Problem 80 - Algebra for College Students

Question

Solve each equation. Identify each equation as a conditional equation, an inconsistent equation, or an identity. State the solution sets to the identities using interval notation. $$\frac{1}{x}+\frac{1}{x^{2}}=\frac{6}{x^{3}}$$

Step-by-Step Solution

Answer

x = -3, 2 (Conditional equation)

Step 1 - Eliminate fractions

Multiply every term by the least common denominator (LCD), which is \(x^3\): \[ x^3 \times \frac{1}{x} + x^3 \times \frac{1}{x^2} = x^3 \times \frac{6}{x^3} \] This simplifies to: \[ x^2 + x = 6 \]

Step 2 - Form a quadratic equation

Rearrange the equation to set it to zero: \[ x^2 + x - 6 = 0 \]

Step 3 - Factor the quadratic equation

Factor the quadratic equation: \[ (x + 3)(x - 2) = 0 \]

Step 4 - Solve for the variable

Set each factor equal to zero and solve: \[ x + 3 = 0 \rightarrow x = -3 \] \[ x - 2 = 0 \rightarrow x = 2 \]

Step 5 - Identify the type of equation

Since this equation has specific solutions (x = -3, x = 2), it is classified as a conditional equation.

Step 6 - State the solution set

The solution set in interval notation is: \[ \{-3, 2\} \]

Key Concepts

solving rational equations

Solving rational equations may seem tricky, but with a few steps, you can solve them easily. A rational equation contains at least one fraction whose numerator and/or denominator involves a variable. To solve these, the goal is to eliminate the fractions. This is usually done by finding the least common denominator (LCD). The LCD is the smallest expression that every denominator in the equation can divide into.

For instance, in the given exercise \(\frac{1}{x}+\frac{1}{x^{2}}=\frac{6}{x^{3}}\), the LCD is \(x^3\). Multiplying every term by the LCD helps to get rid of the fractions. After that, you'll be left with a polynomial equation, which tends to be simpler to solve.

Here, multiplying each term by \(x^3\) yields \(x^2 + x = 6\). Now you can see how eliminating fractions makes the equation much more straightforward.

factoring quadratic equations

Once you've transformed the rational equation into a polynomial, often you'll see a quadratic equation. Quadratic equations are usually in the form \(ax^2 + bx + c = 0\). Factoring is one of the most common methods to solve quadratic equations.

The goal of factoring is to express the quadratic equation as a product of two binomials. For example, after rearranging \(x^2 + x - 6 = 0\), we can factor it to \((x + 3)(x - 2) = 0\). This factorizing tells us that if \((x + 3)\) or \((x - 2)\) equals zero, the entire expression equals zero.

Setting each factor equal to zero, you'll solve for the variable: \(x+3=0\) giving us \(x = -3\) and \(x - 2=0\) giving \(x = 2\). These are your solutions.

Remember, factoring helps in breaking down complex quadratic expressions into simpler, solvable parts.

solution sets

Finally, understanding solution sets is essential. A solution set is a set of all values that satisfy a given equation. Once you solve the equation, you should determine whether the equation is conditional, inconsistent, or an identity.

A conditional equation has specific solutions. For our example, since the solutions are \(x = -3\) and \(x = 2\), this is a conditional equation. An inconsistent equation has no solution, while an identity is always true and has infinite solutions. It's crucial to identify which type of equation it is.

In our exercise, the solution set is expressed in interval notation as \(\text{\textlbrace}-3, 2\text{\textrbrace}\). This means only these two values satisfy the equation.

Understanding and correctly expressing solution sets will help ensure you accurately convey the results of your solution.