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Math Help - Linear equation

  1. #1
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    Linear equation

    Hey hopefully some one can help me with this cuase i cant get it for the life of me

    solve for 'x'

    b-cx over a + a -cx over b + 2 = 0

    also im new so i dont know how to get all signs and stuff lol

    Thanks for the help.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by sinners View Post
    Hey hopefully some one can help me with this cuase i cant get it for the life of me

    solve for 'x'

    b-cx over a + a -cx over b + 2 = 0

    also im new so i dont know how to get all signs and stuff lol

    Thanks for the help.
    \frac {b - cx}a + \frac {a - cx}b + 2 = 0 ..........split up the fractions

    \Rightarrow \frac ba - \frac {cx}a + \frac ab - \frac {cx}b + 2 = 0 .............group like terms and factor out the x

    \Rightarrow \left( - \frac ca - \frac cb \right)x + \frac ba + \frac ab + 2 = 0

    can you continue?


    i have not come to help the math geniuses, but to help sinners do his problem... ...did anyone but me get that?
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  3. #3
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    this is what i did next i dont know if it is right or not.

    (-cb over ab - ca over ba) x + b^2 over ab + a^2 over ba + 2ab = 0 so there is a common denominator then crossed all the common denominators out. So its..

    (-cb - ca)x + b^2 + a^2 + 2ab = 0 i then moved the x part over to the other side so its...

    b^2 + a^2 + 2ab = 0 over ac + bc then i moved the rest of equation over to the other side so it is now...

    b^2 + a^2 + 2ab over ac + bc is this the correct answer? i am really confused because the book had the answer as x = a^2 + b^2 + 2ab over ac + bc = a+b over c. Is the book right or am i? also can some one show me how to actually show up as a over b instead of typing it

    thanks in advance
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by sinners View Post
    this is what i did next i dont know if it is right or not.

    (-cb over ab - ca over ba) x + b^2 over ab + a^2 over ba + 2ab = 0 so there is a common denominator then crossed all the common denominators out. So its..

    (-cb - ca)x + b^2 + a^2 + 2ab = 0 i then moved the x part over to the other side so its...

    b^2 + a^2 + 2ab = 0 over ac + bc then i moved the rest of equation over to the other side so it is now...

    b^2 + a^2 + 2ab over ac + bc is this the correct answer? i am really confused because the book had the answer as x = a^2 + b^2 + 2ab over ac + bc = a+b over c. Is the book right or am i?
    the book is right, it hard to read through your work though because of how you typed it, so i will show you the solution.

    so we were at:

    \left( - \frac ca - \frac cb \right)x + \frac ba + \frac ab + 2 = 0 ............bring the x term on the other side, we are solving for x after all

    \Rightarrow \left( \frac ca + \frac cb \right)x = \frac ba + \frac ab + 2 .............combine all the fractions

    \Rightarrow \left( \frac {ac + bc}{ab}\right)x = \frac {a^2 + b^2 + 2ab}{ab} .........divide both sides by (ac + bc)/ab

    \Rightarrow x = \frac {a^2 + b^2 + 2ab}{ab} \cdot \frac {ab}{ac + bc} .............the ab's cancel, now factorize the other terms

    \Rightarrow x = \frac {(a + b)^2}{ab} \cdot \frac {ab}{c(a + b)}

    \Rightarrow x = \frac {(a + b)^2}{c(a + b)}

     \Rightarrow x = \frac {a + b}{c}

    also can some one show me how to actually show up as a over b instead of typing it
    use LaTex, learn how to here

    or simply type a/b to mean a over b. but please, use parentheses if you have a lot of terms. that is, type (ac - ab)/(b + c) to mean \frac {ac - ab}{b + c}
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