Question

Solve the following for x:
$\dfrac{1}{\left( a+b+x \right)}=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{x},\left[ a\ne 0,b\ne 0,x\ne 0,x\ne -\left( a+b \right) \right]$

Answer 1

First of all take the L.C.M on the right hand side of the given equation and then cross multiply the given equation. After cross multiplying the equation you will find quadratic in x so using the factorization method you can find the values of x.

Complete step-by-step answer:
The equation in which we have to solve the value of x is given as:
$\dfrac{1}{\left( a+b+x \right)}=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{x},\left[ a\ne 0,b\ne 0,x\ne 0,x\ne -\left( a+b \right) \right]$
Taking L.C.M of the denominators of the right hand side of the above equation we get,
$\dfrac{1}{\left( a+b+x \right)}=\dfrac{bx+ax+ab}{abx}$
Now, taking x as common in the numerator of the right hand side of the above equation we get,
$\dfrac{1}{\left( a+b+x \right)}=\dfrac{\left( a+b \right)x+ab}{abx}$
On cross multiplying the above equation we get,
$abx=\left( \left( a+b \right)+x \right)\left( \left( a+b \right)x+ab \right)$
Multiplying the terms written in the first bracket of the right hand side of the above equation with terms written in the second bracket we get,
$\begin{aligned} & abx=\left( a+b \right)\left( a+b \right)x+\left( a+b \right)ab+x\left( a+b \right)x+x\left( ab \right) \\\ & \Rightarrow abx={{\left( a+b \right)}^{2}}x+\left( a+b \right)ab+{{x}^{2}}\left( a+b \right)+x\left( ab \right) \\\ \end{aligned}$
In the above equation “abx” will be cancelled from both the sides of the above equation.
$0={{\left( a+b \right)}^{2}}x+\left( a+b \right)ab+{{x}^{2}}\left( a+b \right)$
As you can see that $\left( a+b \right)$ is common in the all the terms on the right hand side of the above equation so taking $\left( a+b \right)$ out from the terms we get,
$\left( a+b \right)\left[ \left( a+b \right)x+ab+{{x}^{2}} \right]=0$
Now, equating the terms written in the square bracket to 0 we get,
$\left[ \left( a+b \right)x+ab+{{x}^{2}} \right]=0$
The above equation is quadratic in x so we are going to factorize this quadratic equation as follows:
$\begin{aligned} & {{x}^{2}}+\left( a+b \right)x+ab=0 \\\ & \Rightarrow {{x}^{2}}+ax+bx+ab=0 \\\ \end{aligned}$
Now, taking x as common from the first two terms and b as common from the last two terms of the above equation we get,
$x\left( x+a \right)+b\left( x+a \right)=0$
As you can see that $\left( x+a \right)$ is common in both the terms in the above equation so taking $\left( x+a \right)$ out from the two terms will reduce the above equation to:
$\left( x+a \right)\left( x+b \right)=0$
Equating $\left( x+a \right)\And \left( x+b \right)$ to 0 we get,
$\begin{aligned} & x+a=0 \\\ & \Rightarrow x=-a \\\ & x+b=0 \\\ & \Rightarrow x=-b \\\ \end{aligned}$
The values of x from the above solution are –a and –b.
Hence, the solutions of x are –a and –b.

Note: You should check the values of x that we have got above by comparing these values with the conditions on x given in question. The condition on x given in the question are as follows:
$\left[ a\ne 0,b\ne 0,x\ne 0,x\ne -\left( a+b \right) \right]$
The values of x that we have got above are –a and –b. As you can see that these values of x are not coinciding with any of the conditions on x given in the question so both the values of x that we have solved are acceptable.