The Given System of Equations

Oct 13, 2025 by pinkandtrash.com 125 views

Introduction

In this article, we will explore the process of solving a system of nonlinear equations. A system of nonlinear equations is a set of equations that involve nonlinear functions, such as quadratic or polynomial functions. Solving such systems can be challenging, but with the right techniques and tools, we can find the solutions.

The Given System of Equations

The given system of equations is:

{ \begin{align*} 2x + 3y &= 1 \\ y^2 + xy - 2y - 2x &= 0 \end{align*} \}

Our goal is to solve for $x$ and $y$ .

Method 1: Substitution Method

One way to solve this system is to use the substitution method. We can start by solving the first equation for $y$ :

$2x + 3y = 1$

$3y = 1 - 2x$

$y = \frac{1 - 2x}{3}$

Now, we can substitute this expression for $y$ into the second equation:

$y^2 + xy - 2y - 2x = 0$

$\left(\frac{1 - 2x}{3}\right)^2 + x\left(\frac{1 - 2x}{3}\right) - 2\left(\frac{1 - 2x}{3}\right) - 2x = 0$

Simplifying this equation, we get:

$\frac{(1 - 2x)^2}{9} + \frac{x(1 - 2x)}{3} - \frac{2(1 - 2x)}{3} - 2x = 0$

Multiplying both sides by 9 to eliminate the fractions, we get:

$(1 - 2x)^2 + 3x(1 - 2x) - 6(1 - 2x) - 18x = 0$

Expanding and simplifying, we get:

$1 - 4x + 4x^2 + 3x - 6x^2 - 6 + 12x - 18x = 0$

Combine like terms:

$-3x^2 + 3x - 5 = 0$

This is a quadratic equation in $x$ . We can solve it using the quadratic formula:

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

In this case, $a = -3$ , $b = 3$ , and $c = -5$ . Plugging these values into the formula, we get:

$x = \frac{-3 \pm \sqrt{3^2 - 4(-3)(-5)}}{2(-3)}$

$x = \frac{-3 \pm \sqrt{9 - 60}}{-6}$

$x = \frac{-3 \pm \sqrt{-51}}{-6}$

Since the square root of a negative number is not a real number, this equation has no real solutions.

Method 2: Elimination Method

Another way to solve this system is to use the elimination method. We can start by multiplying the first equation by 2 and the second equation by 3 to make the coefficients of $x$ in both equations equal:

$4x + 6y = 2$

$3y^2 + 3xy - 6y - 6x = 0$

Now, we can subtract the first equation from the second equation to eliminate the $x$ term:

$3y^2 + 3xy - 6y - 6x - (4x + 6y) = 0$

Simplifying, we get:

$3y^2 + 3xy - 6y - 10x = 0$

Now, we can factor out a $y$ term:

$y(3y + 3x - 6) - 10x = 0$

Simplifying, we get:

$y(3y + 3x - 6) = 10x$

Now, we can divide both sides by $y$ to get:

$3y + 3x - 6 = \frac{10x}{y}$

Now, we can substitute this expression into the first equation:

$4x + 6y = 2$

$4x + 6\left(\frac{3y + 3x - 6}{3}\right) = 2$

Simplifying, we get:

$4x + 2y + 2x - 12 = 2$

Combine like terms:

$6x + 2y - 14 = 0$

This is a linear equation in $x$ and $y$ . We can solve it using substitution or elimination. Let's use substitution.

We can solve the first equation for $y$ :

$4x + 6y = 2$

$6y = 2 - 4x$

$y = \frac{2 - 4x}{6}$

Now, we can substitute this expression for $y$ into the second equation:

$6x + 2y - 14 = 0$

$6x + 2\left(\frac{2 - 4x}{6}\right) - 14 = 0$

Simplifying, we get:

$6x + \frac{2 - 4x}{3} - 14 = 0$

Multiplying both sides by 3 to eliminate the fraction, we get:

$18x + 2 - 4x - 42 = 0$

Combine like terms:

$14x - 40 = 0$

Add 40 to both sides:

$14x = 40$

Divide both sides by 14:

