Substituting the Linear Equation into the Quadratic Equation

Oct 13, 2025 by wbrum.org 73 views

Introduction

In mathematics, solving systems of equations is a fundamental concept that has numerous applications in various fields, including physics, engineering, and economics. A system of equations consists of two or more equations that involve multiple variables. In this article, we will focus on algebraically solving a system of equations that involves a quadratic equation and a linear equation.

The System of Equations

The given system of equations is:

{ \begin{align*} (x-2)^2 + (y-3)^2 &= 20 \\ y &= -2x + 7 \end{align*} \}

The first equation is a quadratic equation in the form of $(x-h)^2 + (y-k)^2 = r^2$ , where $(h,k)$ is the center of the circle and $r$ is the radius. The second equation is a linear equation in the form of $y = mx + b$ , where $m$ is the slope and $b$ is the y-intercept.

Substituting the Linear Equation into the Quadratic Equation

To solve the system of equations, we can substitute the linear equation into the quadratic equation. This will allow us to eliminate the variable $y$ and solve for the variable $x$ .

{ \begin{align*} (x-2)^2 + ((-2x + 7)-3)^2 &= 20 \\ (x-2)^2 + (-2x + 4)^2 &= 20 \\ (x-2)^2 + 4x^2 - 16x + 16 &= 20 \\ 5x^2 - 16x - 4 &= 0 \end{align*} \}

Solving the Quadratic Equation

Now that we have a quadratic equation in the form of $ax^2 + bx + c = 0$ , we can use the quadratic formula to solve for the variable $x$ .

{ \begin{align*} x &= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\ x &= \frac{16 \pm \sqrt{(-16)^2 - 4(5)(-4)}}{2(5)} \\ x &= \frac{16 \pm \sqrt{256 + 80}}{10} \\ x &= \frac{16 \pm \sqrt{336}}{10} \\ x &= \frac{16 \pm 4\sqrt{21}}{10} \\ x &= \frac{8 \pm 2\sqrt{21}}{5} \end{align*} \}

Finding the Corresponding Values of $y$

Now that we have the values of $x$ , we can substitute them into the linear equation to find the corresponding values of $y$ .

{ \begin{align*} y &= -2x + 7 \\ y &= -2\left(\frac{8 \pm 2\sqrt{21}}{5}\right) + 7 \\ y &= \frac{-16 \pm 4\sqrt{21}}{5} + 7 \\ y &= \frac{-16 \pm 4\sqrt{21} + 35}{5} \\ y &= \frac{19 \pm 4\sqrt{21}}{5} \end{align*} \}

Conclusion

In this article, we have algebraically solved a system of equations that involves a quadratic equation and a linear equation. We have used substitution to eliminate the variable $y$ and solve for the variable $x$ . We have then used the quadratic formula to solve for the variable $x$ and found the corresponding values of $y$ . The solutions to the system of equations are:

{ \begin{align*} x &= \frac{8 \pm 2\sqrt{21}}{5} \\ y &= \frac{19 \pm 4\sqrt{21}}{5} \end{align*} \}

Final Answer

$The final answer is $\boxed{\left(\frac{8 \pm 2\sqrt{21}}{5}, \frac{19 \pm 4\sqrt{21}}{5}\right)}$.<br/> **Algebraically Solving a System of Equations: A Comprehensive Approach**$

Q&A: Algebraically Solving a System of Equations

Q: What is a system of equations?

A: A system of equations is a set of two or more equations that involve multiple variables. In this article, we have focused on a system of equations that involves a quadratic equation and a linear equation.

Q: How do I algebraically solve a system of equations?

A: To algebraically solve a system of equations, you can use substitution or elimination to eliminate one of the variables. In this article, we have used substitution to eliminate the variable $y$ and solve for the variable $x$ .

Q: What is the quadratic formula?

A: The quadratic formula is a mathematical formula that is used to solve quadratic equations. It is given by:

{ \begin{align*} x &= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \end{align*} \}

Q: How do I use the quadratic formula to solve a quadratic equation?

A: To use the quadratic formula to solve a quadratic equation, you need to identify the values of $a$ , $b$ , and $c$ in the equation. You can then plug these values into the quadratic formula to find the solutions to the equation.

Q: What are the solutions to the system of equations?

A: The solutions to the system of equations are:

{ \begin{align*} x &= \frac{8 \pm 2\sqrt{21}}{5} \\ y &= \frac{19 \pm 4\sqrt{21}}{5} \end{align*} \}

Q: How do I check my solutions to a system of equations?

A: To check your solutions to a system of equations, you can plug the values of $x$ and $y$ back into the original equations to see if they are true.

Q: What are some common mistakes to avoid when algebraically solving a system of equations?

A: Some common mistakes to avoid when algebraically solving a system of equations include:

Not following the order of operations
Not simplifying the equations
Not checking the solutions
Not using the correct formula or method

Q: How can I practice algebraically solving systems of equations?

A: You can practice algebraically solving systems of equations by working through examples and exercises in a textbook or online resource. You can also try solving systems of equations on your own and checking your solutions to make sure you are correct.

Conclusion

In this article, we have provided a comprehensive approach to algebraically solving a system of equations. We have used substitution to eliminate the variable $y$ and solve for the variable $x$ . We have also used the quadratic formula to solve the quadratic equation. We hope that this article has been helpful in providing you with a better understanding of how to algebraically solve a system of equations.

Final Answer

The final answer is $\boxed{\left(\frac{8 \pm 2\sqrt{21}}{5}, \frac{19 \pm 4\sqrt{21}}{5}\right)}$ .