Results 1 to 5 of 5

Math Help - Solving for x in terms of other letters (pls help)

  1. #1
    Junior Member
    Joined
    Feb 2007
    Posts
    53

    Solving for x in terms of other letters (pls help)

    My problem is:

    Solve for x: ax + b = bx - a

    I found a similar problem in the book, and wondered if I'm on the right track or not.

    I rewrote:

    ax - bx = -a - b

    x(a - b) = -a - b

    Code:
     
    x = -a  b
         -------
         a  b
    x = -1 ?

    I guess because each letter is only equal to 1 my only possible answer is then 1? Is that how we get this answer? I'm only guessing...
    Follow Math Help Forum on Facebook and Google+

  2. #2
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by lilrhino View Post
    My problem is:

    Solve for x: ax + b = bx - a

    I found a similar problem in the book, and wondered if I'm on the right track or not.

    I rewrote:

    ax - bx = -a - b

    x(a - b) = -a - b

    Code:
     
    x = -a  b
         -------
         a  b
    x = -1 ?

    I guess because each letter is only equal to 1 my only possible answer is then 1? Is that how we get this answer? I'm only guessing...
    no. the answer is x = \frac {-a - b}{a - b} as you have. this is not -1. to have -1 we would need it to be: x = \frac {- (a - b)}{a - b} = \frac {b - a}{a - b}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Feb 2007
    Posts
    53
    Quote Originally Posted by Jhevon View Post
    no. the answer is x = \frac {-a - b}{a - b} as you have. this is not -1. to have -1 we would need it to be: x = \frac {- (a - b)}{a - b} = \frac {b - a}{a - b}
    I thought it was supposed to equal a number not the resulting fraction. It's a little confusing anyway. I'm not sure why I was trying to make x equal a number .

    Can you please explain how this one equals 1 and the other one results to a fraction?

    ax + b = bx + a

    Thanks again! Just trying to understand this, so hopefully I'll retain it.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by lilrhino View Post
    I thought it was supposed to equal a number not the resulting fraction. It's a little confusing anyway. I'm not sure why I was trying to make x equal a number .

    Can you please explain how this one equals 1 and the other one results to a fraction?

    ax + b = bx + a

    Thanks again! Just trying to understand this, so hopefully I'll retain it.
    ok, so there were no instructions that said x must be a number. it said we should simply solve for x in terms of the other letters.

    Now recall, a fraction is equal to 1 if the numerator and denomenator are equal--and of course the denominator can't be zero. We get -1 if the numerator is the negative of the denominator, which was the case in the example i choose that we would need to get -1.

    x = \frac {-(a - b)}{a - b} =-1 \times \frac {a - b}{a - b} = -1 \times 1 = -1

    Now on to the next question:

    ax + b = bx + a ..........get all the x's on one side

    \Rightarrow ax - bx = a - b ........now factor out the common x

    \Rightarrow x(a - b) = a - b .......now divide both sides by a - b

    \Rightarrow x = \frac {a - b}{a - b}

    we have a number divided by itself, provided a \neq b we will get 1

    So, x = 1
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Feb 2007
    Posts
    53
    Quote Originally Posted by Jhevon View Post
    ok, so there were no instructions that said x must be a number. it said we should simply solve for x in terms of the other letters.

    Now recall, a fraction is equal to 1 if the numerator and denomenator are equal--and of course the denominator can't be zero. We get -1 if the numerator is the negative of the denominator, which was the case in the example i choose that we would need to get -1.

    x = \frac {-(a - b)}{a - b} =-1 \times \frac {a - b}{a - b} = -1 \times 1 = -1

    Now on to the next question:

    ax + b = bx + a ..........get all the x's on one side

    \Rightarrow ax - bx = a - b ........now factor out the common x

    \Rightarrow x(a - b) = a - b .......now divide both sides by a - b

    \Rightarrow x = \frac {a - b}{a - b}

    we have a number divided by itself, provided a \neq b we will get 1

    So, x = 1
    Thanks for the explanation Jhevon!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Solving Y in terms of X
    Posted in the Algebra Forum
    Replies: 3
    Last Post: October 28th 2010, 12:26 PM
  2. solving y=tanh(y-x) in terms of x
    Posted in the Calculus Forum
    Replies: 3
    Last Post: March 9th 2010, 02:34 PM
  3. Solving for One Variable in Terms of Others...
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: June 28th 2009, 12:12 PM
  4. solving for y in terms of x
    Posted in the Algebra Forum
    Replies: 7
    Last Post: December 15th 2008, 07:47 PM
  5. combining like terms and solving equation.
    Posted in the Algebra Forum
    Replies: 4
    Last Post: February 27th 2007, 09:16 AM

Search Tags


/mathhelpforum @mathhelpforum