Problem 27 - Precalculus: Mathematical 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: Analyze the Equations

The system consists of two equations: the first one, \(x - y = 4\), is linear, and the second one, \(xy = 12\), is nonlinear. Our goal is to find values for \(x\) and \(y\) that satisfy both equations simultaneously.

Step 2: Express One Variable in Terms of Another

From the first equation, \(x - y = 4\), we can express \(x\) in terms of \(y\): \[x = y + 4\]

Step 3: Substitute the Expression into the Second Equation

Substitute \(x = y + 4\) into the second equation \(xy = 12\): \[(y + 4)y = 12\]Simplify it to get:\[y^2 + 4y = 12\]

Step 4: Rearrange and Simplify the Quadratic Equation

Rearrange the equation from Step 3 to form a standard quadratic equation:\[y^2 + 4y - 12 = 0\]

Step 5: Solve the Quadratic Equation

To solve \(y^2 + 4y - 12 = 0\), use the quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = 4\), and \(c = -12\). Calculate:\[y = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot (-12)}}{2 \cdot 1}\]\[y = \frac{-4 \pm \sqrt{16 + 48}}{2}\]\[y = \frac{-4 \pm \sqrt{64}}{2}\]\[y = \frac{-4 \pm 8}{2}\]Thus, the solutions for \(y\) are:\[y_1 = 2, \quad y_2 = -6\]

Step 6: Find Corresponding Values of x for Each y Solution

For \(y_1 = 2\), substitute back into \(x = y + 4\): \[x_1 = 2 + 4 = 6\]For \(y_2 = -6\), substitute back into \(x = y + 4\): \[x_2 = -6 + 4 = -2\]

Step 7: Write the Solutions as Ordered Pairs

The solutions to the system of equations are the pairs \((x_1, y_1)\) and \((x_2, y_2)\). Thus, the solutions are:\[(6, 2)\] and \[(-2, -6)\]

Key Concepts

Quadratic Equation

A quadratic equation is a type of polynomial equation of the second degree. It generally follows the standard form:

• \( ax^2 + bx + c = 0 \)

where \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \). The presence of the squared term \( x^2 \) makes it a quadratic.
In our problem, after substituting the expression for \( x \) from the linear equation into the nonlinear equation, we derived the quadratic equation:

• \( y^2 + 4y - 12 = 0 \)

This specific equation gives us two potential solutions for \( y \). Quadratic equations can be solved using various methods including:

• Factoring
• Completing the square
• Quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

Quadratics are noteworthy because they often have two solutions, which in our problem resulted in two possible points \((x, y)\) when solved.

Linear Equation

A linear equation is an equation involving two variables that creates a straight line when graphed. It is typically in the form:

• \( ax + by = c \)

where \( a \), \( b \), and \( c \) are constants. It describes a relationship where there is a constant rate of change.
In the system given, the linear equation is:

• \( x - y = 4 \)

This equation indicates that the difference between \( x \) and \( y \) is always 4. It's quite straightforward and crucial because it allows for simplifying the nonlinear equation by expressing one variable in terms of the other.
This linear expression then gets substituted into the quadratic equation, allowing us to focus on a single variable for solving purposes.

Substitution Method

The substitution method is a process applied to solve systems of equations where at least one equation is nonlinear. It involves these steps:

• Solve one of the equations for one variable.
• Substitute this expression into the other equation.

This reduces the system to a single equation in one variable. In our exercise:

• First, we transformed the linear equation \( x - y = 4 \) to express \( x \) as \( y + 4 \).
• Then, we substituted \( x = y + 4 \) into the nonlinear equation \( xy = 12 \).
• This resulted in a quadratic equation \( (y + 4)y = 12 \).

By employing the substitution method, the complexity of solving a system is reduced, as it combines the equations into one and focuses on consistently solving for values of a single variable.

Solution of Equations

The solution of a system of equations refers to the set of values of the variables that satisfy all equations in the system simultaneously. Such solutions can be represented as ordered pairs for two variables.
For our problem, we had two equations:

• \( x - y = 4 \)
• \( xy = 12 \)

After applying the substitution method, we derived the quadratic equation whose solutions were:

• \( y_1 = 2 \)
• \( y_2 = -6 \)

Substituting these back into \( x = y + 4 \) gave us the corresponding \( x \) values:

• For \( y_1 = 2 \), \( x_1 = 6 \)
• For \( y_2 = -6 \), \( x_2 = -2 \)

Thus, the complete solutions for the system of equations are the pairs \((6, 2)\) and \((-2, -6)\). These solutions indicate the points where both equations intersect, satisfying the conditions of each equation concurrently.