Solve For F 6f 9g 3g F

Solving for 'f': A Comprehensive Guide to Algebraic Equations

This article provides a comprehensive guide on how to solve algebraic equations, specifically focusing on solving for the variable 'f' in equations involving 'f' and 'g'. We'll cover various methods, explain the underlying principles, and offer examples to solidify your understanding. This detailed approach will equip you with the skills to tackle similar problems with confidence and precision.

Understanding the Fundamentals: Variables and Equations

Before diving into solving for 'f', let's review the basics. An algebraic equation is a statement that asserts the equality of two expressions. These expressions contain variables (like 'f' and 'g' in our case), constants (numbers), and operators (+, -, ×, ÷). The goal is to find the value(s) of the variable(s) that make the equation true.

A variable, represented by a letter (like 'f' or 'g'), represents an unknown quantity. In our problem, 'f' and 'g' are variables. We need to isolate 'f' to find its value in terms of 'g' (or sometimes, a numerical value if 'g' is given).

Solving the Equation: 6f + 9g = 3g + f

Our target equation is: 6f + 9g = 3g + f

The aim is to isolate 'f' on one side of the equation. We achieve this through a series of algebraic manipulations, ensuring we maintain the equality at every step. This involves applying inverse operations to move terms around.

Step 1: Combine like terms

First, let's gather all the terms containing 'f' on one side and all the terms containing 'g' on the other. To do this, we'll subtract 'f' from both sides of the equation:

6f + 9g - f = 3g + f - f

This simplifies to:

5f + 9g = 3g

Step 2: Isolate the 'f' term

Next, we need to isolate the term with 'f' (which is 5f). We'll subtract 9g from both sides:

5f + 9g - 9g = 3g - 9g

This simplifies to:

5f = -6g

Step 3: Solve for 'f'

Finally, we need to solve for 'f' by itself. Since 'f' is multiplied by 5, we'll divide both sides of the equation by 5:

5f / 5 = -6g / 5

This gives us the solution:

f = -6g/5

This means that the value of 'f' is directly dependent on the value of 'g'. If we know the value of 'g', we can substitute it into this equation to find the corresponding value of 'f'.

Examples with Numerical Values for 'g'

Let's illustrate this with a few examples where 'g' is given a numerical value.

Example 1: g = 5

Substitute g = 5 into the solution:

f = -6(5)/5 = -6

Therefore, if g = 5, then f = -6.

Example 2: g = -10

Substitute g = -10 into the solution:

f = -6(-10)/5 = 12

Therefore, if g = -10, then f = 12.

Example 3: g = 0

Substitute g = 0 into the solution:

f = -6(0)/5 = 0

Therefore, if g = 0, then f = 0.

Verification: Checking Your Solution

It's crucial to verify your solution. Substitute the values you found for 'f' back into the original equation (6f + 9g = 3g + f) to confirm that the equation holds true.

Let's verify our solutions from the examples above:

Example 1 (g = 5, f = -6):

6(-6) + 9(5) = 3(5) + (-6) -36 + 45 = 15 - 6 9 = 9 (The equation holds true)

Example 2 (g = -10, f = 12):

6(12) + 9(-10) = 3(-10) + 12 72 - 90 = -30 + 12 -18 = -18 (The equation holds true)

Example 3 (g = 0, f = 0):

6(0) + 9(0) = 3(0) + 0 0 = 0 (The equation holds true)

This verification step ensures the accuracy of your solution.

Advanced Concepts and Extensions

The approach outlined above is fundamental to solving linear equations. However, we can expand on this by considering:

• Equations with more variables: Solving for one variable when others are present requires similar techniques – isolating the desired variable through algebraic manipulation.

• Systems of equations: If you have multiple equations with 'f' and 'g', methods like substitution or elimination can be used to find the values of both variables.

• Nonlinear equations: If the equation involves exponents or other non-linear terms, more advanced techniques may be needed, such as factoring, the quadratic formula, or other numerical methods.

Practical Applications

Solving algebraic equations like this one has wide-ranging applications in various fields:

• Physics: Solving for unknown forces, velocities, or accelerations.

• Engineering: Calculating dimensions, stresses, or strains in structures.

• Economics: Modeling supply and demand, or determining optimal resource allocation.

• Computer science: Developing algorithms and solving computational problems.

• Data analysis: Creating mathematical models to represent and analyze data.

Conclusion

Solving for 'f' in the equation 6f + 9g = 3g + f involves a systematic approach of combining like terms, isolating the variable, and solving for its value. Understanding the fundamental principles of algebraic manipulation and verifying your solution are crucial for accurate results. This skill is foundational in mathematics and has broad applications across various disciplines. By mastering these techniques, you build a strong foundation for tackling more complex algebraic problems and utilizing mathematical models effectively in your chosen field. Remember to always check your work and practice regularly to reinforce your understanding. The more you practice, the more confident and efficient you'll become at solving algebraic equations.