Problem 52 - Algebra for College Students

Question

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

Step-by-Step Solution

Answer

\[\frac{9 \pm 3\sqrt{33}}{4}\]

Step 1 - Identify a Common Denominator

Identify the common denominator for all fractions in the equation. The denominators are \(2w + 6\), \(w - 1\), and \(w + 3\). The common denominator will be \((2w + 6)(w - 1)(w + 3)\).

Step 2 - Multiply Through by the Common Denominator

Multiply each term in the equation by the common denominator \((2w + 6)(w - 1)(w + 3)\) to eliminate the fractions:\[(2w + 6)(w - 1)(w + 3) \cdot \frac{5}{2 w+6} - (2w + 6)(w - 1)(w + 3) \cdot \frac{1}{w-1} = (2w + 6)(w - 1)(w + 3) \cdot \frac{1}{w+3}\]

Step 3 - Simplify Each Term

When you multiply each fraction by the common denominator, the denominators will cancel out, leaving:\[5(w - 1)(w + 3) - 1(2w + 6)(w + 3) = (2w + 6)(w - 1)\]

Step 4 - Expand and Simplify

Expand each expression:\[5(w^2 + 2w - 3) - (2w + 6)(w + 3) = (2w + 6)(w - 1)\]. First expand nationally:\[5(w^2 + 2w - 3) - (2w^2 + 12w + 18) = 2w^2 - 2w - 6w + 6\], now it simplifies to:\[5w^2 + 10w - 15 - 2w^2 - 12w - 18 = 2w^2 - 2w - 6 \rightarrow 3w - 33 = 2w^2 - 2w - 6 \times 4 {:.w^2 - 5w - 27 = 0} \]

Step 5 - Solve the Quadratic Equation

Rewrite the equation obtained in the standard quadratic form:\(2w^2 - 9w - 27 = 0\). Now use the quadratic formula \(w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 2\), \(b = -9\), and \(c = -27\).Plug in the values:\[w = \frac{9 \pm \sqrt{81 + 216}}{4}\]. Simplified it is:\[\frac{9 \pm \sqrt{297}}{4}\] or \[\frac{9 \pm 3\sqrt{33}}{4}\].

Key Concepts

Common Denominator

When working with rational equations, one of the first steps is finding a common denominator. This simplifies the equation and makes it easier to solve. In the given problem, the denominators are \(2w + 6\), \(w - 1\), and \(w + 3\). The common denominator is found by multiplying these expressions together: \((2w + 6)(w - 1)(w + 3)\). This term includes all the factors necessary to cancel each denominator in the fractions.

Here's why it's important:

• It turns a complex fraction problem into a simpler polynomial equation.
• Allows for the elimination of denominators, making the equation easier to solve.

Remember, correctly identifying the common denominator is crucial for simplifying and eventually solving the equation.

Quadratic Equation

After eliminating the denominators in the rational equation, you generally end up with a quadratic equation. A quadratic equation is written in the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants.

In this case, the equation simplifies to \(2w^2 - 9w - 27 = 0\). To solve it, you can use the quadratic formula:
\(w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).

Plugging in the values \(a = 2\), \(b = -9\), and \(c = -27\), we get:
\(w = \frac{9 \pm \sqrt{81 + 216}}{4}\), which simplifies to:
\(w = \frac{9 \pm \sqrt{297}}{4}\) or
\(w = \frac{9 \pm 3\sqrt{33}}{4}\).

Quadratic equations often have two solutions, which are derived from the \pm symbol in the formula. Ensure you solve for both potential values.

Simplifying Expressions

Simplifying expressions is a fundamental part of solving any algebraic equation. In the context of rational equations, this means breaking down and combining like terms after using the common denominator.

For example, after multiplying each term by the common denominator and simplifying, the equation transformed into:
\[5(w - 1)(w + 3) - 1(2w + 6)(w + 3) = (2w + 6)(w - 1)\].
Expanding these terms methodically is crucial:
\[5w^2 + 10w - 15 - 2w^2 - 12w - 18 = 2w^2 - 2w - 6\].
Then, combining like terms carefully results in:
\[3w^2 - 5w - 33 = 0\].

Thesesteps make the quadratic equation easier to manage. Pay careful attention to arithmetic and proper multiplication to avoid mistakes. Simplifying correctly ensures you can solve the equation efficiently.