$x = \frac{40}{14}$

$x = \frac{20}{7}$

Now, we can substitute this value of $x$ into one of the original equations to solve for $y$ . Let's use the first equation:

$2x + 3y = 1$

$2\left(\frac{20}{7}\right) + 3y = 1$

Simplifying, we get:

$\frac{40}{7} + 3y = 1$

Multiply both sides by 7 to eliminate the fraction:

$40 + 21y = 7$

Subtract 40 from both sides:

$21y = -33$

Divide both sides by 21:

$y = -\frac{33}{21}$

$y = -\frac{11}{7}$

Therefore, the solution to the system of equations is $x = \frac{20}{7}$ and $y = -\frac{11}{7}$ .

Conclusion

$In this article, we have explored the process of solving a system of nonlinear equations. We have used two different methods: the substitution method and the elimination method. The substitution method involves substituting one equation into another, while the elimination method involves eliminating one variable by adding or subtracting the equations. We have also seen how to solve a quadratic equation using the quadratic formula. Finally, we have solved a system of nonlinear equations using both methods and found the solution to be $x = \frac{20}{7}$ and $y = -\frac{11}{7}$.<br/> **Solving a System of Nonlinear Equations: Q&A**$

Introduction

In our previous article, we explored the process of solving a system of nonlinear equations. We used two different methods: the substitution method and the elimination method. In this article, we will answer some common questions related to solving systems of nonlinear equations.

Q: What is a system of nonlinear equations?

A system of nonlinear equations is a set of equations that involve nonlinear functions, such as quadratic or polynomial functions. Nonlinear functions are functions that cannot be written in the form of a linear function, which is a function that can be written in the form of $y = mx + b$ .

Q: How do I know if a system of equations is nonlinear?

A system of equations is nonlinear if it involves nonlinear functions, such as quadratic or polynomial functions. For example, the equation $x^2 + y^2 = 1$ is a nonlinear equation because it involves a quadratic function.

Q: What are the different methods for solving a system of nonlinear equations?

There are two main methods for solving a system of nonlinear equations: the substitution method and the elimination method. The substitution method involves substituting one equation into another, while the elimination method involves eliminating one variable by adding or subtracting the equations.

Q: How do I choose which method to use?

The choice of method depends on the specific system of equations and the variables involved. If the system involves a quadratic equation, the substitution method may be more suitable. If the system involves a linear equation, the elimination method may be more suitable.

Q: What are some common challenges when solving a system of nonlinear equations?

Some common challenges when solving a system of nonlinear equations include:

Nonlinear functions: Nonlinear functions can be difficult to work with, especially when they involve complex expressions.
Multiple solutions: A system of nonlinear equations can have multiple solutions, which can make it difficult to determine the correct solution.
No solution: A system of nonlinear equations can have no solution, which can be frustrating to deal with.

Q: How do I check my work when solving a system of nonlinear equations?

When solving a system of nonlinear equations, it's essential to check your work to ensure that you have found the correct solution. Here are some steps you can follow:

Verify the solution: Verify that the solution you have found satisfies both equations in the system.
Check for extraneous solutions: Check to see if the solution you have found is an extraneous solution, which is a solution that is not actually a solution to the system.
Check for multiple solutions: Check to see if the system has multiple solutions, and if so, determine which solution is the correct one.

Q: What are some real-world applications of solving systems of nonlinear equations?

Solving systems of nonlinear equations has many real-world applications, including:

Physics: Solving systems of nonlinear equations is essential in physics, where it is used to model complex phenomena such as motion and energy.
Engineering: Solving systems of nonlinear equations is essential in engineering, where it is used to design and optimize complex systems such as bridges and buildings.
Economics: Solving systems of nonlinear equations is essential in economics, where it is used to model complex economic systems and make predictions about future economic trends.

Conclusion

In this article, we have answered some common questions related to solving systems of nonlinear equations. We have discussed the different methods for solving a system of nonlinear equations, including the substitution method and the elimination method. We have also discussed some common challenges when solving a system of nonlinear equations, and how to check your work to ensure that you have found the correct solution. Finally, we have discussed some real-world applications of solving systems of nonlinear equations.