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Thread: How do I find the value of $x$ in this equation.

  1. #1
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    How do I find the value of $x$ in this equation.

    The equation is
    $$\dfrac{1}{a}+b+x=\dfrac{1}{a}+\dfrac{1}{b}+ \dfrac{1}{x} $$

    I've tried
    $$\require{cancel}\dfrac{1+ab+ax}{a}=\dfrac{bx+ax+ ab}{abx}\\abx(1+ab+ax)=a(bx+ax+ab)\\\cancel{abx}+a ^2b^2x+a^2bx^2=\cancel{abx}+a^2x+a^2b\\a^2b^2x+a^2 bx^2=a^2x+a^2b$$

    But, don't understand how to solve further. Can somebody show step by step please. Thanks!
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  2. #2
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    Re: How do I find the value of $x$ in this equation.

    cancel $\dfrac1a$ from both the sides<br> now we have $$b+x=\frac{1}{b}+\frac{1}{x}$$ take L.C.M $$x+b=\frac{x+b}{xb}$$ $$x+b\left(1-\frac{1}{xb}\right)=0$$ from here $$x=-b~,~x=\frac1b$$
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  3. #3
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    Re: How do I find the value of $x$ in this equation.

    Quote Originally Posted by deesuwalka View Post
    cancel $\dfrac1a$ from both the sides<br> now we have $$b+x=\frac{1}{b}+\frac{1}{x}$$ take L.C.M $$x+b=\frac{x+b}{xb}$$ $$\color{red}{x+b - \frac{x+b}{xb}=0}$$ $$\color{red}{(x+b)\left(1-\dfrac{1}{xb}\right)=0}$$ from here $$x=-b~,~x=\frac1b$$
    ...
    Thanks from Archie and topsquark
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    Re: How do I find the value of $x$ in this equation.

    $x+b-\dfrac{x+b}{xb}=0$


    $x^2+\bigg(b-\dfrac{1}{b}\bigg)x-1=0$
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