How do you solve the system # [(x)/(x+5)]- [(5)/(5-x)] = (x+31)/(x^2-25)#?

1 Answer
Jan 30, 2016

Answer:

Multiply numerators and denominators to get a common denominator #x^2-25# then equate the numerators to get a quadratic, hence #x = 3# or #x=-2#

Explanation:

#x/(x+5)-5/(5-x)#

#=x/(x+5)+5/(x-5)#

#=(x(x-5))/((x-5)(x+5)) + (5(x+5))/((x-5)(x+5))#

#=(x^2-color(red)(cancel(color(black)(5x)))+color(red)(cancel(color(black)(5x)))+25)/(x^2-25)#

#=(x^2+25)/(x^2-25)#

So we want to solve #x^2+25 = x+31# with exclusion #x != +-5#

Subtract #x+31# from both sides to get:

#0 = x^2-x-6 = (x-3)(x+2)#

So #x=3# or #x=-2#