Problem 7 - A Graphical Approach to Precalculus with Limits

Question

Check that proposed solutions \(\frac{3}{2}\) and \(\frac{5}{3}\) from Example 6 are solutions of \(15 x^{-2}-19 x^{-1}+6=0\).

Step-by-Step Solution

Answer

Both \(x = \frac{3}{2}\) and \(x = \frac{5}{3}\) satisfy the equation.

Step 1: Understand the problem

The task is to verify if the given values \(x = \frac{3}{2}\) and \(x = \frac{5}{3}\) are solutions to the equation \(15 x^{-2} - 19 x^{-1} + 6 = 0\). This means substituting each value into the equation and checking if it satisfies the equation.

Step 2: Substitute \(x = \frac{3}{2}\)

Substitute \(x = \frac{3}{2}\) into the equation. First, calculate \(x^{-1}\) and \(x^{-2}\): \[ x^{-1} = \frac{1}{x} = \frac{2}{3}, \quad x^{-2} = \left(\frac{2}{3}\right)^2 = \frac{4}{9} \] Then substitute these into the equation:\[ 15 \left(\frac{4}{9}\right) - 19 \left(\frac{2}{3}\right) + 6 = 0 \]Calculate each term:\[15 \times \frac{4}{9} = \frac{60}{9} = \frac{20}{3}, \quad 19 \times \frac{2}{3} = \frac{38}{3}\]This simplifies to:\[ \frac{20}{3} - \frac{38}{3} + 6 = 0 \]\[ \frac{-18}{3} + 6 = 0 \]\[ -6 + 6 = 0 \]Thus, \(x = \frac{3}{2}\) is a solution.

Step 3: Substitute \(x = \frac{5}{3}\)

Substitute \(x = \frac{5}{3}\) into the equation. First, calculate \(x^{-1}\) and \(x^{-2}\): \[ x^{-1} = \frac{1}{x} = \frac{3}{5}, \quad x^{-2} = \left(\frac{3}{5}\right)^2 = \frac{9}{25} \] Then substitute these into the equation:\[ 15 \left(\frac{9}{25}\right) - 19 \left(\frac{3}{5}\right) + 6 = 0 \]Calculate each term:\[15 \times \frac{9}{25} = \frac{135}{25} = \frac{27}{5}, \quad 19 \times \frac{3}{5} = \frac{57}{5}\]This simplifies to:\[ \frac{27}{5} - \frac{57}{5} + 6 = 0 \]\[ \frac{-30}{5} + 6 = 0 \]\[ -6 + 6 = 0 \]Thus, \(x = \frac{5}{3}\) is also a solution.

Key Concepts

Solving Equations

When we talk about solving equations in precalculus, the goal is to find the values that make the equation true. In this exercise, we were given a rational equation that contained variables with negative exponents. Our task was to check if these specific values were solutions. To do that, you substitute the proposed solution values back into the equation.

Here’s a simplified guide to solving such equations:

• Understand the Equation: Look at the given mathematical sentence that involves an equals sign, which shows two expressions are equal.
• Identify the Unknowns: Figure out what values need to be found to satisfy the equation.
• Substitution: Replace the variable with the provided value to see if the equation holds true.
• Simplification: Perform operations to simplify both sides of the equation.
• Verification: Check the result to confirm the equality holds.

Successfully solving an equation confirms that the solution satisfies the original equation, validating our substitution.

Verification of Solutions

Verification is a crucial step in problem-solving, especially in mathematics. It ensures the accuracy of your solutions and verifies that no step was incorrectly performed. In our exercise, we verified the potential solutions by substituting back into the equation.

Here are some steps to effectively verify a solution:

• Substitute the Solution: Place the proposed value back into the original equation. For example, using \( x = \frac{3}{2} \), compute \( x^{-1} \) and \( x^{-2} \), then substitute those into the equation.
• Simplify the Expression: Ensure to correctly perform all necessary arithmetic operations.
  Verify each term step by step to avoid calculation errors.
• Check the Result: If the resulting expression satisfies the original equation \( = 0\), the proposed value is verified as a solution.
• Consider All Possible Values: Always verify each potential solution given to ensure completeness. In this case, we checked both \( \frac{3}{2} \) and \( \frac{5}{3} \).

Verification not only confirms correctness but also builds confidence in handling complex equations across various contexts.

Rational Expressions

Rational expressions often appear in precalculus problems. These involve ratios or fractions wherein the numerator and the denominator can have variables. In our exercise, we dealt with rational expressions having negative exponents, such as \( x^{-1} \) and \( x^{-2} \).

Understanding rational expressions is essential:

• Definition: A rational expression is essentially a fraction, where both the top (numerator) and bottom (denominator) are polynomials. They can include variables raised to positive or negative powers.
• Negative Exponents: When working with negative exponents, remember \( x^{-n} = \frac{1}{x^n} \). This is crucial when substituting and simplifying as seen in calculating \( x^{-1} \) and \( x^{-2} \).
• Multiplying Rational Expressions: When substituting values into expressions, it's important to correctly multiply these fractions, simplifying both the multiplication and the resulting rational expression.
• Common Errors: Mistakes often come from forgetting to correctly apply the laws of exponents, or incorrectly simplifying the fractions.

Mastery of rational expressions and understanding their properties is vital for effectively solving and verifying solutions in precalculus.