Problem 52 - Intermediate Algebra

Question

Solve each system of equations for real values of \(x\) and \(y.\) $$ \left\\{\begin{array}{l} x y=-\frac{9}{2} \\ 3 x+2 y=6 \end{array}\right. $$

Step-by-Step Solution

Answer

The solutions are \((x, y) = (-1, \frac{9}{2})\) and \((x, y) = (3, -\frac{3}{2})\).

Step 1: Express one variable in terms of the other

From the second equation, solve for one variable in terms of the other. Let's solve for \(x\):\[3x + 2y = 6\]Subtract \(2y\) from both sides:\[3x = 6 - 2y \]Divide by 3 to solve for \(x\):\[x = 2 - \frac{2}{3}y \]

Step 2: Substitute into the first equation

Use the expression for \(x\) from Step 1 and substitute it into the first equation \(xy = -\frac{9}{2}\):\[(2 - \frac{2}{3}y)y = -\frac{9}{2} \]This simplifies to \[2y - \frac{2}{3}y^2 = -\frac{9}{2}\].

Step 3: Simplify to a standard quadratic form

Rearrange the terms to obtain a quadratic equation in terms of \(y\):\[-\frac{2}{3}y^2 + 2y + \frac{9}{2} = 0\]To simplify this, multiply through by \(-3\) to clear the fraction:\[2y^2 - 6y - \frac{27}{2} = 0\].

Step 4: Solve the quadratic equation for y

Multiply throughout by 2 to clear the denominator, resulting in:\[4y^2 - 12y - 27 = 0\]Apply the quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) with \(a = 4\), \(b = -12\), \(c = -27\):\[y = \frac{12 \pm \sqrt{144 + 432}}{8}\]Calculate the discriminant: \(\sqrt{576} = 24\), so:\[y = \frac{12 \pm 24}{8}\].

Step 5: Calculate specific values for y

This results in two solutions for \(y\):\[y_1 = \frac{12 + 24}{8} = \frac{36}{8} = \frac{9}{2}\]and\[y_2 = \frac{12 - 24}{8} = \frac{-12}{8} = -\frac{3}{2}\]This gives us the potential solutions for \(y\).

Step 6: Solve for x in both scenarios

Substitute \(y = \frac{9}{2}\) back into \(x = 2 - \frac{2}{3}y\):\[x = 2 - \frac{2}{3} \times \frac{9}{2} = 2 - 3 = -1\].Substitute \(y = -\frac{3}{2}\) back into \(x = 2 - \frac{2}{3}y\):\[x = 2 - \frac{2}{3} \times \left(-\frac{3}{2}\right) = 2 + 1 = 3\].

Step 7: Verify the solutions in the original equations

For \((x, y) = (-1, \frac{9}{2})\): Check \((xy = -\frac{9}{2})\) and \(3x + 2y = 6\):\(-1 \times \frac{9}{2} = -\frac{9}{2}\) and \(3(-1) + 2\left(\frac{9}{2}\right) = -3 + 9 = 6\).For \((x, y) = (3, -\frac{3}{2})\):\(3 \times -\frac{3}{2} = -\frac{9}{2}\) and \(3(3) + 2(-\frac{3}{2}) = 9 - 3 = 6\). Both sets satisfy the original equations.

Key Concepts

Quadratic Equations

Quadratic equations are a central concept in algebra that typically involve polynomials of degree 2. They are generally in the form:

• \[ ax^2 + bx + c = 0 \]

where \(a\), \(b\), and \(c\) represent constants, and \(a eq 0\). In these equations, the highest power of the variable, usually denoted as \(x\), is a square term (\(x^2\)).

When solving quadratic equations, the aim is to find the values of \(x\) that satisfy the equation. There are several methods:

• Factoring
• Completing the square
• Using the quadratic formula

In many situations, the quadratic formula \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]is a useful and reliable method, especially when factoring is complex or impossible. It is essential to note that the term under the square root, \(b^2 - 4ac\), called the discriminant, determines the nature of the roots:

• If the discriminant is positive, there are two distinct real roots.
• If it's zero, there's exactly one real root.
• If it's negative, the roots are complex (not real).

This understanding of quadratic equations is crucial because they often appear in real-world problems and are a stepping stone to more advanced algebra concepts.

Substitution Method

The substitution method is a popular technique used to solve systems of equations, particularly when dealing with non-linear systems or mixtures of linear and nonlinear equations. This method involves expressing one variable in terms of another, thereby reducing the system by substituting into the other equation.

Here's how it generally works:

• Select one of the equations and solve for one variable in terms of the other(s).
• Substitute this expression into the other equation(s). This reduces the number of equations and variables.
• Solve the resulting equation for the remaining variable.
• Back-substitute to find the other variable(s).

For example, in a system involving a linear and a quadratic equation like in the exercise provided, solving the linear equation for one variable and then substituting into the quadratic equation simplifies solving significantly. It eventually turns the problem into a single-variable quadratic equation, which you can solve using established techniques.

This method is very powerful for its simplicity and directness, making it particularly useful when one equation is already "solved" for one of the variables or can be easily rearranged. It lays the foundation for understanding more complex methods used in higher-level mathematics.

Solving Simultaneous Equations

Solving simultaneous equations involves finding the values of variables that satisfy all equations in a system at the same time. Such systems can consist of linear equations, nonlinear equations, or a mix of both. Systems with two equations and two unknowns are most commonly encountered at an introductory level.

There are several methods to solve simultaneous equations:

• Graphical Method: Plotting both equations on a coordinate plane and finding their intersection point(s), which represent the solutions.
• Substitution Method: Expressing one variable in terms of the other and substituting this into the second equation, as detailed earlier.
• Elimination Method: Adding or subtracting equations to eliminate one of the variables, simplifying the problem to a single equation.

When solving these equations, consistency and the nature of the system play crucial roles:

• If the equations represent parallel lines, the system has no solution (inconsistent).
• If they represent the same line, an infinite number of solutions exist (dependent).
• If they meet at a point, there is a unique solution (independent).

Understanding and mastering these methods is essential for tackling more complex systems of equations, which frequently occur in engineering, physics, economics, and various scientific disciplines.