Question 1089527
Write the value or values of the variable that make a denominator zero. These are the restrictions on the variable. Part B. Keeping the restrictions in mind, solve the equation. 
 A. {{{1/((x-4)) - 5/((x+2)) = 6/((x^2-2x-8))}}}
Factor
{{{1/((x-4)) - 5/((x+2)) = 6/((x-4)(x+2))}}}
multiply by (x-4)(x+2), cancel the denominators and we have
(x+2) - 5(x-4) = 6
x + 2 - 5x + 20 = 6
-4x + 22 = 6
-4x = 6 - 22
-4x = -16
x = -16/-4
x = +4, makes the denominator 0 in the first and last fractions
:
 B. {{{5/x = 10/((3x + 4))}}}
Cross multiply
5(3x+4) = 10x
15x + 20 = 10x
15x - 10x = -20
5x = -20
 x = -20/5
 x =-4