# How do you solve the rational equation (x+1)/(x-1)=x/3 + 2/(x-1)?

May 4, 2018

$x = 3$

#### Explanation:

Here we go...

$\frac{x + 1}{x - 1}$ = $\frac{x}{3}$ + $\frac{2}{x - 1}$

$\frac{x + 1}{x - 1}$ = $\frac{{x}^{2} - x + 6}{3 x - 3}$

$\left(x + 1\right) \left(3 x - 3\right)$ = $\left(x - 1\right) \left({x}^{2} - x + 6\right)$

$3 {x}^{2} - 3$ = ${x}^{3} - 2 {x}^{2} + 7 x - 6$

${x}^{3} - 5 {x}^{2} + 7 x - 3 = 0$

By factorizing this equation we get :

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

$\therefore$ $x = 1 \mathmr{and} 3$

But as observed in the given question $x - 1$ is in denominator, So $x$ cannot be equal to 1.

Hence, $x = 3$

May 4, 2018

$x = 3$

#### Explanation:

We have an equation with fractions.

It is possible to get rid of the denominators immediately by multiplying each term by the LCM of the denominators, which in this case is $\textcolor{b l u e}{3 \left(x - 1\right)}$

$\frac{\textcolor{b l u e}{3 \cancel{\left(x - 1\right)}} \times \left(x + 1\right)}{\cancel{\left(x - 1\right)}} = \frac{\textcolor{b l u e}{\cancel{3} \left(x - 1\right)} \times x}{\cancel{3}} + \frac{2 \textcolor{b l u e}{\times 3 \cancel{\left(x - 1\right)}}}{\cancel{\left(x - 1\right)}}$

This leaves us with:

$3 \left(x + 1\right) = \left(x - 1\right) x + 6 \text{ } \leftarrow$ no fractions

$3 x + 3 = {x}^{2} - x + 6 \text{ } \leftarrow$ remove the brackets

$0 = {x}^{2} - 4 x + 3 \text{ } \leftarrow$ make the quadratic equal to $0$

$\left(x - 3\right) \left(x - 1\right) = 0 \text{ } \leftarrow$ factorise

If $x - 3 = 0 , \text{ } \Rightarrow x = 3$

If $x - 1 = 0 , \text{ } \Rightarrow x = 1$

However, $x = 1$ is an extraneous solution.
In this equation, $x \ne 1$ because that will make the denominators $0$.

Therefore there is only one solution.

$x = 3$