Question

Solve each system. Use any method you wish. $$ \left\\{\begin{array}{l} 3 x^{2}-2 y^{2}+5=0 \\ 2 x^{2}-y^{2}+2=0 \end{array}\right. $$

Short answer

The solutions are \((1, 2)\), \((1, -2)\), \((-1, 2)\), and \((-1, -2)\).

Step-by-step solution

Label the Equations

Label the given equations for easy reference. Let's call the first equation: \[ 3x^2 - 2y^2 + 5 = 0 \] Equation (1) and the second equation: \[ 2x^2 - y^2 + 2 = 0 \] Equation (2)

Solve One Equation for One Variable

Solve Equation (2) for \( y^2 \). We can isolate \( y^2 \) as follows: \[ 2x^2 - y^2 + 2 = 0 \] Subtract 2 from both sides: \[ 2x^2 - y^2 = -2 \] Add \( y^2 \) to both sides and then divide by 2: \[ y^2 = 2x^2 + 2 \]

Substitute into the First Equation

Substitute the expression for \( y^2 \) from Step 2 into Equation (1): \[ 3x^2 - 2(2x^2 + 2) + 5 = 0 \] Simplify inside the parentheses: \[ 3x^2 - 4x^2 - 4 + 5 = 0 \]

Simplify and Solve for x

Combine like terms to solve for \( x \): \[ 3x^2 - 4x^2 - 4 + 5 = 0 \] Simplify: \[ -x^2 + 1 = 0 \] Rearrange: \[ -x^2 = -1 \] Divide both sides by -1: \[ x^2 = 1 \] Therefore: \( x = 1 \) or \( x = -1 \)

Substitute x Values into Solved Equation

Substitute \( x = 1 \) and \( x = -1 \) into the expression for \( y^2 \) found in Step 2: For \( x = 1 \): \[ y^2 = 2(1)^2 + 2 = 2 + 2 = 4 \] So: \( y = 2 \) or \( y = -2 \) For \( x = -1 \): \[ y^2 = 2(-1)^2 + 2 = 2 + 2 = 4 \] So: \( y = 2 \) or \( y = -2 \)

List the Solution Pairs

From the previous steps, list all solution pairs: \( (1, 2) \), \( (1, -2) \), \( (-1, 2) \), and \( (-1, -2) \)

Key concepts

Quadratic Equations

A quadratic equation is any equation that can be rearranged in standard form as \( ax^2 + bx + c = 0 \). Quadratic equations are key because they often pop up in many areas of math and science. They always graph into a shape called a 'parabola'. The solutions to a quadratic equation are called 'roots', and these can be real or complex numbers. To better understand quadratic equations, focus on identifying the form, solving using various methods such as factoring, completing the square, or using the quadratic formula \( x = \frac{ -b \pm \sqrt{b^2 - 4ac} }{ 2a } \). In our exercise, both equations contain quadratic expressions regarding x and y.

Substitution Method

The substitution method is a way to solve a system of equations where one equation is solved for one variable and then substituted into another equation. This method works well with linear and nonlinear systems. Here is a simple outline:

• Solve one equation for one variable.
• Substitute this solution into the other equation.
• Solve the resulting equation for the remaining variable.
• Substitute back to find the other variable.

In the provided exercise, this method was applied by solving one equation (Equation 2) for \( y^2 \), and then substituting this into the other (Equation 1) to solve for x.

System of Nonlinear Equations

Systems of nonlinear equations involve equations where the variables have an exponent other than 1, such as squared terms. Nonlinear systems are trickier to solve compared to linear ones. The basic steps to solve such a system include:

• Identify the system: Recognize if equations include quadratic, cubic, or other nonlinear forms.
• Choose a method: Substitution, elimination, or graphical methods can be used.
• Solve step by step: Carefully isolate one variable, substitute it into the other equation, and simplify.

In the original exercise, since both equations contained squared variables, we addressed the problem as a system of nonlinear equations. Using substitution, we first simplified the system, then solved for x and y.