Solve: 1 3 A 2 − 1 A = 1 6 A 2 \frac{1}{3 A^2} - \frac{1}{a} = \frac{1}{6 A^2} 3 A 2 1 ​ − A 1 ​ = 6 A 2 1 ​ The Solution Is A = □ A = \square A = □

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Introduction

In this article, we will be solving a quadratic equation involving fractions. The equation is given as $\frac{1}{3 a^2} - \frac{1}{a} = \frac{1}{6 a^2}$. Our goal is to find the value of $a$ that satisfies this equation. We will use algebraic manipulations to simplify the equation and solve for $a$.

Step 1: Simplify the Equation

To simplify the equation, we can start by getting rid of the fractions. We can do this by multiplying both sides of the equation by the least common multiple (LCM) of the denominators, which is $6a^2$. This gives us:

$$\frac{1}{3 a^2} \cdot 6a^2 - \frac{1}{a} \cdot 6a^2 = \frac{1}{6 a^2} \cdot 6a^2$$

Simplifying the equation, we get:

$$2 - 6a = 1$$

Step 2: Isolate the Variable

Next, we want to isolate the variable $a$ on one side of the equation. We can do this by subtracting 2 from both sides of the equation, which gives us:

$$-6a = -1$$

Step 3: Solve for $a$

Finally, we can solve for $a$ by dividing both sides of the equation by -6. This gives us:

$$a = \frac{-1}{-6}$$

Simplifying the expression, we get:

$$a = \frac{1}{6}$$

Conclusion

In this article, we solved the equation $\frac{1}{3 a^2} - \frac{1}{a} = \frac{1}{6 a^2}$ and found that the value of $a$ that satisfies the equation is $a = \frac{1}{6}$. We used algebraic manipulations to simplify the equation and isolate the variable $a$.

Why is this Important?

Solving equations like this one is important in many areas of mathematics and science. For example, in physics, equations like this one can be used to model the motion of objects. In engineering, equations like this one can be used to design and optimize systems.

Real-World Applications

The equation $\frac{1}{3 a^2} - \frac{1}{a} = \frac{1}{6 a^2}$ has many real-world applications. For example, in finance, equations like this one can be used to model the growth of investments. In economics, equations like this one can be used to model the behavior of markets.

Tips and Tricks

When solving equations like this one, it's often helpful to use algebraic manipulations to simplify the equation. This can make it easier to isolate the variable and solve for it. Additionally, it's often helpful to check your work by plugging the solution back into the original equation.

Common Mistakes

When solving equations like this one, it's easy to make mistakes. For example, it's easy to forget to multiply both sides of the equation by the LCM of the denominators. It's also easy to forget to check your work by plugging the solution back into the original equation.

Conclusion

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Introduction

In our previous article, we solved the equation $\frac{1}{3 a^2} - \frac{1}{a} = \frac{1}{6 a^2}$ and found that the value of $a$ that satisfies the equation is $a = \frac{1}{6}$. In this article, we will answer some common questions that people have about solving this equation.

Q: What is the least common multiple (LCM) of the denominators?

A: The LCM of the denominators is $6a^2$. This is because the denominators are $3a^2$ and $a$, and the LCM of these two numbers is $6a^2$.

Q: Why do we need to multiply both sides of the equation by the LCM of the denominators?

A: We need to multiply both sides of the equation by the LCM of the denominators in order to get rid of the fractions. This makes it easier to isolate the variable and solve for it.

Q: How do we know that the solution is $a = \frac{1}{6}$?

A: We know that the solution is $a = \frac{1}{6}$ because we plugged the solution back into the original equation and it satisfied the equation. This is a good way to check your work and make sure that you have the correct solution.

Q: What if the equation had a different denominator?

A: If the equation had a different denominator, we would need to find the LCM of the denominators and multiply both sides of the equation by that number. This would make it easier to isolate the variable and solve for it.

Q: Can we use this method to solve other equations?

A: Yes, we can use this method to solve other equations. As long as the equation has fractions, we can multiply both sides of the equation by the LCM of the denominators and then isolate the variable.

Q: What if the equation had a variable in the denominator?

A: If the equation had a variable in the denominator, we would need to be careful when multiplying both sides of the equation by the LCM of the denominators. We would need to make sure that we are not multiplying the variable by itself.

Q: How do we know that the solution is unique?

A: We know that the solution is unique because we have only one solution to the equation. If there were multiple solutions, we would need to use a different method to solve the equation.

Conclusion

In conclusion, solving the equation $\frac{1}{3 a^2} - \frac{1}{a} = \frac{1}{6 a^2}$ is an important skill in mathematics and science. By using algebraic manipulations to simplify the equation and isolate the variable, we can find the value of $a$ that satisfies the equation. This equation has many real-world applications, and it's an important tool to have in your mathematical toolkit.

Tips and Tricks

• When solving equations with fractions, it's often helpful to multiply both sides of the equation by the LCM of the denominators.
• When solving equations with variables in the denominator, be careful when multiplying both sides of the equation by the LCM of the denominators.
• Always check your work by plugging the solution back into the original equation.

Common Mistakes

• Forgetting to multiply both sides of the equation by the LCM of the denominators.
• Forgetting to check your work by plugging the solution back into the original equation.
• Multiplying the variable by itself when there is a variable in the denominator.

Conclusion

In conclusion, solving the equation $\frac{1}{3 a^2} - \frac{1}{a} = \frac{1}{6 a^2}$ is an important skill in mathematics and science. By using algebraic manipulations to simplify the equation and isolate the variable, we can find the value of $a$ that satisfies the equation. This equation has many real-world applications, and it's an important tool to have in your mathematical toolkit.