Problem 57 - Algebra and Trigonometry Real Mathematics, Real People

Question

Solving an Equation Involving Fractions Find all solutions of the equation. Check your solutions. $$6\left(\frac{s}{s+1}\right)^{2}+5\left(\frac{s}{s+1}\right)-6=0$$

Step-by-Step Solution

Answer

The solutions for the equation are \(s = 1\) and \(s = -0.5\).

Step 1: Simplify the equation

Instead of dealing with the fraction \(\frac{s}{s+1}\), let's substitute it with a simpler variable. Let \(t = \frac{s}{s+1}\). Substituting in the given equation we get: \(6t^{2}+5t-6=0\)

Step 2: Factor the quadratic equation

Now, let's factor this quadratic equation: \(6t^{2}+5t-6 = 0\) becomes \((2t - 1)(3t + 6) = 0\).

Step 3: Solve for t

Setting each factor to zero gives the solutions for \(t\): \(2t - 1 = 0\) becomes \(t = 0.5\), \(3t + 6 = 0\) becomes \(t = -2\).

Step 4: Substitute t back into s

Now, taking the value of \(t\) back to the substituted variable \(s\), we get: For \(t = 0.5\), \(\frac{s}{s+1} = 0.5\) solving this for \(s\), we get \(s = 1\). For \(t = -2\), \(\frac{s}{s+1} = -2\) solving this for \(s\) gives \(s = -0.5\).

Step 5: Verify the solutions

Finally, plug these solutions back into the original equation to verify if they are correct.

Key Concepts

Fractional Equations

Fractional equations involve variables located in the numerator and/or denominator of one or more fractions. Solving these types of equations can be challenging because you often need to manage the fractions separately.

Here is a strategy to tackle them:

• Identify a common denominator if dealing with multiple fractions. This helps simplify the equation.
• Clear the fractions by multiplying every term by the common denominator.
• If possible, substitute the fractional expression with a simpler variable to reduce complexity.

In the given problem, we substitute the fraction \(\frac{s}{s+1}\) with \(t\), leading to a simpler form of equation, which is key in making the problem more manageable.

Quadratic Equations

Quadratic equations are polynomial equations of the form \(ax^2 + bx + c = 0\). Solving quadratic equations can be straightforward once you understand the key methods used.

The equation we derived from our substitution was \(6t^2 + 5t - 6 = 0\), a classic quadratic equation. There are multiple ways to solve quadratics:

• Factoring: Express the quadratic as a product of its factors.
• Completing the Square: Rewriting the quadratic in a perfect square format.
• Quadratic Formula: Using \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) when factoring is difficult.

In this exercise, the quadratic equation was factored directly into \((2t - 1)(3t + 6) = 0\), leading to an easier resolution for the values of \(t\). Not all quadratics can be factored easily, so it's good to be familiar with all methods.

Variable Substitution

Variable substitution is a technique used to simplify equations, particularly in fractional and quadratic forms. It involves replacing a complex expression with a simpler variable, thus reducing the complexity of the original equation.

In the exercise, we substituted \(t = \frac{s}{s+1}\), turning the fractional equation into a quadratic equation. This simplification allows us to apply familiar methods for solving the new equation, without dealing directly with fractions.

• This step is crucial for equations with repeating complex terms.
• Ensure to revert the substitution after solving, to express the solution in terms of the original variables.
• Check all potential solutions to confirm validity in the original expression.

By substituting back, we translated the solutions for \(t\) into solutions for \(s\), leading to our final answers for the equation.

Solution Verification

Verification is a vital step in solving equations to ensure the solutions obtained are valid in the context of the original equation. Sometimes certain transformations can introduce extraneous solutions that are not true for the original problem.

Here’s how to verify your solutions:

• Substitute the solution back into the original equation, not the transformed one.
• Verify if both sides of the equation remain equal when the solution is plugged back.
• Ensure no division by zero or undefined expressions occur when validating the solutions.

In the exercise, after solving for \(s\), it’s crucial to substitute \(s = 1\) and \(s = -0.5\) back in the original equation. These checks confirmed the validity of both solutions, showing they satisfy the original fractional equation.