Problem 21 - Precalculus Mathematics for Calculus

Question

Find all solutions of the system of equations. $$\left\\{\begin{array}{l} x-2 y=2 \\ y^{2}-x^{2}=2 x+4 \end{array}\right.$$

Step-by-Step Solution

Answer

The solution is \((x, y) = (-2, -2)\).

Step 1: Solve the First Equation for x

The first equation is \(x - 2y = 2\). Solve this equation for \(x\) by adding \(2y\) to both sides: \(x = 2 + 2y\).

Step 2: Substitute for x in the Second Equation

Take the expression for \(x\) from Step 1, \(x = 2 + 2y\), and substitute it into the second equation \(y^2 - x^2 = 2x + 4\). This gives: \(y^2 - (2 + 2y)^2 = 2(2 + 2y) + 4\).

Step 3: Expand and Simplify the Equation

Expand \((2 + 2y)^2\) to get \(4 + 8y + 4y^2\), and simplify the right-hand side to \(8 + 4y\). The equation becomes \(y^2 - 4 - 8y - 4y^2 = 8 + 4y\), which simplifies to \(-3y^2 - 8y - 4 = 8 + 4y\).

Step 4: Move All Terms to One Side

Rearrange the equation to get all terms on one side: \(-3y^2 - 12y - 12 = 0\).

Step 5: Simplify the Quadratic Equation

Divide the entire equation by \(-3\) to simplify: \(y^2 + 4y + 4 = 0\).

Step 6: Solve the Quadratic Equation for y

Notice that \(y^2 + 4y + 4\) is a perfect square. It can be factored as \((y+2)^2 = 0\), resulting in the solution \(y = -2\).

Step 7: Substitute y back to Find x

Use the solution \(y = -2\) in the equation \(x = 2 + 2y\) to find \(x\). Substitute \(y = -2\) to get \(x = 2 + 2(-2) = 2 - 4 = -2\).

Step 8: Check the Solution in Both Equations

Substitute \(x = -2\) and \(y = -2\) back into the original equations. For the first equation, \(-2 - 2(-2) = 2\) holds true. For the second equation, \((-2)^2 - (-2)^2 = 2(-2) + 4\) simplifies to \(4 - 4 = -4 + 4\), which is also true. Therefore, both equations are satisfied by \(x = -2\) and \(y = -2\).

Key Concepts

Quadratic Equations

Quadratic equations are a central topic in algebra and are fundamental in understanding systems of equations involving second-degree polynomials. A quadratic equation is one that can be written in the form \(ax^2 + bx + c = 0\), where \(a eq 0\).
The equation typically forms a parabola when graphed on a coordinate plane. Solving quadratic equations often involves finding the "roots" or the points where the parabola crosses the x-axis.

There are several methods to solve quadratic equations. These include:

• Factoring: Expressing the quadratic as a product of two terms to solve for the variables.
• Completing the Square: Reorganizing the equation to form a perfect square trinomial.
• Quadratic Formula: Using the formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) to find the roots.

In the given system of equations, the quadratic created through algebraic manipulation was solved by recognizing it as a perfect square, \((y+2)^2\), leading to the solution \(y = -2\). Understanding these methods is crucial as they provide different strategies in tackling quadratic equations.

Substitution Method

The substitution method is a technique used to solve systems of equations, which involves solving one equation for a particular variable and then substituting that expression into another equation.
This method is especially useful in systems where one of the equations is already solved for one variable, or can easily be manipulated to do so. The goal is to reduce the system into a single equation with one variable, simplifying the solving process.

Here’s how it works in the given exercise:

• The first step was to solve the first linear equation \(x - 2y = 2\) for \(x\), resulting in \(x = 2 + 2y\).
• This expression for \(x\) was then substituted into the second equation \(y^2 - x^2 = 2x + 4\).
• This substitution transformed the problem into a single quadratic equation in terms of \(y\), which was easier to solve.

The substitution method is powerful because it can transform complex, multi-variable systems into simpler forms that are much easier to manage.

Algebraic Manipulation

Algebraic manipulation is the process of using algebraic techniques to rearrange and simplify expressions to solve equations.
This includes adding, subtracting, multiplying, or dividing both sides of the equation, factoring, expanding, and combining like terms to achieve the desired result.

In the provided exercise:

• Algebraic manipulation was used extensively to simplify the equations after substitution.
• After substituting \(x = 2 + 2y\) into the second equation, it required expanding \((2 + 2y)^2\) into \(4 + 8y + 4y^2\) and simplifying both sides to bring all terms together for perfect-square factoring.
• After simplification, the equation \(y^2 + 4y + 4 = 0\) was easily identified as a perfect square, allowing for a straightforward solution for \(y\).

Mastery of algebraic manipulation is key to solving complex systems of equations successfully, as it allows you to see the relationships between terms and create opportunities to factor or simplify.