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

Start with the first equation: \[ x - 2y = 2 \]Rearrange it to express \( x \) in terms of \( y \):\[ x = 2y + 2 \]

Step 2: Substitute x in the Second Equation

Now take the expression for \( x \) from Step 1 and substitute it into the second equation:\[ y^2 - x^2 = 2x + 4 \]Replace \( x \) to get:\[ y^2 - (2y + 2)^2 = 2(2y + 2) + 4 \]

Step 3: Simplify the Substituted Equation

Now simplify the equation:\[ y^2 - (4y^2 + 8y + 4) = 4y + 4 + 4 \]This simplifies to:\[ y^2 - 4y^2 - 8y - 4 = 4y + 8 \]Combine like terms:\[ -3y^2 - 8y - 4 = 4y + 8 \]

Step 4: Rearrange and Solve the Quadratic Equation

Rearrange the terms to form a quadratic equation:\[ -3y^2 - 8y - 4 - 4y - 8 = 0 \]Combine like terms:\[ -3y^2 - 12y - 12 = 0 \]Multiply by -1 to make it standard:\[ 3y^2 + 12y + 12 = 0 \]Divide by 3 to simplify:\[ y^2 + 4y + 4 = 0 \]Factor the quadratic equation:\[ (y+2)^2 = 0 \]So, \( y = -2 \).

Step 5: Solve for x Using y

Substitute \( y = -2 \) back into the expression for \( x \):\[ x = 2(-2) + 2 \]\[ x = -4 + 2 \]\[ x = -2 \]

Step 6: Verify the Solution in Both Equations

Now verify the solution \((x, y) = (-2, -2)\) in the original equations:1. First equation: \( x - 2y = 2 \) \[ -2 - 2(-2) = 2 \rightarrow -2 + 4 = 2 \text{ which is correct.} \]2. Second equation: \( y^2 - x^2 = 2x + 4 \) \[ (-2)^2 - (-2)^2 = 2(-2) + 4 \rightarrow 4 - 4 = -4 + 4 \rightarrow 0 = 0 \text{ which is correct.} \]

Key Concepts

Quadratic Equations

Quadratic equations are essential in mathematics and appear in various applications. A quadratic equation is any equation that can be expressed in the form:

• \( ax^2 + bx + c = 0 \)

where \( a \), \( b \), and \( c \) are constants and \( a eq 0 \). The graph of a quadratic function is a parabola, which can open upwards or downwards depending on the sign of \( a \).

In our exercise, we encountered a quadratic equation when we rearranged and combined terms from the system of equations. The equation:

• \( y^2 + 4y + 4 = 0 \)

is a quadratic equation in standard form. This equation can be solved by factoring, completing the square, or using the quadratic formula. In this case, it was factored as \((y+2)^2 = 0\), leading to a double root \( y = -2 \).

Understanding quadratic equations helps in solving systems that involve second-degree polynomials.

Substitution Method

The substitution method is a common technique for solving systems of equations, especially when one equation is linear. It involves solving one equation for a variable and substituting that expression into the other equation.

Let's apply this method step-by-step as we did in the given exercise. We began with:

• \( x - 2y = 2 \)

We solved for \( x \):

• \( x = 2y + 2 \)

Next, we substituted this expression for \( x \) into the second equation:

• \( y^2 - x^2 = 2x + 4 \)
• Substitute \( x \): \( y^2 - (2y+2)^2 = 2(2y+2) + 4 \)

This way, we reduced the system to a single equation in terms of \( y \), which we simplified and solved as a quadratic equation. The substitution method streamlines solving systems of equations by reducing them to simpler forms.

Verification of Solutions

Verification is a critical step in solving systems of equations to ensure the solutions satisfy all given equations. It involves substituting the found values back into the original equations.

In our exercise, once we determined that \( x = -2 \) and \( y = -2 \), the next step was to verify these solutions in the original equations:

• First equation: \( x - 2y = 2 \). Substituting gives \(-2 - 2(-2) = 2\), which holds true.
• Second equation: \( y^2 - x^2 = 2x + 4 \). Substituting gives \((-2)^2 - (-2)^2 = 2(-2) + 4\), simplifying to \(0 = 0\), which is correct.

Verification not only supports the correctness of your solution but also boosts your confidence that no errors were made during calculations. Ensuring each solution satisfies the original equations is an integral part of solving systems of equations effectively.