Understanding the Given Constraints

Oct 13, 2025 by recoveryshake.com 148 views

Introduction

In this article, we will delve into the world of algebra and solve a complex equation involving variables x, y, and z. The given equation is xy + yz, with the additional constraints that xz = 2 and y = 5. We will break down the problem step by step, using algebraic manipulations to simplify the equation and ultimately find the solution.

Understanding the Given Constraints

Before we begin solving the equation, let's take a closer look at the given constraints. We are told that xz = 2 and y = 5. These constraints provide us with valuable information that we can use to simplify the equation.

The Constraint xz = 2

The constraint xz = 2 tells us that the product of x and z is equal to 2. This means that we can express z in terms of x as z = 2/x.

The Constraint y = 5

The constraint y = 5 tells us that the value of y is fixed at 5. This means that we can substitute y = 5 into the equation xy + yz to simplify it.

Substituting the Constraints into the Equation

Now that we have a better understanding of the given constraints, let's substitute them into the equation xy + yz.

Substituting y = 5 into the Equation

We can substitute y = 5 into the equation xy + yz to get:

5x + 5z

Substituting z = 2/x into the Equation

We can substitute z = 2/x into the equation 5x + 5z to get:

5x + 5(2/x)

Simplifying the Equation

Now that we have substituted the constraints into the equation, let's simplify it further.

Simplifying the Equation 5x + 10/x

We can simplify the equation 5x + 10/x by combining the terms:

(5x^2 + 10)/x

Factoring the Equation

We can factor the equation (5x^2 + 10)/x to get:

(5x^2 + 10)/x = 5x + 10/x

Solving for x

Now that we have simplified the equation, let's solve for x.

Setting the Equation Equal to 0

We can set the equation 5x + 10/x equal to 0 to get:

5x + 10/x = 0

Multiplying Both Sides by x

We can multiply both sides of the equation 5x + 10/x = 0 by x to get:

5x^2 + 10 = 0

Subtracting 10 from Both Sides

We can subtract 10 from both sides of the equation 5x^2 + 10 = 0 to get:

5x^2 = -10

Dividing Both Sides by 5

We can divide both sides of the equation 5x^2 = -10 by 5 to get:

x^2 = -2

Taking the Square Root of Both Sides

We can take the square root of both sides of the equation x^2 = -2 to get:

x = ±√(-2)

Simplifying the Square Root

We can simplify the square root √(-2) to get:

x = ±i√2

Conclusion

In this article, we have solved the equation xy + yz when xz = 2 and y = 5. We have used algebraic manipulations to simplify the equation and ultimately find the solution. The final solution is x = ±i√2.

Final Answer

The final answer is x = ±i√2.

Discussion

The equation xy + yz is a complex equation that involves variables x, y, and z. The given constraints xz = 2 and y = 5 provide valuable information that we can use to simplify the equation. By substituting the constraints into the equation and simplifying it further, we can solve for x.

References

[1] Algebra textbook by Michael Artin
[2] Calculus textbook by James Stewart
[3] Online resources for algebra and calculus

Introduction

In our previous article, we solved the equation xy + yz when xz = 2 and y = 5. We used algebraic manipulations to simplify the equation and ultimately find the solution. In this article, we will answer some frequently asked questions related to the equation and its solution.

Q: What is the equation xy + yz?

A: The equation xy + yz is a complex equation that involves variables x, y, and z. It is a quadratic equation in terms of x and z.

Q: What are the given constraints?

A: The given constraints are xz = 2 and y = 5. These constraints provide valuable information that we can use to simplify the equation.

Q: How do we simplify the equation?

A: We simplify the equation by substituting the constraints into the equation and using algebraic manipulations to simplify it further.

Q: What is the final solution?

A: The final solution is x = ±i√2.

Q: Why do we get a complex solution?

A: We get a complex solution because the equation is a quadratic equation in terms of x and z. The quadratic formula involves the square root of a negative number, which results in a complex solution.

Q: Can we solve the equation using other methods?

A: Yes, we can solve the equation using other methods such as substitution or elimination. However, the method we used in this article is the most straightforward and efficient way to solve the equation.

Q: What are some real-world applications of this equation?

A: This equation has many real-world applications in fields such as physics, engineering, and computer science. For example, it can be used to model the motion of objects in a two-dimensional space.

Q: Can we generalize this equation to higher dimensions?

A: Yes, we can generalize this equation to higher dimensions. However, the solution will be more complex and will involve more variables.

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

A: Some common mistakes to avoid when solving this equation include:

Not substituting the constraints into the equation
Not simplifying the equation correctly
Not using the correct method to solve the equation

Q: How can we verify the solution?

A: We can verify the solution by plugging it back into the original equation and checking if it satisfies the equation.

Q: What are some related equations that we can solve?

A: Some related equations that we can solve include:

xy + yz = 0
xz + yz = 0
xy + xz = 0

Q: Can we solve these equations using the same method?

A: Yes, we can solve these equations using the same method. However, the solution will be different for each equation.

Conclusion

In this article, we have answered some frequently asked questions related to the equation xy + yz when xz = 2 and y = 5. We have also discussed some related equations and how to solve them using the same method.

Final Answer

The final answer is x = ±i√2.

Discussion

The equation xy + yz is a complex equation that involves variables x, y, and z. It has many real-world applications and can be solved using algebraic manipulations. We have also discussed some related equations and how to solve them using the same method.

References

[1] Algebra textbook by Michael Artin
[2] Calculus textbook by James Stewart
[3] Online resources for algebra and calculus