Problem 25 - Algebra and Trigonometry

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 to the system is \( x = -2, y = -2 \).

Step 1: Solve the first equation for x

The first equation is \( x - 2y = 2 \). To solve for \( x \), add \( 2y \) to both sides: \( x = 2y + 2 \).

Step 2: Substitute x in the second equation

Substitute \( x = 2y + 2 \) into the second equation \( y^2 - x^2 = 2x + 4 \). This gives:\[ y^2 - (2y + 2)^2 = 2(2y + 2) + 4 \].

Step 3: Expand and simplify the equation

Expand \( (2y + 2)^2 \) to get \( 4y^2 + 8y + 4 \), and expand \( 2(2y + 2) + 4 \) to get \( 4y + 4 + 4 = 4y + 8 \). Thus, the equation becomes:\[ y^2 - 4y^2 - 8y - 4 = 4y + 8 \] which simplifies to:\[ -3y^2 - 8y - 4 = 4y + 8 \].

Step 4: Rearrange to form a quadratic equation

Move all terms to one side of the equation:\[ -3y^2 - 12y - 12 = 0 \].

Step 5: Simplify the quadratic equation

Divide the entire equation by -3 to simplify:\[ y^2 + 4y + 4 = 0 \]. This can be factored as \( (y + 2)^2 = 0 \).

Step 6: Solve for y

Solve the equation \( (y + 2)^2 = 0 \), which gives \( y = -2 \).

Step 7: Solve for x using the value of y

Substitute \( y = -2 \) back into the equation \( x = 2y + 2 \) to find \( x \):\[ x = 2(-2) + 2 = -4 + 2 = -2 \].

Step 8: Conclude the solution

The solutions to the system of equations are \( x = -2 \) and \( y = -2 \). The ordered pair is \((-2, -2)\).

Key Concepts

Quadratic Equations

Quadratic equations are equations of the form \( ax^2 + bx + c = 0 \), where \( a, b, \) and \( c \) are constants and \( a eq 0 \). These equations graph as parabolas in the coordinate plane, which can open upwards or downwards.
Solving quadratic equations is a frequent task in algebra, and they can have up to two real solutions. Several methods can be used to find these solutions:

• **Factoring**: This involves expressing the quadratic equation as a product of its linear factors.
• **Completing the Square**: This method transforms the quadratic into a perfect square trinomial.
• **Quadratic Formula**: This formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), provides direct solutions for any quadratic equation that might not be easily factorable.

In the given problem, the quadratic part arises in Step 5, where we factorized \( y^2 + 4y + 4 = 0 \) into \((y+2)^2 = 0\). This demonstrated the equation only has one solution, \(y = -2\), indicating a parabola that touches the x-axis at only one point.

Substitution Method

The substitution method is a useful technique for solving systems of equations. It involves solving one equation for a single variable and substituting this expression into the other equation.
The benefit is that it reduces two-variable problems to one variable, simplifying the solving process.
In the original exercise, we used the substitution method successfully:

• First, we solved the linear equation \( x - 2y = 2 \) for \( x \) and got \( x = 2y + 2 \).
• Then, we substituted \( x = 2y + 2 \) into the second quadratic equation \( y^2 - x^2 = 2x + 4 \).

This substitution transformed a more complex system into simpler terms that provided a clear path to the solution of the system.

Factoring

Factoring is a technique used to simplify polynomials into a product of simpler expressions, making it easier to solve equations. When dealing with quadratic equations, factoring is often an effective strategy when the trinomial can be expressed as the square of a binomial or the product of two distinct binomials.
In the step-by-step solution, we reached a quadratic equation \( y^2 + 4y + 4 = 0 \), which we recognized as a perfect square trinomial. This allowed us to factor it as \((y+2)^2 = 0\).
The process of factoring generally involves:

• Finding two numbers that multiply to the constant term \( c \) and add up to the coefficient of the linear term \( b \).
• Expressing the quadratic as the product of two binomials.

In cases where the quadratic is in a perfect square form like in our example, it can be directly expressed as \((y+k)^2 = 0\). Solving \((y+k)^2 = 0\) is straightforward, as it directly gives the solution \(y = -k\). In the context of our problem, this approach led us directly to the solution \( y = -2 \).
Factoring simplifies quadratic equations and often makes it quicker to find the roots, especially for straightforward cases.