Question

Find the roots of the equation
$\dfrac{3}{x+1}-\dfrac{1}{2}=\dfrac{2}{3x-1}$, $x\ne -1,\dfrac{1}{3}$

Answer 1

We solve this question by first considering the LHS of the given equation. Then we find the LCM of denominators and multiply them to make the denominators equal to LCM. Then we simplify it and substitute the value obtained in the LHS of the given equation. Then we apply cross multiplication and simplify it to find an equation. Then we factorize the obtained equation and find the roots of the given equation.

Complete step-by-step solution
Let us consider the given equation.
$\dfrac{3}{x+1}-\dfrac{1}{2}=\dfrac{2}{3x-1}............\left( 1 \right)$
Now let us consider the left-hand side of the above equation.
$\Rightarrow \dfrac{3}{x+1}-\dfrac{1}{2}$
As we see in the above equation both the fractions have different denominators. So, let us find the LCM of the denominators.
Here, $LCM=2\left( x+1 \right)$
So, multiplying the numerator and denominator to make the denominator of both the fractions equal to LCM.
So, we get,
$\begin{aligned} & \Rightarrow \left( \dfrac{3}{x+1}\times \dfrac{2}{2} \right)-\left( \dfrac{1}{2}\times \dfrac{x+1}{x+1} \right) \\\ & \Rightarrow \dfrac{6}{2\left( x+1 \right)}-\dfrac{x+1}{2\left( x+1 \right)} \\\ \end{aligned}$
Now the denominators of both the fractions are the same. So, let us subtract the fractions.
$\Rightarrow \dfrac{5-x}{2\left( x+1 \right)}$
So, we get that
$\Rightarrow \dfrac{3}{x+1}-\dfrac{1}{2}=\dfrac{5-x}{2\left( x+1 \right)}$
Substituting this in equation (1) we get,
$\Rightarrow \dfrac{5-x}{2\left( x+1 \right)}=\dfrac{2}{3x-1}$
As we are given that $x\ne -1,\dfrac{1}{3}$, let us apply cross multiplication. Then we get,
$\begin{aligned} & \Rightarrow \left( 5-x \right)\left( 3x-1 \right)=2\left( 2\left( x+1 \right) \right) \\\ & \Rightarrow \left( 5-x \right)\left( 3x-1 \right)=4\left( x+1 \right) \\\ & \Rightarrow 15x-5-3{{x}^{2}}+x=4x+4 \\\ & \Rightarrow 16x-5-3{{x}^{2}}=4x+4 \\\ & \Rightarrow 4x+4-16x+5+3{{x}^{2}}=0 \\\ & \Rightarrow 3{{x}^{2}}-12x+9=0 \\\ \end{aligned}$
So, we get the equation
$\Rightarrow 3{{x}^{2}}-12x+9=0$
Now we can take 3 common from the above equation.
$\begin{aligned} & \Rightarrow 3\left( {{x}^{2}}-4x+3 \right)=0 \\\ & \Rightarrow {{x}^{2}}-4x+3=0 \\\ \end{aligned}$
Now we can factorise the above equation as,
$\Rightarrow {{x}^{2}}-x-3x+3=0$
Now we can take $x$ common from the first two terms and -3 common from the last two terms. Then we get,
$\begin{aligned} & \Rightarrow x\left( x-1 \right)-3\left( x-1 \right)=0 \\\ & \Rightarrow \left( x-3 \right)\left( x-1 \right)=0 \\\ & \Rightarrow x=1,3 \\\ \end{aligned}$
So, the roots of the equation $\dfrac{3}{x+1}-\dfrac{1}{2}=\dfrac{2}{3x-1}$ as $x=1,3$.
Hence the answer is $x=1,3$.

Note: We can also solve the equation obtained above, that is ${{x}^{2}}-4x+3=0$ by using the formula for the roots of the quadratic equation $a{{x}^{2}}+bx+c=0$.
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Using it we get the values of x as,
$\begin{aligned} & \Rightarrow x=\dfrac{-\left( -4 \right)\pm \sqrt{{{\left( -4 \right)}^{2}}-4\left( 1 \right)\left( 3 \right)}}{2\left( 1 \right)} \\\ & \Rightarrow x=\dfrac{4\pm \sqrt{16-12}}{2} \\\ & \Rightarrow x=\dfrac{4\pm \sqrt{4}}{2} \\\ \end{aligned}$
Taking the square root of 4 we get,
$\begin{aligned} & \Rightarrow x=\dfrac{4\pm 2}{2} \\\ & \Rightarrow x=\dfrac{4-2}{2},\dfrac{4+2}{2} \\\ & \Rightarrow x=\dfrac{2}{2},\dfrac{6}{2} \\\ & \Rightarrow x=1,3 \\\ \end{aligned}$
Hence answer is $x=1,3$.