Solve $\frac{1}{6 x^2}-\frac{1}{2 x}=\frac{1}{12 x^2}$
A. $x=-1 / 6$
B. $x=6$
C. $x=-6$
D. $x=1 / 6$
Answer (1)
Multiply both sides of the equation by 12 x 2 to eliminate the fractions: 2 − 6 x = 1 .
Solve for x : − 6 x = − 1 , which gives x = 6 1 .
Verify the solution by substituting x = 6 1 into the original equation.
The solution is x = 6 1 .
Explanation
Problem Analysis We are given the equation 6 x 2 1 − 2 x 1 = 1 1 12 x 2 and we need to solve for x .
Eliminating Fractions To eliminate the fractions, we can multiply both sides of the equation by the least common multiple of the denominators, which is 12 x 2 . This gives us: 12 x 2 ( 6 x 2 1 − 2 x 1 ) = 12 x 2 ( 12 x 2 1 ) Distributing 12 x 2 on the left side, we get: 6 x 2 12 x 2 − 2 x 12 x 2 = 12 x 2 12 x 2 Simplifying each term, we have: 2 − 6 x = 1
Solving for x Now, we solve for x . Subtracting 2 from both sides of the equation, we get: − 6 x = 1 − 2 − 6 x = − 1 Dividing both sides by -6, we find: x = − 6 − 1 = 6 1
Verification To verify the solution, we substitute x = 6 1 back into the original equation: 6 ( 6 1 ) 2 1 − 2 ( 6 1 ) 1 = 6 ( 36 1 ) 1 − 3 1 1 = 6 1 1 − 3 = 6 − 3 = 3 12 ( 6 1 ) 2 1 = 12 ( 36 1 ) 1 = 3 1 1 = 3 Since both sides of the equation are equal when x = 6 1 , our solution is correct.
Examples
Consider a scenario where you are adjusting the resistance in an electrical circuit. The equation 6 x 2 1 − 2 x 1 = 12 x 2 1 could represent a simplified model of the circuit's behavior. Solving for x helps you determine the specific resistance value needed to achieve a desired circuit performance. This type of algebraic problem is fundamental in electrical engineering for designing and troubleshooting circuits.
