Question

Solve for x: $ \dfrac{{(x - 1)}}{{(x - 2)}} + \dfrac{{(x - 3)}}{{(x - 4)}} = \dfrac{{10}}{3} $

Answer 1

Hint: Perform cross multiplication to take LCM (least common multiple) for the denominators of the terms and simplify to get the values of “x”. Since, “x” is multiplied with “x” , a simplified equation is formed in the form of a quadratic equation and uses the split of the middle term to get the required values.

Complete step-by-step answer:
Take the given expression: $ \dfrac{{(x - 1)}}{{(x - 2)}} + \dfrac{{(x - 3)}}{{(x - 4)}} = \dfrac{{10}}{3} $
Take LCM (least common multiple) for the above expression –
 $ \Rightarrow \dfrac{{(x - 1)(x - 4) + (x - 3)(x - 2)}}{{(x - 2)(x - 4)}} = \dfrac{{10}}{3} $
Expand the brackets in the above expression finding the product of the two binomials.
 $ \Rightarrow \dfrac{{{x^2} - x - 4x + 4 + {x^2} - 3x - 2x + 6}}{{{x^2} - 2x - 4x + 8}} = \dfrac{{10}}{3} $
Combine the like terms and simplify the expressions –
 $ \Rightarrow \dfrac{{{x^2} + {x^2} - x - 4x - 3x - 2x + 4 + 6}}{{{x^2} - 2x - 4x + 8}} = \dfrac{{10}}{3} $
Combine the like terms, when there are two negative terms you have to add and then give a negative sign to the resultant value.
 $ \Rightarrow \dfrac{{2{x^2} - 10x + 10}}{{{x^2} - 6x + 8}} = \dfrac{{10}}{3} $
Perform cross multiplication, when the denominator of one side is multiplied with the numerator of the opposite side.
 $ \Rightarrow 3(2{x^2} - 10x + 10) = 10({x^2} - 6x + 8) $
Open the brackets in the above expression by multiplying the term outside the bracket with the term inside the brackets.
 $ \Rightarrow 6{x^2} - 30x + 30 = 10{x^2} - 60x + 80 $
The above expression can be re-written as –
 $ \Rightarrow 10{x^2} - 60x + 80 = 6{x^2} - 30x + 30 $
Move all the terms on one side of the equation. When you move any term from one side of the equation to the opposite side then the sign of the terms also changes. Positive term becomes negative and vice-versa.
 $ \Rightarrow 10{x^2} - 60x + 80 - 6{x^2} + 30x - 30 = 0 $
Combine the like terms –
 $ \Rightarrow \underline {10{x^2} - 6{x^2}} - \underline {60x + 30x} + \underline {80 - 30} = 0 $
When you combine two terms with the bigger negative term and smaller positive term do subtraction and give negative sign to the resultant value.
 $ \Rightarrow 4{x^2} - 30x + 50 = 0 $
Take the common multiple common –
 $ \Rightarrow 2(2{x^2} - 15x + 25) = 0 $
When zero is multiplied with any number it gives zero as the resultant value.
 $ \Rightarrow 2{x^2} - 15x + 25 = 0 $
Split the middle term in such a way that its product is equal to the product of the first and last term.
 $ \Rightarrow 2{x^2} - 10x - 5x + 25 = 0 $
Make the pair of first two and last two terms-
 $ \Rightarrow \underline {2{x^2} - 10x} - \underline {5x + 25} = 0 $
Take common factors common –
 $ \Rightarrow 2x(x - 5) - 5(x - 5) = 0 $
 $ \Rightarrow (x - 5)(2x - 5) = 0 $
Here, we have two cases –
 $
   \Rightarrow x - 5 = 0 \\
   \Rightarrow x = 5 \;
  $
and
 $
   \Rightarrow 2x - 5 = 0 \\
   \Rightarrow 2x = 5 \\
   \Rightarrow x = \dfrac{5}{2} \;
  $
Hence, $ x = 2,\dfrac{5}{2} $
So, the correct answer is “ $ x = 2,\dfrac{5}{2} $ ”.

Note: Always be careful about the sign convention when you move any term from one side to the other. When you combine terms with the same sign add the values and give the common sign to the resultant value. Term multiplicative on one side if moved to the opposite side then it goes to the numerator.