How do you solve #x/2+12/x=5# and check for extraneous solutions?

1 Answer
Sep 6, 2016

Answer:

Multiply by the common denominator of 2x, move all terms to one side, factor, solve and check for extraneous solutions (dividing by zero).

Explanation:

The common denominator is 2x, which is obtained by multiplying the denominators.

Multiple both sides of the equation by the common denominator.

#2x(x/2 + 12/x) = 2x (5)#

The result is
#x^2 +24=10x#

Subtract #10x# from both sides of the equation to get all terms on one side.

#x^2 -10x+24 = 0#

Factor
#(x-6)(x-4)=0#

Set each factor equal to zero and solve.
#x-6=0# and #x-4=0#
#x=6# and #x=4#

To check for extraneous solutions, substitute each answer into the original equation. The answers need to "check", and should not result in division by zero in any of the terms of the original equation. Division by zero means the solution is "extraneous". Neither of these solutions is extraneous though, and both satisfy the equation.