Problem 23 - Precalculus Mathematics for Calculus

Question

Find all solutions of the system of equations. $$\left\\{\begin{aligned} x-y &=4 \\ x y &=12 \end{aligned}\right.$$

Step-by-Step Solution

Answer

The solutions are \((6, 2)\) and \((-2, -6)\).

Step 1 (Express x in terms of y)

From the first equation, we have \(x - y = 4\). Rearrange this equation to express \(x\) in terms of \(y\): \(x = y + 4\).

Step 2 (Substitute x into the second equation)

Take the expression for \(x\) from Step 1, \(x = y + 4\), and substitute it into the second equation of the system: \(x y = 12\). This yields \((y + 4)y = 12\).

Step 3 (Solve the resulting quadratic equation)

The equation from Step 2 becomes \(y^2 + 4y = 12\). Subtract 12 from both sides to form the quadratic equation: \(y^2 + 4y - 12 = 0\). Use the quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), with \(a = 1\), \(b = 4\), and \(c = -12\): \[ y = \frac{-4 \pm \sqrt{16 + 48}}{2} = \frac{-4 \pm \sqrt{64}}{2} \].

Step 4 (Calculate y values)

Evaluate the solutions for \(y\): \(y = \frac{-4 \pm 8}{2}\). This gives two solutions: \(y_1 = 2\) and \(y_2 = -6\).

Step 5 (Determine corresponding x values)

Substitute \(y_1 = 2\) into \(x = y + 4\) to get \(x_1 = 6\). Similarly, substitute \(y_2 = -6\) into \(x = y + 4\) to get \(x_2 = -2\). Thus, the solutions \((x_1, y_1)\) is \((6, 2)\), and \((x_2, y_2)\) is \((-2, -6)\).

Key Concepts

Quadratic Equations

Quadratic equations are equations of the second degree, typically expressed in the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \). These equations are called quadratic because of the term "quad" hinting at squares, the highest power linked to the variable here. A unique feature of quadratic equations is their graphical representation, which is a parabola.

When dealing with quadratic equations, the most well-known method to find the solutions is the quadratic formula:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

This formula helps us find the roots or solutions to any quadratic equation. The expression inside the square root, \( b^2 - 4ac \), is called the discriminant. It determines the nature of the solutions (roots):

• If it's positive, there are two real solutions.
• If it's zero, there is exactly one real solution.
• If it's negative, the solutions are complex or imaginary.

In the given problem, after rearranging and simplifying the terms, we applied the quadratic formula to derive the solutions for \( y \). This set the foundation for finding the corresponding values of \( x \).

Substitution Method

The substitution method is a technique used to solve systems of equations where one variable is replaced with an expression containing the other variables. This approach simplifies the problem, making it easier to find the solution.

Let's break it down step-by-step:

• First, solve one of the equations for one variable in terms of the others. In this problem, the first equation gives us \( x = y + 4 \).
• Next, substitute this expression into the other equation. Therefore, the original system reduces the two separate equations into a single equation in one variable.

This method transforms a two-variable system into a single-variable equation, like the quadratic equation we derived: \( y^2 + 4y - 12 = 0 \). With our simpler equation, solving for \( y \) became a straightforward task using the quadratic formula. Once one variable is determined, the corresponding values for the other variable can easily be found by substituting back into one of the original equations.

Solution of Equations

Solving systems of equations involves finding the set of values (solutions) that satisfy all given equations in the system. These solutions are often expressed as ordered pairs (x, y) in the context of two-variable systems.

In this problem, we aimed to solve the system:

• \( x - y = 4 \)
• \( xy = 12 \)

By applying the substitution method, we simplified the system into a single quadratic equation. Solving this equation revealed two possible values for \( y \), namely \( y = 2 \) and \( y = -6 \). Each value of \( y \) corresponds to a specific value of \( x \) determined through substitution:

• For \( y = 2 \), substituting into \( x = y + 4 \), we found \( x = 6 \).
• For \( y = -6 \), similarly, we derived \( x = -2 \).

Hence, the complete solutions for the system of equations are \((6, 2)\) and \((-2, -6)\). This demonstrates how two different pairs of numbers can simultaneously satisfy both equations in the system, providing a complete understanding of the potential solutions.