside what Plato said, for example take the equation: x+2-x+2=4. What are the solutions (and why)?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    Jul 2011
    Posts
    121
    Thanks
    1

    Re: Absolute values 2

    Quote Originally Posted by Siron View Post
    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
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,567
    Thanks
    1409

    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").
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Member
    Joined
    Jul 2011
    Posts
    121
    Thanks
    1

    Re: Absolute values 2

    Quote Originally Posted by HallsofIvy View Post
    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
    Follow Math Help Forum on Facebook and Google+

  11. #11
    MHF Contributor Siron's Avatar
    Joined
    Jul 2011
    From
    Norway
    Posts
    1,250
    Thanks
    20

    Re: Absolute values 2

    Quote Originally Posted by psolaki View Post
    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?
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Member
    Joined
    Jul 2011
    Posts
    121
    Thanks
    1

    Re: Absolute values 2

    Quote Originally Posted by Siron View Post
    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??
    Follow Math Help Forum on Facebook and Google+

  13. #13
    MHF Contributor Siron's Avatar
    Joined
    Jul 2011
    From
    Norway
    Posts
    1,250
    Thanks
    20

    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.
    Follow Math Help Forum on Facebook and Google+

  14. #14
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,567
    Thanks
    1409

    Re: Absolute values 2

    Quote Originally Posted by psolaki View Post
    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.
    Follow Math Help Forum on Facebook and Google+

  15. #15
    Member
    Joined
    Jul 2011
    Posts
    121
    Thanks
    1

    Re: Absolute values 2

    Quote Originally Posted by HallsofIvy View Post
    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
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: September 7th 2011, 07:11 AM
  2. Replies: 1
    Last Post: June 1st 2011, 01:47 AM
  3. Replies: 8
    Last Post: May 23rd 2010, 10:59 PM
  4. Absolute values
    Posted in the Calculus Forum
    Replies: 2
    Last Post: April 29th 2010, 04:03 PM
  5. Replies: 2
    Last Post: November 8th 2009, 01:52 PM

Search Tags


/mathhelpforum @mathhelpforum