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

But, don't understand how to solve further. Can somebody show step by step please. Thanks!

• My first step would be to subtract $\frac{1}{a}$ from both sides... – JMoravitz Mar 4 '17 at 6:53

You are on the right track, then you have

$bx^2+x(b^2-1)-b=0$

You solve the quadratic equation for x and you get: $x=\dfrac{1}{b}$ or $x=-b$

Since

$$\dfrac{1}{a}+b+x=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{x}\quad\iff\quad b+x=\frac1b+\frac1x\quad\iff\quad x^2+\left(b-\frac1b\right)x-1=0$$ Factorization give us $$\left(x+b\right)\left(x-\frac1b\right)=0$$ Thus $x=-b$ or $x=\frac1b$.

subtracting $$\frac{1}{a}$$ on both sides we have $$b+x=\frac{1}{b}+\frac{1}{x}$$ then we have $$x-\frac{1}{x}=\frac{1-b^2}{b}$$ and after multiplying by $$x$$ we have $$x^2-\frac{1-b^2}{b}x-1=0$$ you Need to use the quadratic equation formula additionally it must be $$a,b,x\ne 0$$

$$\dfrac{1}{a}+b+x=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{x}$$ Take the LCM at right side $$\Rightarrow\dfrac{1}{a}+b+x=\dfrac{ab+ax+bx}{abx}$$ $$\Rightarrow abx\bigg(\dfrac{1}{a}+b+x\bigg)=ab+ax+bx$$ $$\Rightarrow bx+ab^2x+abx^2=ab+ax+bx$$ Take $x$ common from first two terms and last two terms $$\Rightarrow x(b+ab^2)+abx^2=ab+x(a+b)$$ Subtract $ab+x(a+b)$ from both sides $$\Rightarrow -ab-x(a+b)+x(b+ab^2)+abx^2=0$$ Simplifying it we get $$\Rightarrow -ab+x(ab^2-a)+abx^2=0$$ Rewrite it $$\Rightarrow abx^2+(ab^2-a)x-ab=0$$ It’s a quadratic equation so, factoring out it we get $$\Rightarrow a(bx^2+b^2x-x-b)=0$$ $$\Rightarrow a(bx(x+b)-1(x+b))=0$$ $$\Rightarrow a(b+x)(bx-1)=0$$ $$x=\dfrac{1}{b}$$ $$x=-b$$

cancel $\dfrac1a$ from both the sides
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 $$x=-b~,~x=\frac1b$$

\begin{align} \dfrac{1}{a}+b+x&=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{x}\\ b+x&=\frac 1b+\frac 1x\\ x-\frac 1x&=\frac 1b-b=-\left(b-\frac 1b\right)\\ \color{red}{x}&\color{red}{=\frac 1b, -b}\end{align}