Problem 33 - Introductory Algebra for College Students

Question

Solve each rational equation. $$\frac{3}{x-1}+\frac{8}{x}=3$$

Step-by-Step Solution

Answer

The solutions to the equation are \(x = 4\) and \(x = \frac{2}{3}\).

Step 1: Combine the fractions

To combine the fractions, find a common denominator for \(x-1\) and \(x\), which is \(x(x-1)\). Rewrite the equation using the common denominator: \[\frac{3x}{x(x-1)}+\frac{8(x-1)}{x(x-1)} = \frac{3x(x-1)}{x(x-1)}.\] This simplifies to: \[3x + 8(x-1) = 3x^2 - 3x.\]

Step 2: Distribute and simplify

Expand equation to isolate terms: \[3x + 8x - 8 = 3x^2 - 3x.\]Then, rearrange to get all terms on one side: \[0 = 3x^2 - 14x + 8.\]

Step 3: Solve for x

The equation is now a quadratic equation and can be solved by various methods like factoring, completing the square, or using the quadratic formula. However, this equation can be easily factored: \[0 = 3x^2 - 12x - 2x + 8 => 0 = 3x(x - 4) - 2(x - 4) => 0 = (3x - 2)(x - 4).\]Setting each factor equal to zero gives the solutions: \[x = 4, \frac{2}{3}.\]

Key Concepts

Quadratic Equations

Understanding quadratic equations is crucial when solving rational equations resulting in a quadratic form. A quadratic equation typically takes the form \(ax^2 + bx + c = 0\). Here, \(a\), \(b\), and \(c\) are constants, with \(a\) not equal to zero. Quadratic equations are called so because the highest exponent of the variable \(x\) is 2. These types of equations can have zero, one, or two real solutions.

There are several methods to solve quadratic equations, each useful in different scenarios:

• Factoring
• Using the quadratic formula
• Completing the square

In this exercise, the rational equation was transformed into a quadratic equation to facilitate the solution process. This happens because after finding a common denominator and simplifying the equation, the terms naturally arrange into a standard quadratic form. Once in this form, it becomes easier to tackle the problem with the right method.

Factoring

Factoring is a method used to break down an equation into simpler expressions, making it easier to solve. When you factor a quadratic equation, you're looking to express it as a product of two binomials.

For instance, consider the equation \(3x^2 - 14x + 8 = 0\). We want to express this in the form \((px + q)(rx + s) = 0\). By doing so, each of the parts can be easily solved independently.

Factoring is often used because it can quickly yield solutions if the quadratic equation can be decomposed neatly into whole-numbered binomial factors. In our solution, the equation is factored as \((3x - 2)(x - 4) = 0\), leading us to solutions by setting each factor to zero \((3x - 2 = 0 \text{ or } x - 4 = 0)\). These directly give us the possible values of \(x\) as potential solutions.

Common Denominator

Finding a common denominator is an essential step when dealing with equations involving fractions. It allows us to combine terms more effectively. When fractions have unlike denominators, it becomes difficult to manipulate the equation without this step.

In the exercise, the fractions \(\frac{3}{x-1}\) and \(\frac{8}{x}\) required a common denominator to be combined. This involved using the least common multiple (LCM) of the denominators, \(x\) and \(x-1\). The LCM in this case is \(x(x-1)\).

Once both fractions are rewritten with this common denominator, the numerators can be combined to form a single, more manageable fraction. This simplifies the rest of the equation-solving process. It is an important skill to develop when handling any rational equations.

Quadratic Formula

The quadratic formula is a powerful tool that provides the solutions to any quadratic equation that might not be easily factored. It's given by the formula: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

Here, \(a\), \(b\), and \(c\) are coefficients from the quadratic equation \(ax^2 + bx + c = 0\). This formula is especially useful when the quadratic cannot be factored neatly or solved easily using other methods.

Substituting the values of \(a\), \(b\), and \(c\) into the formula gives the roots of the equation. It is systematic and guarantees a solution by offering a straightforward approach. In scenarios where factoring doesn't result in easy solutions, or when working with complex numbers, the quadratic formula is indispensable. For this particular exercise, while we factored the equation to find solutions, knowing the formula provides a broader toolkit for tackling varied quadratic equations.