How do you solve #1/(6a) + 1/(3a^2) = 1/(3a)# and find any extraneous solutions?

Question

Abigail Allen · Accepted Answer

#a/(6a^2) + 2/(6a^2) =(2a)/(6a^2)# #a+ 2 = 2a# #2 = 2a - a# #a = 2# It's worth noting that #a!=0#. I hope this is useful!

Kennedy Adams · Answer

undefined To solve the equation 1/(6a) + 1/(3a^2) = 1/(3a) and find any extraneous solutions, we can follow these steps:

1. Find a common denominator for the fractions on the left side of the equation. In this case, the common denominator is 6a^2.

2. Rewrite each fraction with the common denominator:
   - 1/(6a) becomes (a)/(6a^2)
   - 1/(3a^2) becomes (2)/(6a^2)

3. Combine the fractions on the left side of the equation:
   (a)/(6a^2) + (2)/(6a^2) = 1/(3a)

4. Combine the numerators over the common denominator:
   (a + 2)/(6a^2) = 1/(3a)

5. Cross-multiply to eliminate the denominators:
   3a * (a + 2) = 6a^2 * 1

6. Simplify the equation:
   3a^2 + 6a = 6a^2

7. Move all terms to one side of the equation:
   6a^2 - 3a^2 - 6a = 0

8. Combine like terms:
   3a^2 - 6a = 0

9. Factor out common terms:
   3a(a - 2) = 0

10. Set each factor equal to zero and solve for 'a':
   a = 0 or a - 2 = 0

11. Solve for 'a':
   a = 0 or a = 2

12. Check for extraneous solutions by substituting each value back into the original equation:
   - For a = 0: 1/(6*0) + 1/(3*0^2) = 1/(3*0) -> Division by zero is undefined, so a = 0 is an extraneous solution.
   - For a = 2: 1/(6*2) + 1/(3*2^2) = 1/(3*2) -> 1/12 + 1/12 = 1/6 -> 1/6 = 1/6, which is true.

Therefore, the only valid solution is a = 2, and a = 0 is an extraneous solution.