Problem 14 - Beginning and Intermediate Algebra

Question

Solve each system. $$\begin{array}{c} 2 x^{2}-y^{2}=7 \\ 2 y^{2}-3 x^{2}=2 \end{array}$$

Step-by-Step Solution

Answer

The short answer is: There are four possible solutions to the given system of equations: 1. \((4, \sqrt{23})\) 2. \((4, -\sqrt{23})\) 3. \((-4, \sqrt{23})\) 4. \((-4, -\sqrt{23})\)

Step 1: Express y in terms of x using the first equation

From the first equation \(2x^2 - y^2 = 7\), we can express y in terms of x: \(y^2 = 2x^2 - 7\)

Step 2: Substitute the expression for y in the second equation

Now we have an expression for \(y^2\): \(2x^2 - 7\). We will substitute this into the second equation \(2y^2 - 3x^2 = 2\): \(2(2x^2 - 7) - 3x^2 = 2\)

Step 3: Solve for x

Now we'll solve the equation for x: \(4x^2 - 14 - 3x^2 = 2\) Simplifying further, we get: \(x^2 - 14 = 2\) Now, add 14 to both sides: \(x^2 = 16\) This equation has two solutions for x: \(x = \pm 4\)

Step 4: Solve for y for each value of x

Now, let's find the corresponding values of y for both x values using the expression for y that we found in step 1: When \(x = 4\): \(y^2 = 2(4^2) - 7 = 23\) \(y = \pm \sqrt{23}\) So, for \(x = 4\), there are two possible solutions: \((4, \sqrt{23})\) and \((4, -\sqrt{23})\). When \(x = -4\): \(y^2 = 2(-4^2) - 7 = 23\) \(y = \pm \sqrt{23}\) So, for \(x = -4\), there are two possible solutions: \((-4, \sqrt{23})\) and \((-4, -\sqrt{23})\).

Step 5: Write the final solutions

In conclusion, there are four possible solutions to the given system of equations: 1. \((4, \sqrt{23})\) 2. \((4, -\sqrt{23})\) 3. \((-4, \sqrt{23})\) 4. \((-4, -\sqrt{23})\)

Key Concepts

Quadratic Equations

Quadratic equations are mathematical expressions that involve terms where the variable is squared. These appear frequently in algebra and are usually of the form:\[ ax^2 + bx + c = 0 \]where \(a\), \(b\), and \(c\) are constants, and \(a eq 0\). In our exercise, we deal with two quadratic equations in a system:

• \(2x^2 - y^2 = 7\)
• 2y^2 - 3x^2 = 2

Characteristics of Quadratic Equations:

• They can curve upwards or downwards based on the leading coefficient \(a\).
• May have two, one, or no real solutions depending on the discriminant \(b^2 - 4ac\).
• Often solved using methods like factoring, completing the square, or the quadratic formula.

In systems involving quadratic equations, like ours, multiple solutions are common due to the shape and nature of parabolas formed by such equations.

Substitution Method

The substitution method is a technique used to solve systems of equations where one variable is expressed in terms of another. By substituting this expression into another equation, you can reduce the number of variables and solve for a specific variable.In our exercise, we express \(y\) in terms of \(x\) from the first equation: \[ y^2 = 2x^2 - 7 \]We then substitute this expression for \(y^2\) into the second equation: \[ 2(2x^2 - 7) - 3x^2 = 2 \]Why Use the Substitution Method?

• Reduces complex systems to simpler forms by focusing on one variable at a time.
• Useful when one equation is easier to manipulate than others.
• Allows converting systems with nonlinear components, such as quadratic equations, into single-variable equations.

This approach transforms the problem, making it more approachable while maintaining the integrity of the original system.

Algebraic Solutions

Algebraic solutions involve solving equations using algebraic manipulations and methods. For systems of equations, these solutions include finding exact answers for variables involved. Our exercise includes tasks like solving for \(x\) after substitution: \[ x^2 - 14 = 2 \]This simplifies our calculation significantly. Once \(x\) is determined, \(y\) can be found easily using previously substituted equations. Key Steps in Finding Algebraic Solutions:

• Isolating terms: Begin by rewriting equations to isolate the target variable.
• Substituting values: Plug in known values or expressions from earlier steps.
• Solving for one variable first: Allows you to focus on reduce complexity and risk error.

These solutions unveil the possible pairs \((x, y)\) that satisfy both equations, which are essential in verifying the accuracy of solutions for systems of quadratic equations.

Real Number Solutions

When solving systems of equations, it's crucial to identify solutions that fall within the real number set, which includes all rational and irrational numbers. In our example, the solution involves real numbers as the equations form parabolas on a coordinate plane. Real numbers consist of both perfect squares and irrational numbers, as demonstrated in the exercise where \(y\) is solved in terms of square roots and leads to:\[ y = \pm \sqrt{23} \]Characteristics of Real Number Solutions:

• Include integers, fractions, decimals, and surds.
• May require approximation when dealing with irrational numbers.
• Can be positive or negative, reflecting the duality of real solutions.

Understanding that quadratic systems can yield multiple real solutions helps in appreciating problem-solving in algebra, ensuring all possible solutions are considered.