Which System Is Equivalent To ${ \left{ \begin{array}{c} y = 8x^2 \ x + Y = 5 \end{array} \right. }$?A. ${ \left{ \begin{aligned} 5-x &= 9x^2 \ y &= 5-x \end{aligned} \right. }$B. $[ \left{ \begin{array}{l} y = 9y^3 -

Introduction

In mathematics, a system of equations is a set of equations that are related to each other through a common variable or variables. Solving a system of equations involves finding the values of the variables that satisfy all the equations in the system. In this article, we will explore a system of equations and determine which of the given options is equivalent to it.

The Given System of Equations

The given system of equations is:

$$\left\{ \begin{array}{c} y = 8x^2 \\ x + y = 5 \end{array} \right.$$

This system consists of two equations: $y = 8x^2$ and $x + y = 5$. Our goal is to find the values of $x$ and $y$ that satisfy both equations.

Option A

Option A is given by:

$$\left\{ \begin{aligned} 5-x &= 9x^2 \\ y &= 5-x \end{aligned} \right.$$

To determine if this option is equivalent to the given system, we need to analyze each equation separately.

Equation 1: $5-x = 9x^2$

We can rewrite this equation as:

$$9x^2 + x - 5 = 0$$

This is a quadratic equation in the form $ax^2 + bx + c = 0$, where $a = 9$, $b = 1$, and $c = -5$. We can solve this equation using the quadratic formula:

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

Substituting the values of $a$, $b$, and $c$, we get:

$$x = \frac{-1 \pm \sqrt{1 + 180}}{18}$$

Simplifying further, we get:

$$x = \frac{-1 \pm \sqrt{181}}{18}$$

Equation 2: $y = 5-x$

This equation is already in a simple form, and we can substitute the value of $x$ from the first equation to find the value of $y$.

Option B

Option B is given by:

$$\left\{ \begin{array}{l} y = 9y^3 - 5 \\ x + y = 5 \end{array} \right.$$

To determine if this option is equivalent to the given system, we need to analyze each equation separately.

Equation 1: $y = 9y^3 - 5$

This equation is a cubic equation in the form $y = ay^3 + by + c$, where $a = 9$, $b = 0$, and $c = -5$. We can rewrite this equation as:

$$9y^3 - y - 5 = 0$$

This is a cubic equation, and solving it analytically is not straightforward. However, we can try to find a solution by inspection.

Equation 2: $x + y = 5$

This equation is already in a simple form, and we can substitute the value of $y$ from the first equation to find the value of $x$.

Comparing the Options

Now that we have analyzed both options, we can compare them to the given system of equations.

Option A

From our analysis, we can see that Option A has a quadratic equation in the form $ax^2 + bx + c = 0$, where $a = 9$, $b = 1$, and $c = -5$. This equation can be solved using the quadratic formula, and we get two possible values for $x$. We can then substitute these values into the second equation to find the corresponding values of $y$.

Option B

From our analysis, we can see that Option B has a cubic equation in the form $y = ay^3 + by + c$, where $a = 9$, $b = 0$, and $c = -5$. This equation is more complex than the quadratic equation in Option A, and solving it analytically is not straightforward.

Conclusion

Based on our analysis, we can conclude that Option A is equivalent to the given system of equations. The quadratic equation in Option A can be solved using the quadratic formula, and we get two possible values for $x$. We can then substitute these values into the second equation to find the corresponding values of $y$.

In contrast, Option B has a cubic equation that is more complex and difficult to solve analytically. Therefore, we can conclude that Option A is the correct answer.

Final Answer

The final answer is:

Q: What is a system of equations?

A: A system of equations is a set of equations that are related to each other through a common variable or variables. Solving a system of equations involves finding the values of the variables that satisfy all the equations in the system.

Q: How do I solve a system of equations?

A: There are several methods to solve a system of equations, including substitution, elimination, and graphing. The method you choose will depend on the type of equations and the number of variables involved.

Q: What is the substitution method?

A: The substitution method involves solving one equation for one variable and then substituting that expression into the other equation. This method is useful when one equation is linear and the other equation is quadratic or higher degree.

Q: What is the elimination method?

A: The elimination method involves adding or subtracting the equations to eliminate one variable. This method is useful when the equations have the same coefficient for one variable.

Q: What is the graphing method?

A: The graphing method involves graphing the equations on a coordinate plane and finding the point of intersection. This method is useful when the equations are linear and the system has a unique solution.

Q: How do I determine if a system of equations has a unique solution, no solution, or infinitely many solutions?

A: To determine the type of solution, you can use the following criteria:

• If the system has a unique solution, the equations must be consistent and independent.
• If the system has no solution, the equations must be inconsistent.
• If the system has infinitely many solutions, the equations must be consistent and dependent.

Q: What is the difference between a consistent and an inconsistent system of equations?

A: A consistent system of equations has at least one solution, while an inconsistent system of equations has no solution.

Q: What is the difference between an independent and a dependent system of equations?

A: An independent system of equations has a unique solution, while a dependent system of equations has infinitely many solutions.

Q: How do I solve a system of equations with three variables?

A: To solve a system of equations with three variables, you can use the substitution method or the elimination method. You can also use the method of matrices to solve the system.

Q: What is the method of matrices?

A: The method of matrices involves representing the system of equations as a matrix and using row operations to solve the system.

Q: How do I use the method of matrices to solve a system of equations?

A: To use the method of matrices, you can follow these steps:

1. Represent the system of equations as a matrix.
2. Use row operations to transform the matrix into row-echelon form.
3. Solve the system by back-substitution.

Q: What is row-echelon form?

A: Row-echelon form is a matrix that has the following properties:

• All the entries below the leading entry in each row are zero.
• The leading entry in each row is to the right of the leading entry in the previous row.
• The leading entry in each row is a 1.

Q: How do I solve a system of equations using row-echelon form?

A: To solve a system of equations using row-echelon form, you can follow these steps:

1. Represent the system of equations as a matrix.
2. Use row operations to transform the matrix into row-echelon form.
3. Solve the system by back-substitution.

Conclusion

Solving systems of equations is an important skill in mathematics and science. By understanding the different methods and techniques, you can solve a wide range of problems and apply them to real-world situations. Remember to always check your work and verify your solutions to ensure accuracy.

Final Tips

• Practice, practice, practice! Solving systems of equations takes practice, so make sure to work on a variety of problems to build your skills.
• Use technology to your advantage! Graphing calculators and computer software can help you visualize and solve systems of equations.
• Don't be afraid to ask for help! If you're struggling with a problem, don't hesitate to ask your teacher or a classmate for assistance.