Not multiplying both sides of the equation by 5

Oct 13, 2025 by recoveryshake.com 147 views

Introduction

In mathematics, solving for a variable means finding the value of that variable that makes the equation true. In this article, we will focus on solving for z in the equation 1 = z/5. This equation is a simple linear equation, and we will use algebraic methods to solve for z.

Understanding the Equation

The equation 1 = z/5 is a simple linear equation. It states that the value of z divided by 5 is equal to 1. To solve for z, we need to isolate the variable z on one side of the equation.

Isolating the Variable

To isolate the variable z, we can multiply both sides of the equation by 5. This will cancel out the division by 5 on the left-hand side of the equation.

Multiplying Both Sides by 5

1 = z/5
5 * 1 = 5 * (z/5)
5 = z

Solving for z

Now that we have isolated the variable z, we can see that z is equal to 5. This is the solution to the equation 1 = z/5.

Conclusion

Solving for z in the equation 1 = z/5 is a simple process that involves isolating the variable z using algebraic methods. By multiplying both sides of the equation by 5, we can cancel out the division by 5 and solve for z. The solution to the equation is z = 5.

Example Use Cases

Solving for z in the equation 1 = z/5 has many practical applications in mathematics and real-world problems. Here are a few examples:

Finance: In finance, solving for z can help us calculate the value of an investment or a loan. For example, if we know that the interest rate on a loan is 5% and the loan amount is $100, we can use the equation 1 = z/5 to calculate the value of z, which represents the total amount paid back to the lender.
Science: In science, solving for z can help us calculate the value of a physical quantity, such as the speed of an object or the distance between two points. For example, if we know that the speed of an object is 5 meters per second and the time it takes to travel a certain distance is 1 second, we can use the equation 1 = z/5 to calculate the value of z, which represents the distance traveled.
Engineering: In engineering, solving for z can help us design and optimize systems, such as electrical circuits or mechanical systems. For example, if we know that the resistance of a circuit is 5 ohms and the voltage across the circuit is 1 volt, we can use the equation 1 = z/5 to calculate the value of z, which represents the current flowing through the circuit.

Tips and Tricks

Here are a few tips and tricks to help you solve for z in the equation 1 = z/5:

Use algebraic methods: To solve for z, use algebraic methods, such as multiplying both sides of the equation by 5.
Check your work: Always check your work to make sure that the solution is correct.
Use real-world examples: Use real-world examples to help you understand the concept of solving for z.

Conclusion

Frequently Asked Questions

In this article, we will answer some of the most frequently asked questions about solving for z in the equation 1 = z/5.

Q: What is the value of z in the equation 1 = z/5?

A: The value of z in the equation 1 = z/5 is 5. This is because when we multiply both sides of the equation by 5, we get 5 = z.

Q: How do I solve for z in the equation 1 = z/5?

A: To solve for z in the equation 1 = z/5, you can multiply both sides of the equation by 5. This will cancel out the division by 5 on the left-hand side of the equation, and you will be left with z = 5.

Q: What if the equation is more complex, such as 2 = z/5?

A: If the equation is more complex, such as 2 = z/5, you can still solve for z by multiplying both sides of the equation by 5. This will give you 10 = z.

Q: Can I use other methods to solve for z, such as adding or subtracting numbers?

A: No, you cannot use other methods to solve for z, such as adding or subtracting numbers. The only way to solve for z in the equation 1 = z/5 is to multiply both sides of the equation by 5.

Q: What if I get a different answer when I solve for z?

A: If you get a different answer when you solve for z, it may be because you made a mistake in your calculations. Double-check your work to make sure that you are using the correct method to solve for z.

Q: Can I use the equation 1 = z/5 to solve for other variables?

A: No, the equation 1 = z/5 is only used to solve for the variable z. You cannot use this equation to solve for other variables.

Q: How do I apply the concept of solving for z to real-world problems?

A: To apply the concept of solving for z to real-world problems, you can use the equation 1 = z/5 to calculate the value of z in a variety of situations. For example, you can use this equation to calculate the value of an investment or a loan, or to determine the distance between two points.

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

A: Some common mistakes to avoid when solving for z include:

Not multiplying both sides of the equation by 5: This will result in an incorrect solution for z.
Adding or subtracting numbers: This will not give you the correct solution for z.
Not checking your work: This can lead to errors in your calculations.

Q: How can I practice solving for z?

A: To practice solving for z, you can try solving a variety of equations that involve the variable z. You can also use online resources or practice problems to help you improve your skills.

Q: What are some advanced topics related to solving for z?

A: Some advanced topics related to solving for z include:

Solving systems of equations: This involves solving multiple equations that involve the variable z.
Using algebraic methods to solve for z: This involves using more complex algebraic methods to solve for z.
Applying the concept of solving for z to real-world problems: This involves using the equation 1 = z/5 to solve for z in a variety of real-world situations.

Conclusion

Solving for z in the equation 1 = z/5 is a simple process that involves multiplying both sides of the equation by 5. By following the steps outlined in this article, you can become proficient in solving for z and apply this skill to a wide range of mathematical and real-world problems.