For Each Value Of $w$, Determine Whether It Is A Solution To $3=\frac{w}{6}-3$.$[ \begin{array}{|c|c|c|} \hline w & \text{Yes} & \text{No} \ \hline -12 & \text{0} & \text{0} \ \hline -48 & \text{0} & \text{0} \ \hline 30 &

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Introduction

In this article, we will delve into the world of algebra and solve a given equation to determine whether a specific value of ww is a solution. The equation in question is 3=w6βˆ’33=\frac{w}{6}-3. We will explore the steps involved in solving this equation and provide a clear understanding of the process.

Understanding the Equation

The given equation is 3=w6βˆ’33=\frac{w}{6}-3. To begin solving this equation, we need to isolate the variable ww. The first step is to add 33 to both sides of the equation, which will eliminate the negative term on the right-hand side.

3 + 3 = \frac{w}{6} - 3 + 3
6 = \frac{w}{6}

Simplifying the Equation

Next, we need to simplify the equation by multiplying both sides by 66 to eliminate the fraction.

6 \times 6 = \frac{w}{6} \times 6
36 = w

Determining the Solution

Now that we have simplified the equation, we can determine whether a specific value of ww is a solution. We will examine the given values of ww and determine whether they satisfy the equation.

Value of ww: -12

To determine whether w=βˆ’12w = -12 is a solution, we substitute this value into the simplified equation.

36 = -12

As we can see, the equation is not satisfied, and therefore, w=βˆ’12w = -12 is not a solution.

Value of ww: -48

Next, we will examine whether w=βˆ’48w = -48 is a solution. We substitute this value into the simplified equation.

36 = -48

Again, the equation is not satisfied, and therefore, w=βˆ’48w = -48 is not a solution.

Value of ww: 30

Finally, we will examine whether w=30w = 30 is a solution. We substitute this value into the simplified equation.

36 = 30

As we can see, the equation is not satisfied, and therefore, w=30w = 30 is not a solution.

Conclusion

In this article, we have solved the equation 3=w6βˆ’33=\frac{w}{6}-3 and determined whether specific values of ww are solutions. We have examined the values w=βˆ’12w = -12, w=βˆ’48w = -48, and w=30w = 30 and found that none of them satisfy the equation.

Discussion

The equation 3=w6βˆ’33=\frac{w}{6}-3 is a simple linear equation that can be solved using basic algebraic techniques. However, the values of ww that we examined were not solutions to the equation. This highlights the importance of carefully checking the values of variables in equations to ensure that they satisfy the equation.

Final Thoughts

Q: What is the first step in solving an equation?

A: The first step in solving an equation is to isolate the variable. This can be done by adding, subtracting, multiplying, or dividing both sides of the equation by the same value.

Q: How do I simplify an equation?

A: To simplify an equation, you can combine like terms, eliminate fractions by multiplying both sides by a common denominator, and rearrange the equation to isolate the variable.

Q: What is the difference between a linear equation and a quadratic equation?

A: A linear equation is an equation in which the highest power of the variable is 1, whereas a quadratic equation is an equation in which the highest power of the variable is 2.

Q: How do I determine whether a value is a solution to an equation?

A: To determine whether a value is a solution to an equation, you can substitute the value into the equation and check if the equation is satisfied. If the equation is true, then the value is a solution.

Q: What is the importance of checking the values of variables in equations?

A: Checking the values of variables in equations is crucial to ensure that they satisfy the equation. If a value does not satisfy the equation, it is not a solution.

Q: Can you provide an example of a simple equation and its solution?

A: Here is an example of a simple equation and its solution:

Equation: 2x + 5 = 11

Solution:

2x + 5 = 11

Subtract 5 from both sides:

2x = 6

Divide both sides by 2:

x = 3

Therefore, the solution to the equation is x = 3.

Q: What are some common mistakes to avoid when solving equations?

A: Some common mistakes to avoid when solving equations include:

  • Not isolating the variable
  • Not simplifying the equation
  • Not checking the values of variables
  • Not following the order of operations (PEMDAS)

Q: Can you provide an example of a more complex equation and its solution?

A: Here is an example of a more complex equation and its solution:

Equation: x^2 + 4x + 4 = 0

Solution:

x^2 + 4x + 4 = 0

Factor the equation:

(x + 2)(x + 2) = 0

Set each factor equal to 0:

x + 2 = 0

x = -2

Therefore, the solution to the equation is x = -2.

Conclusion

In this article, we have answered some frequently asked questions about solving equations. We have covered topics such as simplifying equations, determining whether a value is a solution, and avoiding common mistakes. By following the steps outlined in this article, you can become more confident in your ability to solve equations and tackle more complex problems.