Problem 42 - Algebra A Combined Function

Question

Solve each equation. $$ \frac{2 t+3}{t-1}-\frac{2}{t+3}=\frac{5-6 t}{t^{2}+2 t-3} $$

Step-by-Step Solution

Answer

The solutions are \(t = -0.5\) and \(t = -6\).

Step 1: Identify the common denominator

In the given equation, we see three different denominators: \(t-1\), \(t+3\), and \(t^2+2t-3\). The expression \(t^2+2t-3\) factors into \((t-1)(t+3)\). Therefore, the common denominator for all terms is \((t-1)(t+3)\).

Step 2: Rewrite all terms with the common denominator

To eliminate the fractions, rewrite each term using the common denominator \((t-1)(t+3)\):\[ \frac{(2t+3)(t+3)}{(t-1)(t+3)} - \frac{2(t-1)}{(t-1)(t+3)} = \frac{5-6t}{(t-1)(t+3)} \]

Step 3: Simplify each term in the equation

Expand the numerators to simplify:1. Expand \((2t+3)(t+3)\): \[ 2t^2 + 6t + 3t + 9 = 2t^2 + 9t + 9 \]2. Expand \(2(t-1)\): \[ 2t - 2 \]Now rewrite the equation:\[ \frac{2t^2 + 9t + 9 - (2t - 2)}{(t-1)(t+3)} = \frac{5-6t}{(t-1)(t+3)} \]

Step 4: Simplify the left side of the equation

Combine the numerators on the left:\[ 2t^2 + 9t + 9 - 2t + 2 = 2t^2 + 7t + 11 \]Thus, the equation becomes:\[ \frac{2t^2 + 7t + 11}{(t-1)(t+3)} = \frac{5-6t}{(t-1)(t+3)} \]

Step 5: Set numerators equal and solve for t

Since the denominators are the same, set the numerators equal to each other:\[ 2t^2 + 7t + 11 = 5 - 6t \]Rearrange to form a quadratic equation:\[ 2t^2 + 13t + 6 = 0 \]

Step 6: Solve the quadratic equation

Use the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \(a = 2\), \(b = 13\), and \(c = 6\):\[ t = \frac{-13 \pm \sqrt{13^2 - 4 \cdot 2 \cdot 6}}{2 \cdot 2} \]Calculate the discriminant: \[ 169 - 48 = 121 \]So, \( t = \frac{-13 \pm 11}{4} \)Thus, \( t = -0.5 \) or \( t = -6 \).

Step 7: Verify each solution with the original equation

Substitute both \(t = -0.5\) and \(t = -6\) back into the original equation to verify that neither value makes any denominator zero:1. For \(t = -0.5\), none of \(t-1\), \(t+3\), or \(t^2+2t-3\) are zero.2. For \(t = -6\), none of \(t-1\), \(t+3\), or \(t^2+2t-3\) are zero.Both values are valid solutions.

Key Concepts

Common Denominator

When solving rational equations like the one given, the first step is to find a common denominator. Rational equations are equations that involve fractions, and a common denominator will help us combine these fractions into a single term.
Think about the denominators in the given equation: \(t-1\), \(t+3\), and \(t^2+2t-3\). The expression \(t^2+2t-3\) can be factored into \((t-1)(t+3)\). Therefore, the common denominator for all terms in the equation is \((t-1)(t+3)\).

• This step is crucial because it allows us to rewrite the original equation so that all terms have the same denominator.
• Having the same denominator simplifies the process of combining the terms because we can focus on the numerators, which is the next step in solving the equation.

By rewriting each term with this common denominator, we can then eliminate the fractions and solve the equation more easily.

Quadratic Formula

Once we have simplified the rational equation using a common denominator and manipulated the terms, we often end up with a quadratic equation. In our problem, after setting the numerators equal, we obtained the equation: \(2t^2 + 13t + 6 = 0\).
At this point, we can apply the quadratic formula, which is a staple method for solving quadratic equations. The quadratic formula is given by:\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

• Here, \(a\), \(b\), and \(c\) are the coefficients from the quadratic equation \(ax^2 + bx + c = 0\).
• The formula provides the solutions for \(t\) by calculating the expressions inside the square root, known as the discriminant, and considering both the plus and minus roots.

In our exercise, substituting the values for \(a = 2\), \(b = 13\), and \(c = 6\) into the formula gives us the roots \( t = -0.5 \) and \( t = -6 \), showing us where the quadratic equation intersects the x-axis.

Factoring Polynomials

Understanding how to factor polynomials is key in solving many algebraic equations, including rational equations. In our exercise, the polynomial \(t^2+2t-3\) was factorable, helping us find the common denominator.
Factoring involves rewriting a polynomial as a product of two or more simpler polynomials. The expression \(t^2 + 2t - 3\) was factored into \((t-1)(t+3)\). This factorization works because:

• The product of the two expressions \((t-1)\) and \((t+3)\) equals the original quadratic polynomial.
• Checking by distribution, \((t-1)(t+3)\) results in the original polynomial.

This technique not only aids in finding common denominators but also becomes essential when solving quadratic equations by the factoring method. If a quadratic can be factored, it often eliminates the need for the quadratic formula, leading to quicker solutions.

Verifying Solutions

After finding potential solutions for a rational equation, it's crucial to verify them. Verifying ensures that solutions don't create undefined expressions in the original equation.
In rational equations, certain values for the variable can make the denominator zero, which is undefined. Thus, a solution must be checked against this criterion.
For example, after finding \( t = -0.5 \) and \( t = -6 \) in our exercise, we substitute these back into the denominators of the original rational equation:

• Neither \(t =-0.5\) nor \(t = -6\) result in any denominators equalling zero in \(t-1\), \(t+3\), or \(t^2+2t-3\).
• This confirms that both values are valid solutions for the equation.

Always check your solutions with the original equation to ensure they are legitimate, especially when dealing with rational expressions.