In Exercises 91–100, find all values of x satisfying the given conditions. y1 = 6(2x/(x - 3))^2, y2 = 5(2x/(x - 3)), and y1 exceeds y2 by 6

Verified step by step guidance

1. Start by setting up the equation based on the condition that \( y_1 \) exceeds \( y_2 \) by 6: \( y_1 = y_2 + 6 \).
2. Substitute the expressions for \( y_1 \) and \( y_2 \) into the equation: \( 6\left(\frac{2x}{x - 3}\right)^2 = 5\left(\frac{2x}{x - 3}\right) + 6 \).
3. Let \( z = \frac{2x}{x - 3} \) to simplify the equation: \( 6z^2 = 5z + 6 \).
4. Rearrange the equation to form a quadratic equation: \( 6z^2 - 5z - 6 = 0 \).
5. Solve the quadratic equation \( 6z^2 - 5z - 6 = 0 \) using the quadratic formula or factoring, and then substitute back to find the values of \( x \).

Key Concepts

Rational Functions

Rational functions are expressions formed by the ratio of two polynomials. In this question, both y1 and y2 are rational functions involving the variable x. Understanding how to manipulate and analyze these functions is crucial for solving the problem, particularly in identifying their behavior and points of intersection.

Exceeding Condition

The condition that y1 exceeds y2 by 6 translates mathematically to the equation y1 - y2 = 6. This requires setting up an equation based on the given functions and solving for x. Recognizing how to express and manipulate this condition is essential for finding the values of x that satisfy the problem.

Solving Quadratic Equations

The expression for y1 involves squaring a rational function, which can lead to a quadratic equation when simplified. Solving quadratic equations is a fundamental skill in algebra, as it allows us to find the roots or solutions for x. Familiarity with methods such as factoring, completing the square, or using the quadratic formula is necessary to effectively solve the resulting equation.