Question

How do you solve: $\dfrac{{x - 1}}{{x - 2}} + \dfrac{{x - 3}}{{x - 4}} = \dfrac{{10}}{3}$?

Answer 1

To solve this question, we will first take the LCM, then we will cross multiply the denominators. Further we will expand the terms in the brackets. Then we will shift all the terms to LHS and make the RHS equal to zero. After that we will solve the quadratic equation to obtain the value of $x$.

Complete step by step solution:
We have;
$\dfrac{{x - 1}}{{x - 2}} + \dfrac{{x - 3}}{{x - 4}} = \dfrac{{10}}{3}$
Now we take the LCM of the denominators and multiply correspondingly. So, we get;
$\Rightarrow \dfrac{{\left( {x - 1} \right)\left( {x - 4} \right) + \left( {x - 3} \right)\left( {x - 2} \right)}}{{\left( {x - 2} \right)\left( {x - 4} \right)}} = \dfrac{{10}}{3}$
Now we will multiply the terms and expand the brackets. So, we have;
$\Rightarrow \dfrac{{{x^2} - 4x - x + 4 + {x^2} - 2x - 3x + 6}}{{{x^2} - 4x - 2x + 8}} = \dfrac{{10}}{3}$
Adding and subtracting the like terms in the numerator and denominator we get;
$\Rightarrow \dfrac{{2{x^2} - 10x + 10}}{{{x^2} - 6x + 8}} = \dfrac{{10}}{3}$
Now we will cross multiply the denominators. So, we get;
$\Rightarrow 3\left( {2{x^2} - 10x + 10} \right) = 10\left( {{x^2} - 6x + 8} \right)$
Now we will expand the brackets. So, we have;
$\Rightarrow 6{x^2} - 30x + 30 = 10{x^2} - 60x + 80$
Now we will shift all the terms to RHS and add and subtract the like terms. So, we get;
$\Rightarrow - 4{x^2} + 30x - 50 = 0$
Multiplying both sides with minus one we get;
$\Rightarrow 4{x^2} - 30x + 50 = 0$
Now we will solve this quadratic by middle term splitting. So, we get;
$\Rightarrow 4{x^2} - 20x - 10x + 50 = 0$
Now we will group the terms and take out the common. So, we have;
$\Rightarrow 4x\left( {x - 5} \right) - 10\left( {x - 5} \right) = 0$
Now we will take $\left( {x - 5} \right)$ common. So, we have;
$\Rightarrow \left( {4x - 10} \right)\left( {x - 5} \right) = 0$
Now we will equate each term separately to zero.
$\therefore 4x - 10 = 0$
$\Rightarrow x = \dfrac{{10}}{4}$
On simplification we get;
$\Rightarrow x = \dfrac{5}{2}$
Also,
$x - 5 = 0$
$\Rightarrow x = 5$
So, we have $\Rightarrow x = 5,\dfrac{5}{2}$.

Note: One important point to note is that in the solution we have used the middle term splitting method to solve the quadratic equation but it is not always possible to use this method. In those cases, we can use the direct formula for finding the roots of a quadratic equation which is given by:
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Where, $b$ is the coefficient of $x$, $a$ is the coefficient of ${x^2}$ and $c$ is the constant term.