Problem 35 - Algebra A Combined Function

Question

Solve each equation. $$ \text { 5. } \frac{t}{t-4}=\frac{t+4}{6} $$

Step-by-Step Solution

Answer

The solutions are \(t = 8\) and \(t = -2\).

Step 1: Understand the Equation

The given equation is \(\frac{t}{t-4} = \frac{t+4}{6}\). We have two rational expressions that we want to make equal.

Step 2: Cross-Multiply

To eliminate the fractions, we cross-multiply the terms. This gives us: \(t \cdot 6 = (t+4)(t-4)\).

Step 3: Simplify Each Side

On the left side of the equation, simplifying gives us: \(6t\). On the right side, we apply the distributive property: \((t+4)(t-4) = t^2 - 16\). Now our equation is \(6t = t^2 - 16\).

Step 4: Rearrange to Standard Form

Rearrange the equation to standard quadratic form: \(t^2 - 6t - 16 = 0\).

Step 5: Solve the Quadratic Equation

Now we solve the quadratic equation \(t^2 - 6t - 16 = 0\) using the quadratic formula: \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -6\), and \(c = -16\).

Step 6: Calculate the Discriminant

Calculate the discriminant: \(b^2 - 4ac = (-6)^2 - 4(1)(-16) = 36 + 64 = 100\).

Step 7: Apply the Quadratic Formula

Substitute the values into the quadratic formula: \(t = \frac{6 \pm \sqrt{100}}{2}\). This gives two possible solutions: \(t = \frac{6 + 10}{2}\) and \(t = \frac{6 - 10}{2}\).

Step 8: Simplify the Solutions

Simplify the solutions: \(t = \frac{16}{2} = 8\) and \(t = \frac{-4}{2} = -2\).

Step 9: Check for Extraneous Solutions

Check the original equation to see if any solution makes the denominator zero. If \(t = 4\), the denominator becomes zero and is not allowed. However, neither 8 nor -2 makes the denominator zero in the original equation.

Key Concepts

Cross-Multiplication

When faced with an equation involving fractions, like the rational equation \(\frac{t}{t-4} = \frac{t+4}{6}\), cross-multiplication is a reliable technique to eliminate the fractions and make the equation easier to solve. Cross-multiplying means multiplying the numerator of each fraction by the denominator of the opposite fraction.

• Left Side: multiply \(t\) by 6, giving \(6t\).
• Right Side: multiply \(t+4\) by \(t-4\), which expands to \(t^2 - 16\) using the distributive property.

Once each side is multiplied out, you have a polynomial equation without fractions. Cross-multiplication simplifies solving the rational equation by creating the expression \(6t = t^2 - 16\). This is much easier to handle and gives a clear way forward to further simplify and solve the equation.

Quadratic Formula

After cross-multiplying and simplifying, the equation \(6t = t^2 - 16\) transforms into a standard quadratic form by rearranging terms: \(t^2 - 6t - 16 = 0\). This can be solved using the quadratic formula, a powerful tool for finding solutions to any quadratic equation in the form \(ax^2 + bx + c = 0\). The quadratic formula is given by:\[t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]In this problem, \(a = 1\), \(b = -6\), and \(c = -16\). Plug these values into the formula:

• Calculate the discriminant: \(b^2 - 4ac = 36 + 64 = 100\).
• Since the discriminant is positive, the quadratic equation has two distinct real solutions.

Substitute the values into the quadratic formula to get:

• \(t = \frac{6 \pm \sqrt{100}}{2}\).
• Simplify the solutions: \(t = 8\) and \(t = -2\).

Using the quadratic formula helps to efficiently find solutions for quadratic equations, especially when factoring is difficult or impossible.

Checking Solutions

Once solutions are derived from the quadratic equation, it's crucial to verify them within the context of the original equation. Rational equations can have extraneous solutions, which are values that don't satisfy the original equation or make the denominator zero. For our solutions \(t = 8\) and \(t = -2\), we need to ensure they don't result in a zero denominator in the original fractions.

• Check \(t = 8\): Neither the numerator nor the denominator becomes zero.
• Check \(t = -2\): Again, neither produces zero in the denominator.

Always remember that solutions making any part of the rational expression undefined (like zero in the denominator) must be discarded. However, in this case, both solutions \(t = 8\) and \(t = -2\) are valid, as they do not violate the constraints of the original equation. Checking solutions ensures correctness and avoids including any extraneous or invalid answers.