Problem 40 - Intermediate Algebra

Question

Solve each system. See Using Your Calculator: Solving Systems of Equations. $$ \left\\{\begin{array}{l} x^{2}-y^{2}=-5 \\ 3 x^{2}+2 y^{2}=30 \end{array}\right. $$

Step-by-Step Solution

Answer

The solutions are (2, 3), (-2, 3), (2, -3), and (-2, -3).

Step 1: Express One Equation

Begin by choosing one of the equations to manipulate. Let's express the first equation in terms of one variable. We choose the first equation: \( x^2 - y^2 = -5 \). Rearrange to express \( x^2 \) in terms of \( y^2 \): \[ x^2 = y^2 - 5 \]

Step 2: Substitute for Variable

Substitute the expression from Step 1 into the second equation. Replace \( x^2 \) in \( 3x^2 + 2y^2 = 30 \) with \( y^2 - 5 \):\[ 3(y^2 - 5) + 2y^2 = 30 \]

Step 3: Simplify and Solve for \( y^2 \)

Expand and simplify the equation from Step 2:\[ 3y^2 - 15 + 2y^2 = 30 \]Combine like terms:\[ 5y^2 - 15 = 30 \]Add 15 to both sides:\[ 5y^2 = 45 \]Divide by 5:\[ y^2 = 9 \]

Step 4: Solve for \( y \)

Since \( y^2 = 9 \), take the square root of both sides to solve for \( y \):\[ y = \pm 3 \]

Step 5: Solve for \( x \) using \( y = 3 \)

Use \( y = 3 \) in the expression \( x^2 = y^2 - 5 \):\[ x^2 = 3^2 - 5 = 9 - 5 = 4 \]Therefore, \( x = \pm 2 \).

Step 6: Solve for \( x \) using \( y = -3 \)

Use \( y = -3 \) in the same expression \( x^2 = y^2 - 5 \):\[ x^2 = (-3)^2 - 5 = 9 - 5 = 4 \]Thus, \( x = \pm 2 \).

Step 7: Write Solution Sets

Using Steps 5 and 6, the solutions to the system of equations are the pairs \((2, 3)\), \((-2, 3)\), \((2, -3)\), and \((-2, -3)\).

Key Concepts

Understanding the Substitution Method

The substitution method is a powerful tool for solving systems of equations. Here's how it works in simple terms. You start by solving one of the equations for one of the variables. This means you express one variable in terms of the other.
Next, you take this expression and substitute it into the other equation. This replaces one variable and transforms the system of equations into a single equation in terms of one variable.
Why is this helpful? Because solving for one variable is much easier! Once you solve this new equation, you get a numeric value for one of the variables. Finally, you substitute this numeric value back into one of the original equations to find the value of the other variable.

• Start by rewriting one equation to isolate a variable.
• Substitute the expression into the second equation.
• Solve the resulting equation for the other variable.
• Use the solution to find the second variable.

This step-by-step approach can make complex systems much more manageable.

Understanding Quadratic Equations in Systems

Quadratic equations are equations that include terms with variables raised to the power of two, like \( x^2 \) or \( y^2 \). These types of equations often occur in various systems when dealing with curves and parabolic paths. Solving these can sometimes be tricky due to the multiple solutions they may present.
For example, consider the equation \( x^2 = 9 \). Unlike linear equations, where solutions are straightforward, this can have two solutions: \( x = 3 \) and \( x = -3 \). This happens because when squared, both positive and negative numbers can produce the same result.
In systems of equations, quadratic expressions are transformed and plugged into other equations. This helps simplify the process since solving one quadratic then allows you to substitute back to find solutions for other variables.

• Quadratic terms appear as squared variables.
• Multiple solutions, positive and negative, are possible.
• Simplifying systems often involves expressing quadratics in forms involving linear equations.

Understanding how these quadratic terms interact with other elements of systems is key to mastery.

Defining the Solution Set for a System of Equations

When you solve a system of equations, especially one involving quadratics, the results can include multiple pairs of solutions. A solution set is a collection of all possible solutions that satisfy the equations.
Each element of the solution set is a pair of values for each variable that works in every equation of the system. In our exercise, for example, we found solutions like \((2, 3)\) and \((-2, -3)\). Each pair consists of \(x\) and \(y\) values that you can substitute back into both original equations to verify they hold true.
Interestingly, systems with quadratics can produce unique solution sets compared to linear systems. These combinations depend on the nature and intersection of parabolas or other curves in the system.

• A solution set includes all valid solution pairs.
• Each solution must satisfy every equation in the system.
• Quadratics can result in symmetrical or multiple valid solutions, enriching the possibilities.

The richness of the solution set comes from all the different ways the equations can hold true.