How do you solve #(4-8x)/(x+1)=8/(x+1)# and find any extraneous solutions?

1 Answer
May 21, 2016

Answer:

Solve using the rule #a/b = m/n -> a xx n = bxxm#

Explanation:

#8(x + 1) = (4 - 8x)(x + 1)#

#8x + 8 = 4x - 8x^2 - 8x + 4#

#8x^2 + 12x + 4 = 0#

#4(2x^2 + 3x + 1) = 0#

#4(2x^2 + 2x + x + 1) = 0#

#4(2x(x + 1) + 1(x + 1)) = 0#

#4(2x + 1)(x + 1) = 0#

#x = -1/2 and -1#

Since #x = -1# in the original équation makes the denominator equal to 0, and division by 0 in mathematics is undefined, this is an extraneous solution.

Therefore, our solution set is #{x = -1/2}#.

Hopefully this helps!