Problem 14 - Algebra and Trigonometry

Question

Use the elimination method to find all solutions of the system of equations. \(\left\\{\begin{aligned} 2 x^{2}+4 y &=13 \\ x^{2}-y^{2} &=\frac{7}{2} \end{aligned}\right.\)

Step-by-Step Solution

Answer

Solutions are approximately \((4.46, 1.25)\) and \((-4.46, 1.25)\).

Step 1: Express One Variable in Terms of Another

Start by solving one of the equations for one variable. Let's take the simpler form: from the first equation, solve for \( y \). Rearrange it as follows: \( 4y = 13 - 2x^2 \). Then, divide both sides by 4 to express \( y \) in terms of \( x \): \( y = \frac{13 - 2x^2}{4} \).

Step 2: Substitute in the Second Equation

Now substitute the expression for \( y \) from Step 1 into the second equation. Start with the equation: \( x^2 - y^2 = \frac{7}{2} \). Substitute \( y = \frac{13 - 2x^2}{4} \) into the equation, giving: \( x^2 - \left(\frac{13 - 2x^2}{4}\right)^2 = \frac{7}{2} \).

Step 3: Expand and Simplify the Equation

Expand the squared term: \( \left(\frac{13 - 2x^2}{4}\right)^2 = \frac{(13 - 2x^2)^2}{16} \). Simplify: \((13 - 2x^2)^2 = 169 - 52x^2 + 4x^4\). Substituting back, the equation becomes: \( x^2 - \frac{169 - 52x^2 + 4x^4}{16} = \frac{7}{2} \).

Step 4: Clear the Fractions

Multiply the entire equation by 16 to eliminate fractions: \( 16x^2 - (169 - 52x^2 + 4x^4) = 56 \). Solve the resulting equation: \( 16x^2 - 169 + 52x^2 - 4x^4 = 56 \). Rearrange it: \( -4x^4 + 68x^2 - 169 = 56 \).

Step 5: Solve the Quadratic in Terms of \( x^2 \)

Bring all terms to one side to form a quadratic equation: \( -4x^4 + 68x^2 - 225 = 0 \). Let \( z = x^2 \), turning the equation into a quadratic: \( -4z^2 + 68z - 225 = 0 \).

Step 6: Factor or Use the Quadratic Formula

Given the complexity in factoring, use the quadratic formula: \[ z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]where \( a = -4 \), \( b = 68 \), and \( c = -225 \). Calculate the discriminant: \( b^2 - 4ac = 4624 + 3600 = 8224 \). So, \( z = \frac{-68 \pm \sqrt{8224}}{-8} \).

Step 7: Calculate the Roots for \( z = x^2 \)

Compute the square root: \( \sqrt{8224} = 90.66 \approx 91 \). So, \( z = \frac{-68 \pm 91}{-8} \). Calculate each possibility:1. \( z_1 = \frac{-68 + 91}{-8} = \frac{23}{-8} ot\) possible for a square.2. \( z_2 = \frac{-68 - 91}{-8} = \frac{-159}{-8} \approx 19.875 \).

Step 8: Substitute Back to Find \( x \) and \( y \)

For \( z_2 = 19.875 \), since \( z = x^2 \), solve for \( x \): \( x = \pm \sqrt{19.875} \approx \pm 4.46 \). Use \( x \) values in \( y = \frac{13 - 2x^2}{4} \) to find corresponding \( y \) values. Calculate \( y_1 \approx -5.25 \) or \( y_2 \approx 1.25 \).

Step 9: Validate Solutions in Original Equations

Check each pair \((x, y)\) in the original equations to ensure they satisfy both equations. Substitute values back into both equations and ensure they hold true.

Key Concepts

Systems of Equations

Systems of equations consist of two or more equations with multiple variables, seeking solutions that satisfy all given equations simultaneously. In this exercise, we encounter a system involving two equations with two variables, typically labeled as \( x \) and \( y \). Solving these systems is essential in various fields like engineering, physics, and economics, where multiple conditions must be satisfied at once.

The main goal when solving systems of equations is to find values for the variables that satisfy every condition presented. There are several methods to approach this, such as graphing, substitution, and elimination. Each method has its strengths, and the choice often depends on the nature of the equations involved.

• Graphical Method: Involves plotting each equation on a graph to find their intersection point(s). However, this method can be less precise.
• Substitution Method: Involves solving one equation for a variable and substituting that expression in the other equations, a process we'll explore in this discussion.
• Elimination Method: Also called the addition method, this involves adding or subtracting equations to cancel out variables, thus simplifying the system.

Understanding these methods provides a useful toolkit for tackling systems of equations effectively.

Quadratic Equations

Quadratic equations are equations of the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants. These equations represent parabolas when graphed and can have up to two solutions, also known as roots. In our system, we encounter quadratic terms such as \( 2x^2 \) and \( -y^2 \), making them quadratic equations when considered separately.

Solving quadratic equations can be done using several methods:

• Factoring: This involves expressing the equation as a product of its factors, if possible. This method is usually straightforward but is limited to equations that can be easily factorized.
• Completing the Square: A more general method that involves rewriting the equation in a form that reveals the roots.
• Quadratic Formula: A powerful tool given by the formula \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]. This method works for all quadratic equations, making it very versatile.

In the exercise, the quadratic formula was used to find solutions due to the complexity of the resulting equation, highlighting the utility of this method when other approaches aren't feasible.

Substitution Method

The substitution method is a technique for solving systems of equations where one equation is solved for a variable, and that expression is substituted into the other equation(s). This method is especially effective when dealing with non-linear systems, such as those involving quadratic equations.

The process begins with isolating one variable. In our exercise, the given system was:
\[\begin{aligned}2x^2 + 4y &= 13 \x^2 - y^2 &= \frac{7}{2}\end{aligned}\]
To use substitution, we first solved the first equation for \( y \), expressing it in terms of \( x \): \( y = \frac{13 - 2x^2}{4} \). By substituting this expression into the second equation, a more manageable equation in terms of \( x \) was obtained. This simplification allowed us to focus on solving a single quadratic equation.

The substitution method is advantageous because it usually reduces the problem to dealing with one variable. However, it requires careful algebraic manipulation to avoid errors. This method is often contrasted with elimination, which might be more efficient for linear systems with straightforward coefficients.