# How do you solve (x+1)/(x-1)=2/(2x-1)+2/(x-1)?

$x = \frac{3}{2} \text{ }$ the root

$x = 1 \text{ " }$ the extraneous root

#### Explanation:

From the give equation, we multiply both sides by the
LCD$= \left(2 x - 1\right) \left(x - 1\right)$

$\frac{x + 1}{x - 1} = \frac{2}{2 x - 1} + \frac{2}{x - 1}$

$\left(2 x - 1\right) \left(x - 1\right) \cdot \frac{x + 1}{x - 1} = \left(2 x - 1\right) \left(x - 1\right) \left[\frac{2}{2 x - 1} + \frac{2}{x - 1}\right]$

We simplify to obtain

$\left(2 x - 1\right) \cancel{x - 1} \cdot \frac{x + 1}{\cancel{x - 1}} = \left(2 x - 1\right) \left(x - 1\right) \left[\frac{2}{2 x - 1} + \frac{2}{x - 1}\right]$

$\left(2 x - 1\right) \left(x + 1\right) = \left(2 x - 1\right) \left(x - 1\right) \left(\frac{2}{2 x - 1}\right) + \left(2 x - 1\right) \left(x - 1\right) \left(\frac{2}{x - 1}\right)$

$\left(2 x - 1\right) \left(x + 1\right) = \cancel{\left(2 x - 1\right)} \left(x - 1\right) \left(\frac{2}{\cancel{2 x - 1}}\right) + \left(2 x - 1\right) \cancel{\left(x - 1\right)} \left(\frac{2}{\cancel{x - 1}}\right)$

$\left(2 x - 1\right) \left(x + 1\right) = \left(x - 1\right) \left(2\right) + \left(2 x - 1\right) \left(2\right)$

We expand by multiplication

$2 {x}^{2} - x + 2 x - 1 = 2 x - 2 + 4 x - 2$

Transpose all terms to the left side of the equation

$2 {x}^{2} - x + 2 x - 1 - 2 x + 2 - 4 x + 2 = 0$

$2 {x}^{2} - 5 x + 3 = 0$

We can solve this now by Factoring method

$\left(2 x - 3\right) \left(x - 1\right) = 0$

Equate each factor to 0 and solve for the values of x

First factor
$2 x - 3 = 0$

$x = \frac{3}{2}$

Second Factor
$x - 1 = 0$
$x = 1$

By checking these values in the original equation, we will find that $x = 1$ is an extraneous root because it will have a division by zero

The value $x = \frac{3}{2}$ is a Root.

God bless....I hope the explanation is useful.

Mar 26, 2016

$x = \frac{3}{2}$

#### Explanation:

Before starting, it is important to note the restrictions in this equation. When the denominator of each fraction is set to not equal $0$, the restrictions are:

$x - 1 \ne 0 \textcolor{w h i t e}{X X X X X X X X} 2 x - 1 \ne 0$

$x \ne 1 \textcolor{w h i t e}{X X X X X X X X X X} 2 x \ne 1$

$\textcolor{w h i t e}{X X X X X X X X X X X X X} x \ne \frac{1}{2}$

Thus, the restrictions are $\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} x \ne 1 , \frac{1}{2} \textcolor{w h i t e}{\frac{a}{a}} |}}}$.

Solving the Equation
$1$. Start by finding the L.C.D. (lowest common denominator) for the right side of the equation. Then rewrite the fractions on the right side with the inclusion of the L.C.D.

$\frac{x + 1}{x - 1} = \frac{2}{2 x - 1} + \frac{2}{x - 1}$

$\frac{x + 1}{x - 1} = \frac{2 \textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{\left(x - 1\right)}}{\left(2 x - 1\right) \textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{\left(x - 1\right)}} + \frac{2 \textcolor{b l u e}{\left(2 x - 1\right)}}{\left(x - 1\right) \textcolor{b l u e}{\left(2 x - 1\right)}}$

$\frac{x + 1}{x - 1} = \frac{2 \left(x - 1\right) + 2 \left(2 x - 1\right)}{\left(2 x - 1\right) \left(x - 1\right)}$

$2$. Simplify the right side of the equation.

$\frac{x + 1}{x - 1} = \frac{2 x - 2 + 4 x - 2}{\left(2 x - 1\right) \left(x - 1\right)}$

$\frac{x + 1}{x - 1} = \frac{6 x - 4}{\left(2 x - 1\right) \left(x - 1\right)}$

$3$. Multiply both sides of the equation by $\textcolor{t e a l}{\left(x - 1\right)}$ and $\textcolor{t e a l}{\left(2 x - 1\right)}$ to get rid of the denominator.

$\frac{x + 1}{\textcolor{red}{\cancel{\textcolor{b l a c k}{\left(x - 1\right)}}}} \times \textcolor{t e a l}{\left(2 x - 1\right) \textcolor{red}{\cancel{\textcolor{t e a l}{\left(x - 1\right)}}}} = \frac{6 x - 4}{\textcolor{red}{\cancel{\textcolor{b l a c k}{\left(2 x - 1\right) \left(x - 1\right)}}}} \times \textcolor{red}{\cancel{\textcolor{t e a l}{\left(2 x - 1\right) \left(x - 1\right)}}}$

$\left(x + 1\right) \left(2 x - 1\right) = 6 x - 4$

$4$. Expand the brackets on the left side of the equation.

$2 {x}^{2} + x - 1 = 6 x - 4$

$5$. Bring all the terms to the left side of the equation.

$\textcolor{v i o \le t}{2} {x}^{2}$ $\textcolor{b r o w n}{- 5} x$ $\textcolor{t u r q u o i s e}{+ 3} = 0$

$6$. Use the quadratic formula to solve for the values of $x$.

$\textcolor{v i o \le t}{a = 2} \textcolor{w h i t e}{X X X X X X} \textcolor{b r o w n}{b = - 5} \textcolor{w h i t e}{X X X X X X} \textcolor{t u r q u o i s e}{c = 3}$

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$x = \frac{- \left(\textcolor{b r o w n}{- 5}\right) \pm \sqrt{{\left(\textcolor{b r o w n}{- 5}\right)}^{2} - 4 \left(\textcolor{v i o \le t}{2}\right) \left(\textcolor{t u r q u o i s e}{3}\right)}}{2 \left(\textcolor{v i o \le t}{2}\right)}$

$x = \frac{5 \pm \sqrt{1}}{4}$

$x = \frac{6}{4} \textcolor{w h i t e}{X} , \textcolor{w h i t e}{X} \frac{4}{4}$

$x = \frac{3}{2} \textcolor{w h i t e}{X} , \textcolor{w h i t e}{X} 1$

However, looking back at the restrictions, ( $\textcolor{red}{x \ne 1 , \frac{1}{2}}$ ), $x = 1$ is not a valid solution. Therefore:

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} x = \frac{3}{2} \textcolor{w h i t e}{\frac{a}{a}} |}}}$