Solve The Following System Of Equations And Show All Work.$\[ \begin{array}{l} y = 2x^2 \\ y = -3x - 1 \end{array} \\]

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Introduction

Solving a system of equations is a fundamental concept in mathematics that involves finding the values of variables that satisfy multiple equations simultaneously. In this article, we will focus on solving a system of two equations, where one equation is quadratic and the other is linear. We will use algebraic methods to solve the system and provide a step-by-step guide on how to approach this type of problem.

The System of Equations

The given system of equations is:

y=2x2y=βˆ’3xβˆ’1\begin{array}{l} y = 2x^2 \\ y = -3x - 1 \end{array}

Our goal is to find the values of xx and yy that satisfy both equations simultaneously.

Step 1: Set the Equations Equal to Each Other

To solve the system of equations, we need to set the two equations equal to each other. This is because both equations are equal to yy, so we can set them equal to each other.

2x2=βˆ’3xβˆ’12x^2 = -3x - 1

Step 2: Simplify the Equation

Now, we need to simplify the equation by combining like terms.

2x2+3x+1=02x^2 + 3x + 1 = 0

Step 3: Factor the Quadratic Equation

The equation is a quadratic equation, and we can try to factor it. However, in this case, the equation does not factor easily, so we will use the quadratic formula to solve for xx.

The Quadratic Formula

The quadratic formula is:

x=βˆ’bΒ±b2βˆ’4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

In our case, a=2a = 2, b=3b = 3, and c=1c = 1. Plugging these values into the formula, we get:

x=βˆ’3Β±32βˆ’4(2)(1)2(2)x = \frac{-3 \pm \sqrt{3^2 - 4(2)(1)}}{2(2)}

Step 4: Simplify the Expression Under the Square Root

Now, we need to simplify the expression under the square root.

x=βˆ’3Β±9βˆ’84x = \frac{-3 \pm \sqrt{9 - 8}}{4}

x=βˆ’3Β±14x = \frac{-3 \pm \sqrt{1}}{4}

Step 5: Simplify the Square Root

The square root of 1 is 1, so we can simplify the expression.

x=βˆ’3Β±14x = \frac{-3 \pm 1}{4}

Step 6: Solve for xx

Now, we have two possible values for xx:

x=βˆ’3+14=βˆ’24=βˆ’12x = \frac{-3 + 1}{4} = \frac{-2}{4} = -\frac{1}{2}

x=βˆ’3βˆ’14=βˆ’44=βˆ’1x = \frac{-3 - 1}{4} = \frac{-4}{4} = -1

Step 7: Find the Corresponding Values of yy

Now that we have the values of xx, we can find the corresponding values of yy by plugging the values of xx into one of the original equations. We will use the first equation:

y=2x2y = 2x^2

Step 8: Plug in the Values of xx

Plugging in the values of xx, we get:

y=2(βˆ’12)2=2(14)=12y = 2\left(-\frac{1}{2}\right)^2 = 2\left(\frac{1}{4}\right) = \frac{1}{2}

y=2(βˆ’1)2=2(1)=2y = 2(-1)^2 = 2(1) = 2

Conclusion

In this article, we solved a system of two equations using algebraic methods. We set the equations equal to each other, simplified the resulting equation, and used the quadratic formula to solve for xx. We then found the corresponding values of yy by plugging the values of xx into one of the original equations. The final solutions to the system of equations are x=βˆ’12x = -\frac{1}{2} and y=12y = \frac{1}{2}, and x=βˆ’1x = -1 and y=2y = 2.

Final Answer

The final answer is:

\boxed{\left(-\frac{1}{2}, \frac{1}{2}\right), \left(-1, 2\right)}$<br/> **Solving a System of Equations: A Q&A Guide** ===================================================== **Introduction** --------------- In our previous article, we solved a system of two equations using algebraic methods. We set the equations equal to each other, simplified the resulting equation, and used the quadratic formula to solve for $x$. We then found the corresponding values of $y$ by plugging the values of $x$ into one of the original equations. In this article, we will answer some common questions related to solving systems of equations. **Q: What is a system of equations?** ----------------------------------- A system of equations is a set of two or more equations that are related to each other. In a system of equations, each equation is equal to a different value, and we need to find the values of the variables that satisfy all the equations simultaneously. **Q: How do I know if a system of equations has a solution?** --------------------------------------------------- A system of equations has a solution if and only if the two equations are consistent with each other. In other words, if the two equations have the same solution, then the system has a solution. **Q: What is the difference between a linear equation and a quadratic equation?** ------------------------------------------------------------------- A linear equation is an equation in which the highest power of the variable is 1. For example, $y = 2x + 3$ is a linear equation. A quadratic equation is an equation in which the highest power of the variable is 2. For example, $y = 2x^2 + 3x + 4$ is a quadratic equation. **Q: How do I solve a system of linear equations?** ------------------------------------------------ To solve a system of linear equations, we can use the method of substitution or the method of elimination. The method of substitution involves solving one equation for one variable and then substituting that expression into the other equation. The method of elimination involves adding or subtracting the two equations to eliminate one variable. **Q: How do I solve a system of quadratic equations?** ------------------------------------------------ To solve a system of quadratic equations, we can use the quadratic formula or the method of substitution. The quadratic formula involves solving the quadratic equation for one variable and then substituting that expression into the other equation. The method of substitution involves solving one equation for one variable and then substituting that expression into the other equation. **Q: What is the quadratic formula?** ----------------------------------- The quadratic formula is a formula that is used to solve quadratic equations. The quadratic formula is: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Q: How do I use the quadratic formula?

To use the quadratic formula, we need to identify the values of aa, bb, and cc in the quadratic equation. We then plug these values into the quadratic formula and simplify the expression.

Q: What is the difference between a system of equations and a system of inequalities?

A system of equations is a set of two or more equations that are related to each other. A system of inequalities is a set of two or more inequalities that are related to each other. In a system of inequalities, each inequality is either true or false, and we need to find the values of the variables that satisfy all the inequalities simultaneously.

Q: How do I solve a system of inequalities?

To solve a system of inequalities, we can use the method of substitution or the method of elimination. The method of substitution involves solving one inequality for one variable and then substituting that expression into the other inequality. The method of elimination involves adding or subtracting the two inequalities to eliminate one variable.

Conclusion

In this article, we answered some common questions related to solving systems of equations. We discussed the difference between a system of equations and a system of inequalities, and we provided a step-by-step guide on how to solve a system of linear equations and a system of quadratic equations. We also discussed the quadratic formula and how to use it to solve quadratic equations.

Final Answer

The final answer is:

  • A system of equations is a set of two or more equations that are related to each other.
  • A system of equations has a solution if and only if the two equations are consistent with each other.
  • A linear equation is an equation in which the highest power of the variable is 1.
  • A quadratic equation is an equation in which the highest power of the variable is 2.
  • The quadratic formula is a formula that is used to solve quadratic equations.
  • The quadratic formula is: x=βˆ’bΒ±b2βˆ’4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}