Problem 58 - Elementary Algebra

Question

For the following problems, solve the rational equations. $$ \frac{1}{x}=\frac{2}{x+4}-\frac{3}{2} $$

Step-by-Step Solution

Answer

Answer: The possible values of x that satisfy the given rational equation are: $$ x = \frac{14 + \sqrt{148}}{6} $$ and $$ x = \frac{14 - \sqrt{148}}{6} $$

Step 1: Find a common denominator

To find a common denominator for the fractions in the equation, we need to consider the denominators x, (x+4), and 2. In this case, the common denominator will be the least common multiple (LCM) of these denominators, which is 2x(x+4).

Step 2: Multiply both sides by the common denominator

Now, we'll multiply both sides of the equation by the common denominator 2x(x+4). This helps us to eliminate the fractions: $$ 2x(x+4) \cdot \frac{1}{x}=2x(x+4) \cdot \frac{2}{x+4}-2x(x+4) \cdot \frac{3}{2} $$

Step 3: Simplify the equation and solve for x

After multiplying both sides of the equation by the common denominator, we'll simplify and solve for x: $$ 2(x+4) = 2(2x) - 3x(x+4) $$ Distribute the numbers: $$ 2x + 8 = 4x - 3x^2 - 12x $$ Move all the terms to one side of the equation to get a quadratic polynomial: $$ 3x^2 - 14x + 8 = 0 $$ Now we need to solve the quadratic equation. We can use the quadratic formula: $$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$ For our equation, a = 3, b = -14, and c = 8: $$ x = \frac{14 \pm \sqrt{(-14)^2 - 4(3)(8)}}{2(3)} $$ Simplify: $$ x = \frac{14 \pm \sqrt{148}}{6} $$ Thus, there are two possible solutions for x: $$ x = \frac{14 + \sqrt{148}}{6} $$ and $$ x = \frac{14 - \sqrt{148}}{6} $$

Key Concepts

Common Denominator

When working with rational equations, identifying a common denominator is key. A common denominator allows us to combine the fractions in an equation. This makes it easier to solve.

The common denominator is found by identifying the least common multiple (LCM) of the denominators involved. In our given problem, the denominators were \(x\), \(x+4\), and \(2\).

**Steps to Find a Common Denominator:**

• List the denominators in the equation.
• Find the LCM of these denominators.

For the problem at hand, the LCM is \(2x(x+4)\). This unifies the denominators into a single expression, allowing you to multiply through the equation to eliminate fractions entirely.

Quadratic Formula

The quadratic formula is a powerful tool for solving quadratic equations of the form \(ax^2 + bx + c = 0\).

**Formula:** \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

This formula provides the solutions for \(x\) by substituting the coefficients \(a\), \(b\), and \(c\). In our solved equation, we had \(a = 3\), \(b = -14\), and \(c = 8\).

**Steps to Use the Quadratic Formula:**

• Substitute the values of \(a\), \(b\), and \(c\) into the formula.
• Solve for \(x\) by calculating the discriminant \(b^2 - 4ac\).
• Find the two possible values for \(x\) using the \(\pm\) symbol.

This method ensures that you can find all possible roots of any quadratic equation, no matter how complex.

Least Common Multiple

The Least Common Multiple (LCM) is crucial for solving equations that involve multiple fractions. It helps find a consistent scale to combine these fractions.

**Understanding LCM:**

• The LCM is the smallest number that is evenly divisible by all denominators.
• For example, in the problem, we calculated the LCM of \(x\), \(x+4\), and \(2\), resulting in \(2x(x+4)\).

Why use the LCM? It allows us to clear fractions by ensuring that each part of the equation is multiplied by an equivalent denominator, making further simplification possible.

Simplifying Equations

Simplifying equations involves making them easier to solve by performing operations that eliminate unnecessary complexity.

**Process of Simplifying:**

• Multiply through by the common denominator to remove fractions.
• Distribute terms to remove parentheses.
• Combine like terms to minimize the equation's complexity.

In our exercise, after multiplying and distributing, we moved all terms to one side to form a standard quadratic equation: \(3x^2 - 14x + 8 = 0\). Simplifying equations is a critical step toward solving them efficiently and accurately.