Problem 60 - Intermediate Algebra

Question

Solve each system of equations for real values of \(x\) and \(y.\) $$ \left\\{\begin{array}{l} x y=\frac{1}{12} \\ y+x=7 x y \end{array}\right. $$

Step-by-Step Solution

Answer

The solutions are \(\left(x, y\right) = \left(\frac{1}{3}, \frac{1}{4}\right)\) and \(\left(x, y\right) = \left(\frac{1}{4}, \frac{1}{3}\right)\).

Step 1: Solve for one variable

From the first equation, \(xy = \frac{1}{12}\), solve for \(y\) in terms of \(x\). Divide both sides by \(x\) to get \(y = \frac{1}{12x}\).

Step 2: Substitute in the second equation

Substitute \(y = \frac{1}{12x}\) into the second equation \(y + x = 7xy\). This gives \(\frac{1}{12x} + x = 7x \cdot \frac{1}{12x}\).

Step 3: Simplify the equation

Simplify the equation \(\frac{1}{12x} + x = \frac{7}{12}\). Multiply every term by \(12x\) to clear the fraction: \(1 + 12x^2 = 7x\).

Step 4: Rearrange into a quadratic equation

Rearrange the simplified equation into a standard quadratic form: \(12x^2 - 7x + 1 = 0\).

Step 5: Solve the quadratic equation

Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 12\), \(b = -7\), and \(c = 1\). Substitute these values to find \(x\).

Step 6: Calculate the discriminant

Calculate the discriminant \(b^2 - 4ac\) as follows: \((-7)^2 - 4 \cdot 12 \cdot 1 = 49 - 48 = 1\). Since the discriminant is positive, there are two real solutions.

Step 7: Find the roots using the quadratic formula

Solve using the quadratic formula: \(x = \frac{7 \pm 1}{24}\). This gives two solutions for \(x\): \(x = \frac{8}{24} = \frac{1}{3}\) and \(x = \frac{6}{24} = \frac{1}{4}\).

Step 8: Solve for corresponding \(y\) values

For \(x = \frac{1}{3}\), substitute back into \(y = \frac{1}{12x}\) to find \(y = \frac{1}{12 \cdot \frac{1}{3}} = \frac{1}{4}\).For \(x = \frac{1}{4}\), substitute back to get \(y = \frac{1}{12 \cdot \frac{1}{4}} = \frac{1}{3}\).

Step 9: Solution validation

Verify the solutions: For \((x, y) = \left(\frac{1}{3}, \frac{1}{4}\right)\) and \((x, y) = \left(\frac{1}{4}, \frac{1}{3}\right)\), both pairs satisfy the original equations. Therefore, these are the solutions.

Key Concepts

Quadratic Equations

A quadratic equation is a type of polynomial equation of the second degree. Its standard form is \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are real numbers, and \( a \) is not equal to zero. These equations are pivotal in algebra due to their wide applications and unique properties. Quadratic equations can have:

• Two real roots
• One real root (a double root)
• No real roots (when solutions are complex numbers)

The number of real roots is determined by the discriminant \( b^2 - 4ac \). If the discriminant is:

• Positive, there are two distinct real roots.
• Zero, there is exactly one real root.
• Negative, no real roots exist, and solutions are complex.

Solving quadratic equations often involves factoring, completing the square, or using the quadratic formula. In the provided exercise, we used the quadratic formula to solve \( 12x^2 - 7x + 1 = 0 \) and found two real solutions for \( x \). Thus, showing how these types of equations play a critical role in finding variable values in systems.

Real Numbers

Real numbers are all the numbers that can be found on the number line. This vast set includes both rational numbers (such as fractions and integers) and irrational numbers (like \( \pi \) or \( \sqrt{2} \)). In the context of solving systems of equations, such as the one given, we look for solutions that fall within the set of real numbers. When the discriminant of a quadratic equation is non-negative, it ensures that the solution exists within the real numbers.
In the exercise provided, after calculating the discriminant \( 1 \), we confirmed its positivity, indicating two real solutions for \( x \): \( \frac{1}{3} \) and \( \frac{1}{4} \). Real numbers are foundational in algebra and other mathematical disciplines because solutions must be practical and applicable in real-world scenarios.

Algebraic Manipulation

Algebraic manipulation involves rearranging and simplifying equations to make them easier to solve. It requires a strong understanding of algebraic rules, such as distributing and combining like terms. In this exercise, we started with a system of equations and employed algebraic manipulation by rearranging terms and simplifying expressions. For example, when given \( xy = \frac{1}{12} \), we isolated \( y \) by dividing both sides by \( x \), obtaining \( y = \frac{1}{12x} \).
We substituted this expression for \( y \) in the second equation \( y + x = 7xy \) and further manipulated the equation \( \frac{1}{12x} + x = \frac{7}{12} \). By multiplying through by \( 12x \), we eliminated fractions leading to \( 1 + 12x^2 = 7x \), simplifying to a quadratic equation \( 12x^2 - 7x + 1 = 0 \). This example shows how algebraic manipulation is a key skill, allowing us to transform complex systems into solvable formats.