Problem 58 - Intermediate Algebra

Question

Solve equation. If a solution is extraneous, so indicate. \(\frac{4 t^{2}+36}{t^{2}-9}-\frac{4 t}{t+3}=\frac{-12}{t-3}\)

Step-by-Step Solution

Answer

The equation has no valid solution; \( t = -3 \) is extraneous.

Step 1: Identify the Domain Restrictions

The denominators in the given equation are \( (t^2 - 9) \), \( (t + 3) \), and \( (t - 3) \). These cause restrictions since they cannot be zero. Thus, \( t^2 - 9 = 0 \) implies \( t = \pm 3 \), and for the other denominators, \( t eq 3 \) and \( t eq -3 \). Therefore, the values \( t = 3 \) and \( t = -3 \) are excluded from the domain.

Step 2: Simplify the Equation

The term \( \frac{4t^2 + 36}{t^2 - 9} \) can be simplified since \( t^2 - 9 = (t-3)(t+3) \). So the whole equation becomes \( \frac{4t^2 + 36}{(t-3)(t+3)} - \frac{4t}{t+3} = \frac{-12}{t-3} \). We need the common denominator which is \( (t-3)(t+3) \).

Step 3: Rewrite All Terms with the Common Denominator

The term \( \frac{4t}{t+3} \) needs to be rewritten with denominator \( (t-3)(t+3) \). So it becomes \( \frac{4t(t-3)}{(t-3)(t+3)} \) = \( \frac{4t^2 - 12t}{(t-3)(t+3)} \). The right-hand side remains \( \frac{-12(t+3)}{(t-3)(t+3)} \).

Step 4: Combine and Simplify Terms

The equation \( \frac{4t^2 + 36 - (4t^2 - 12t)}{(t-3)(t+3)} = \frac{-12(t+3)}{(t-3)(t+3)} \) simplifies to \( \frac{4t^2 + 36 - 4t^2 + 12t}{(t-3)(t+3)} = \frac{-12t - 36}{(t-3)(t+3)} \). Cancel out common factors to simplify both sides.

Step 5: Solve the Resulting Equation

Now we have \( 12t + 36 = -12t - 36 \), so adding \( 12t \) to each side gives \( 24t + 36 = -36 \). Subtract 36 from both sides to get \( 24t = -72 \). Divide by 24 to solve for \( t \), resulting in \( t = -3 \).

Step 6: Check to See if Solution Is Extraneous

Recall from Step 1 that \( t = -3 \) is excluded from the domain. Hence, \( t = -3 \) is an extraneous solution, meaning that it does not satisfy the original equation due to division by zero.

Key Concepts

Rational Expressions

A rational expression is essentially a fraction that has polynomials in its numerator and denominator. In simple terms, it's like any fraction you already know, but instead of just numbers, we have algebraic expressions. Understanding rational expressions is vital when solving equations involving them because they behave similarly to regular fractions.
To perform operations on rational expressions, such as addition or subtraction, we look for common denominators, just like with regular fractions. In our original exercise, we dealt with the rational expressions \( \frac{4t^2 + 36}{t^2 - 9}\), \(\frac{4t}{t+3}\), and \(\frac{-12}{t-3}\). Each term represents a rational expression with its unique numerator and denominator.
Understanding these expressions allows us to combine and simplify them to solve equations systematically. Never forget that these expressions are sensitive to changes, especially when the denominator equals zero, which leads us to consider certain restrictions.

Domain Restrictions

When working with rational expressions, domain restrictions are critical. These are specific values that the variable cannot take, as they make the denominator zero. A zero denominator makes the expression undefined because you can't divide by zero. This is why identifying these restrictions is a crucial early step in solving equations with rational expressions.
In the given problem, we have the denominators \(t^2 - 9\), \(t + 3\), and \(t - 3\). By setting these equal to zero, we find that:

• \(t^2 - 9 = 0\) results in \(t = \pm 3\)
• \(t + 3 = 0\) results in \(t = -3\)
• \(t - 3 = 0\) results in \(t = 3\)

These values must be excluded from the domain, meaning the solutions can't include \(t = 3\) or \(t = -3\), as they would lead to division by zero and make the rational expression undefined. Becoming comfortable identifying and applying these restrictions is key in approaching problems with rational equations effectively.

Equation Simplification

Simplifying equations with rational expressions requires getting all parts of the equation to a common denominator. Once that's done, the numerators can be combined just like normal algebraic expressions. This process clears up the equation, allowing us to isolate the variable and find its potential solutions.
For example, in our exercise, we initially have the equation \(\frac{4t^2 + 36}{t^2 - 9} - \frac{4t}{t+3} = \frac{-12}{t-3}\). The focus is on rewriting each term so they all share the denominator \((t-3)(t+3)\). Simplifying the first term, \(t^2 - 9\), factored as \((t-3)(t+3)\), becomes the common denominator. Next, the other rational expressions are rewritten to align under this denominator.

• \(\frac{4t}{t+3}\) becomes \(\frac{4t(t-3)}{(t-3)(t+3)}\)
• \(\frac{-12}{t-3}\) becomes \(\frac{-12(t+3)}{(t-3)(t+3)}\)

This common foundation allows us to compare and simplify effectively, leading to further steps in the solution.

Identifying Denominator Restrictions

Identifying denominator restrictions means spotting which values will make the denominator zero, rendering the expression undefined. This is non-negotiable in rational expressions since any logical math operations hinge on division being valid.
In the equation from the step by step solution, we isolate denominators \((t-3)\), \((t+3)\), and \(t^2 - 9\). Each contributes potential restrictions:

• \(t^2 - 9 = 0\) gives \(t = 3\) and \(t = -3\). So, both values \(t = -3\) and \(t = 3\) must be excluded to avoid making \(t^2 - 9\), and thus \( (t-3)(t+3) \), zero.
• The single variable denominators follow suit providing further restrictions on the domain

By avoiding these values, we prevent extraneous solutions, which are nominal answers that don't meet the criteria of the original equation due to these restrictions. Going through this process early on keeps us from chasing down dead-end solutions.