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

1. ## 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!

2. ## 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 $$x=-b~,~x=\frac1b$$

3. ## Re: How do I find the value of $x$ in this equation.

Originally Posted by deesuwalka
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 $$x=-b~,~x=\frac1b$$
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4. ## Re: How do I find the value of $x$ in this equation.

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

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