In Exercises 91–100, find all values of x satisfying the given conditions.$y_1 = 6 \left( \frac{2x}{x - 3} \right)^2, \quad y_2 = 5 \left( \frac{2x}{x - 3} \right), \quad \text{and} \quad y_1 \text{ exceeds } y_2 \text{ by } 6.$

Verified step by step guidance

1. Start by translating the condition "y1 exceeds y2 by 6" into an equation. This means that y1 is equal to y2 plus 6, so write: \(y_1 = y_2 + 6\).
2. Substitute the given 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. To simplify the equation, let \(t = \frac{2x}{x - 3}\). Rewrite the equation in terms of \(t\): \$6t^2 = 5t + 6$.
4. Rearrange the equation to standard quadratic form: \$6t^2 - 5t - 6 = 0$.
5. Solve the quadratic equation for \(t\) using the quadratic formula: \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a=6\), \(b=-5\), and \(c=-6\). After finding the values of \(t\), substitute back \(t = \frac{2x}{x - 3}\) and solve for \(x\).

Key Concepts

Rational Expressions

Rational expressions are fractions where the numerator and denominator are polynomials. Understanding how to simplify, manipulate, and evaluate these expressions is essential, especially when variables appear in denominators, as restrictions on the domain must be considered to avoid division by zero.

Setting Up and Solving Equations

To find values of x that satisfy a condition, translate the problem into an equation. Here, expressing 'y1 exceeds y2 by 6' as y1 = y2 + 6 allows you to set up an equation involving rational expressions, which you then solve by clearing denominators and simplifying.

Quadratic Equations

After simplifying the equation, you often get a quadratic equation in terms of x. Knowing how to solve quadratics using factoring, completing the square, or the quadratic formula is crucial to find all possible solutions that satisfy the original problem.