Question

First find all x-values that are excluded from the solution set, then solve the rational equation for x.


$\frac{5}{x+1}$ + $\frac{1}{x-3}$ = $\frac{-6}{x^{2} -2x-3}$


Thanks in advance.

Answer 1

Answer:

• excluded: x = -1, 3
• solution: x = 4/3

Step-by-step explanation:

You want the values excluded from the domain of the rational equation 5/(x+1)+1/(x-3)=-6/(x²-2x-3), and the solutions to that equation.

Setup

It often works well to cast rational equations into the form f(x)=0 by subtracting the expression found on one side of the equation. When we do that, we get ...

  $\dfrac{5}{x+1}+\dfrac{1}{x-3}=\dfrac{-6}{x^2-2x-3}\qquad\text{given}\\\\\dfrac{5}{x+1}+\dfrac{1}{x-3}+\dfrac{6}{x^2-2x-3}=0\qquad\text{subtract right side}\\\\\dfrac{5(x-3)+1(x+1)+6}{(x+1)(x-3)}=0\qquad\text{use common denominator}\\\\\dfrac{5x-15+x+1+6}{(x+1)(x-3)}=0\qquad\text{eliminate numerator parentheses}\\\\\dfrac{6x-8}{(x+1)(x-3)}=0\qquad\text{collect terms}$

Solution

The values excluded from the solution set are the values of x that make the denominator factors zero:

  x = -1  and  x = 3  are excluded from the solution set

The fraction is zero when its numerator is zero:

  6x -8 = 0

  6x = 8 . . . . . . add 8

  x = 8/6 = 4/3 . . . . . divide by 6 and reduce the fraction

The solution is x = 4/3.

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Additional comment

By approaching the solution of rational equations in this way, we can cancel any common factors from numerator and denominator. This avoids extraneous solutions.

Here, there are no common factors and no extraneous solutions. The attached graph shows the one point of intersection between the left-side expression and the right-side expression is at x=4/3, as above.