Problem 38 - College Algebra Essentials

Question

In Exercises \(29-42,\) solve each system by the method of your choice. $$ \left\\{\begin{array}{l} x^{2}-y^{2}-4 x+6 y-4=0 \\ x^{2}+y^{2}-4 x-6 y+12=0 \end{array}\right. $$

Step-by-Step Solution

Answer

The solution to the given system of equations is x = 2 and y = -4.

Step 1: Subtract one equation from the other

Subtract the second equation from the first. \[ x^{2} - y^{2} - 4x + 6y - 4 - (x^{2} + y^{2} - 4x + 6y - 12) = 0\]Solving this will give 2y + 8 = 0.

Step 2: Solve for y

Solving the equation from Step 1 for y will give: \[ y = -4\]

Step 3: Substitute y value into one of the original equations

Substitute y = -4 into one of the original equations to solve for x. Let's substitute y in the first equation, giving:\[ x^{2} - (-4)^{2} - 4x + 6*(-4) - 4 = 0.\]Solve this equation to find the value(s) of x.

Step 4: Solve the equation for x

The equation from Step 3 simplifies to: \[ x^{2} - 16 - 4x + 24 - 4 = 0, \]which further simplifies to: \[ x^{2} - 4x +4 = 0.\]This is a perfect square trinomial, and it can be factored to: \[ (x - 2) ^ 2 = 0.\]The solution is x = 2.

Key Concepts

Solving Quadratic Equations

Solving quadratic equations involves finding the values of the unknown variable that satisfy the equation. A quadratic equation is generally of the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( x \) is the variable. To solve these equations, we can use various methods:

• Factoring: Express the quadratic equation as a product of two binomials.
• Completing the square: Rewriting the equation to form a perfect square trinomial.
• Quadratic formula: Using the formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).

Choosing the right method depends on the specific equation. In problems with simple factors, factoring is quick. If the trinomial structure isn't obvious, use the quadratic formula or complete the square.
Each method helps in efficiently finding the roots of the equation.

Perfect Square Trinomial

A perfect square trinomial is a special type of quadratic expression. It can be expressed as the square of a binomial. The general form of a perfect square trinomial is \( (x + p)^2 = x^2 + 2px + p^2 \).
Identifying a perfect square trinomial helps because it simplifies the equation-solving process. You immediately know that it factors into a square. For instance, in the equation \( x^2 - 4x + 4 = 0 \), notice that it is a perfect square trinomial because:

• \((x - 2)^2 = x^2 - 4x + 4\)

By recognizing this format, you can directly write the factors without extra calculations, allowing us to quickly solve for \( x \).
This special pattern reduces the time and effort needed to solve quadratic equations.

Factoring

Factoring is the process of breaking down an expression into products of simpler expressions. When dealing with quadratic equations, factoring involves writing the quadratic as a product of two binomial expressions, if possible.
For example, let's consider the quadratic equation \( x^2 - 4x + 4 = 0 \). Look for two numbers whose product equals the constant term and whose sum equals the linear coefficient.

• Here, the numbers \(-2\) and \(-2\) multiply to give \(4\) and add to give \(-4\).

This makes \((x - 2)(x - 2) = 0\). Thus, the quadratic is factored. Once factored, use the zero-product property which states that if \( ab = 0 \), then \( a = 0 \) or \( b = 0 \).
Solving gives you the value of \( x \) that satisfies the equation. Factoring is quick with practice and when polynomials are factorable, offering immediate solutions.

Substitution Method

The substitution method is a technique used to solve systems of equations. It involves solving one of the equations for one variable and then substituting this expression into the other equation.
In our original exercise, after simplifying the system, we found \( y = -4 \).
Next, substitute \( y = -4 \) into one of the original equations to solve for \( x \). This transforms our complicated system of equations into a single equation in terms of \( x \) alone.

• This allows us to focus on solving one variable at a time, making the process simpler.

Once one variable is determined, substitute it back into the modified equation to find the second variable.
The substitution method is beneficial when the system of equations makes it straightforward to isolate a variable in one of the equations.