# How do you solve (x-1)/(x-2)+(x-3)/(x-4)=10/3?

Jun 20, 2016

There is no solution to the equation.

#### Explanation:

When we are solving an EQUATION with fractions, it is possible to GET RID of the denominators altogether by multiplying each term by the LCM (same as LCD) of the denominators.

The LCM in this example is $3 \left(x - 2\right) \left(x - 4\right)$

Multiply each term by $\textcolor{red}{3 \left(x - 2\right) \left(x - 4\right)}$

$\textcolor{red}{3 \left(x - 2\right) \left(x - 4\right)} \times \frac{x - 1}{x - 2} + \frac{\textcolor{red}{3 \left(x - 2\right) \left(x - 4\right)} \times \left(x - 3\right)}{x - 4} = \frac{10}{3} \textcolor{red}{3 \left(x - 2\right) \left(x - 4\right)}$

Cancel the denominators:

$\textcolor{red}{3 \cancel{\left(x - 2\right)} \left(x - 4\right)} \frac{x - 1}{\cancel{x - 2}} + \frac{\textcolor{red}{3 \left(x - 2\right) \cancel{\left(x - 4\right)}} \left(x - 3\right)}{\cancel{x - 4}} = \frac{10}{\cancel{3}} \textcolor{red}{\cancel{3} \left(x - 2\right) \left(x - 4\right)}$

3(x -4)(x-1)+3(x-2)(x-3) =10(x-2)(x -4))
$3 \left({x}^{2} - 5 x + 1\right) + 3 \left({x}^{2} - 5 x + 6\right) = 10 \left({x}^{2} - 6 x + 8\right)$

$3 {x}^{2} - 15 x + 3 + 3 {x}^{2} - 15 x + 18 = 10 {x}^{2} - 60 x + 80$

$0 = 4 {x}^{2} - 30 x + 59$

Use the quadratic formula: $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$a = 4 , b = - 30 , c = 59$

$x = \frac{- \left(- 30\right) \pm \sqrt{{30}^{2} - 4 \times 4 \times 59}}{2 \times 4}$

$x = \frac{30 \pm \sqrt{900 - 944}}{8}$

As this gives $\sqrt{- 44}$, which is a non-real value, it means there is no solution to the equation.