Question

How do you solve \[\dfrac{{x - 4}}{{x - 7}} + \dfrac{{x - 5}}{{x - 6}} = \dfrac{{x + 164}}{{{x^2} - 13x + 42}}\] and check for extraneous solutions?

Answer 1

Hint: In this question, we are given an algebraic expression, we have to first simplify this expression and eliminate all the terms in the denominator so that the equation can be further solved for the values of x. After simplifying the expression, we will get a polynomial equation; a polynomial expression has exactly as many roots as the degree of the equation, so on observing the degree, we can find the roots of the equation by a suitable method. If the obtained equation is quadratic, we can solve it by factorization. In case we are not able to factorize the equation, we use the quadratic formula.

Complete step-by-step answer:
Given,
\[
  \dfrac{{x - 4}}{{x - 7}} + \dfrac{{x - 5}}{{x - 6}} = \dfrac{{x + 164}}{{{x^2} - 13x + 42}} \\
  \dfrac{{(x - 4)(x - 6) + (x - 5)(x - 7)}}{{(x - 7)(x - 6)}} = \dfrac{{x + 164}}{{{x^2} - 13x + 42}} \\
   \Rightarrow \dfrac{{{x^2} - 10x + 24 + {x^2} - 12x + 35}}{{{x^2} - 13x + 42}} = \dfrac{{x + 164}}{{{x^2} - 13x + 42}} \\
   \Rightarrow 2{x^2} - 22x + 59 = x + 164 \\
   \Rightarrow 2{x^2} - 23x - 105 = 0 \;
 \]
The equation obtained is a quadratic equation that can be solved by factorization as follows –
 $
  2{x^2} - 30x + 7x - 105 = 0 \\
   \Rightarrow 2x(x - 15) + 7(x - 15) = 0 \\
   \Rightarrow (2x + 7)(x - 15) = 0 \\
   \Rightarrow x = - \dfrac{7}{2},\,x = 15 \;
  $
Hence the factors of \[\dfrac{{x - 4}}{{x - 7}} + \dfrac{{x - 5}}{{x - 6}} = \dfrac{{x + 164}}{{{x^2} - 13x + 42}}\] are $ x + \dfrac{7}{2} = 0 $ and $ x - 15 = 0 $
So, the correct answer is “$x = - \dfrac{7}{2},\,x = 15$”.

Note: There are different ways to solve the equations of different degrees, if a polynomial has a degree greater than 2, then we find out a solution by hit and trial method and then divide the whole equation by that factor and get an equation of degree 1 less than the original equation. We continue this procedure until the degree of the equation obtained is two whose solutions can be obtained by factoring or quadratic formula or completing the square method (After finding one root, we can solve the equations of higher degree by factorization too but the process is confusing and long so we simply follow this method).