Question

How do you solve $\dfrac{{x - 5}}{{x - 8}} + \dfrac{{x - 6}}{{x - 7}} = \dfrac{{x + 97}}{{{x^2} - 15x + 56}}$ and check for extraneous solutions?

Answer 1

In this type of question we simplify the left hand side of the equation by taking LCM. After equating the simplified form and simplifying we will have a quadratic equation. We can solve the quadratic equation by factorization. While finding the extraneous solution we concentrate on the left hand side of the equation.

Complete step by step solution:
Given,
$\dfrac{{x - 5}}{{x - 8}} + \dfrac{{x - 6}}{{x - 7}} = \dfrac{{x + 97}}{{{x^2} - 15x + 56}}$.
Now taking left hand side of the equation
$= \dfrac{{x - 5}}{{x - 8}} + \dfrac{{x - 6}}{{x - 7}}$
Taking LCM we have
$= \dfrac{{(x - 5)(x - 7) + (x - 6)(x - 8)}}{{(x - 8)(x - 7)}}$
$= \dfrac{{\left( {x(x - 7) - 5(x - 7)} \right) + \left( {x(x - 8) - 6(x - 8)} \right)}}{{x(x - 7) - 8(x - 7)}}$
$= \dfrac{{\left( {{x^2} - 7x - 5x + 35} \right) + \left( {{x^2} - 8x - 6x + 48} \right)}}{{{x^2} - 7x - 8x + 56}}$
Separating the like terms and adding we have,
$= \dfrac{{{x^2} + {x^2} - 7x - 5x - 8x - 6x + 48 + 35}}{{{x^2} - 7x - 8x + 56}}$
$= \dfrac{{2{x^2} - 26x + 83}}{{{x^2} - 15x + 56}}$
Now the original problem will become
$\Rightarrow \dfrac{{2{x^2} - 26x + 83}}{{{x^2} - 15x + 56}} = \dfrac{{x + 97}}{{{x^2} - 15x + 56}}$
Cancelling the denominator term we have
$2{x^2} - 26x + 83 = x + 97$
Shifting all the terms to the left hand side of the equation,
$2{x^2} - 26x - x + 83 - 97 = 0$
$2{x^2} - 27x + - 14 = 0$
Now we can split the middle term,
$2{x^2} - 28x + x - 14 = 0$
Taking ‘2x’ common in the first two terms and taking 1 common in the remaining terms ,
$2x(x - 14) + 1(x - 14) = 0$
Again taking $(x - 14)$ as common,
$(x - 14)(2x + 1) = 0$
Using zero product principle we have
$(x - 14) = 0$ and $(2x + 1) = 0$
$x = 14$ and $2x = - 1$
$x = 14$ and $x = - \dfrac{1}{2}$.
Hence, the solution of $\dfrac{{x - 5}}{{x - 8}} + \dfrac{{x - 6}}{{x - 7}} = \dfrac{{x + 97}}{{{x^2} - 15x + 56}}$ is $x = 14$ and $x = - \dfrac{1}{2}$ .
Here there is no extraneous solution as none of the roots is equal to 7 and 8.

Note: Since we have a polynomial of degree 2. Hence we have two roots. We know that roots are also called zeros. If we are unable to solve the obtained quadratic equation by factorization we use a quadratic formula to solve the equation. If we have $x = 7$ and $x = 8$ then 7 and 8 are extraneous solutions. But we have $x = 14$ and $x = - \dfrac{1}{2}$if we substitute this in the given equation it satisfies.
Let’s check for $x = 14$,
$\dfrac{{14 - 5}}{{14 - 8}} + \dfrac{{14 - 6}}{{14 - 7}} = \dfrac{{14 + 97}}{{{{(14)}^2} - 15(14) + 56}}$
$\dfrac{9}{6} + \dfrac{8}{7} = \dfrac{{111}}{{196 - 210 + 56}}$
$\dfrac{9}{6} + \dfrac{8}{7} = \dfrac{{111}}{{42}}$
$\dfrac{{63 + 48}}{{42}} = \dfrac{{111}}{{42}}$
$\Rightarrow \dfrac{{111}}{{42}} = \dfrac{{111}}{{42}}$. It will satisfy for $x = - \dfrac{1}{2}$.
In other words extraneous solutions are solutions to an equation that seem to be right, but when checked by substituting it into the original equation turns out not to be right.