How do you solve the equation and identify any extraneous solutions for #(x-4)/x + x/3=6#?

Question

How do you solve the equation and identify any extraneous solutions for #(x-4)/x    + x/3=6#?

Kennedy Adams · Accepted Answer

Multiply through by #x# to get a quadratic in #x#, then use quadratic formula to get:

#x = (15+-sqrt(273))/2# Multiply all the terms by #3x# to get: #3(x-4)+x^2 = 18x# In theory this could introduce an extraneous #x=0# solution, but it does not... Subtract #18x# from both sides to get: #x^2-15x-12 = 0# This quadratic is in the standard #ax^2+bx+c# form, with #a=1#, #b=-15# and #c=-12# Solve using the quadratic formula: #x = (-b+-sqrt(b^2-4ac))/(2a)# #=(15+-sqrt(15^2-(4xx1xx-12)))/(2xx1)# #=(15+-sqrt(225+48))/2# #=(15+-sqrt(273))/2#

Avery Alexander · Answer

undefined To solve the equation (x-4)/x + x/3 = 6 and identify any extraneous solutions, we can follow these steps:

1. Multiply every term in the equation by the common denominator, which is 3x. This eliminates the fractions.
   (3x)((x-4)/x) + (3x)(x/3) = (3x)(6)

2. Simplify each term by canceling out common factors.
   3(x-4) + x(x) = 18x

3. Distribute and combine like terms.
   3x - 12 + x^2 = 18x

4. Rearrange the equation to bring all terms to one side and set it equal to zero.
   x^2 + 15x - 12 = 0

5. Factor the quadratic equation or use the quadratic formula to solve for x.
   (x + 16)(x - 3) = 0
   x = -16 or x = 3

6. Check each solution by substituting it back into the original equation.
   For x = -16: (-16-4)/(-16) + (-16)/3 = 6
   For x = 3: (3-4)/(3) + (3)/3 = 6

7. Identify any extraneous solutions by checking if they make any denominators zero or result in undefined values.
   None of the solutions make any denominators zero or result in undefined values.

Therefore, the solutions to the equation are x = -16 and x = 3, and there are no extraneous solutions.