Problem 88 - Elementary Algebra

Question

For exercises 87-90, the completed problem has one mistake. (a) Describe the mistake in words, or copy down the whole problem and highlight or circle the mistake. (b) Do the problem correctly. Problem: Solve: \(\frac{2}{x-2}+\frac{8}{x}=\frac{x}{x-2}\) Incorrect Answer: \(\frac{2}{x-2}+\frac{8}{x}=\frac{x}{x-2}\) \(x(x-2)\left(\frac{2}{x-2}+\frac{8}{x}\right)=x(x-2)\left(\frac{x}{x-2}\right)\) \(x(x-2)\left(\frac{2}{x-2}\right)+x(x-2)\left(\frac{8}{x}\right)=x^{2}\) \(2 x+(x-2) 8=x^{2}\) \(2 x+8 x-16=x^{2}\) \(10 x-16=x^{2}\) \(0=x^{2}-10 x+16\) \(0=(x-8)(x-2)\) \(x-8=0 \quad\) or \(\quad x-2=0\) \(x=8 \quad\) or \(\quad x=2\)

Step-by-Step Solution

Answer

The mistake was improper distribution of terms. Correct solutions are x = 8, x = 2 is extraneous.

Step 1: Identify the mistake

The mistake is in handling the denominators while equating both sides. When clearing the denominators, there was an improper distribution of terms on the left side.

Step 2: Clear the denominators correctly

Multiply both sides of the original equation by the common denominator, which is \(x(x-2)\):\[x(x-2)\left(\frac{2}{x-2} + \frac{8}{x}\right) = x(x-2)\left(\frac{x}{x-2}\right)\]

Step 3: Distribute the terms correctly

The left side becomes: \[x\cdot2 + (x-2)\cdot8 = x^2\] which simplifies to:\[2x + 8x - 16 = x^2\]

Step 4: Simplify the equation

Combine like terms and set the equation to zero: \[10x - 16 = x^2\] \[0 = x^2 - 10x + 16\]

Step 5: Factor the quadratic equation

Factor the quadratic equation to find the solutions: \[0 = (x-8)(x-2)\]

Step 6: Solve for x

Set each factor equal to zero to find the solutions: \[x - 8 = 0 \quad\Rightarrow\quad x = 8\] \[x - 2 = 0 \quad\Rightarrow\quad x = 2\]

Step 7: Check for extraneous solutions

Verify each solution in the original equation to check for extraneous solutions. Substituting x = 2 in the original equation makes a denominator zero, suggesting it's an extraneous solution.

Key Concepts

rational expressions

A rational expression is a fraction where both the numerator and the denominator are polynomials. For example, in the equation \(\frac{2}{x-2}+\frac{8}{x}=\frac{x}{x-2}\), each fraction is a rational expression. To solve rational expressions, you often need to find a common denominator or manipulate the equation to eliminate fractions. This process helps to simplify the expressions and find the variable's values.

common denominators

Finding a common denominator is essential when adding or subtracting rational expressions. It allows you to combine the fractions into a single rational expression. In the equation \(\frac{2}{x-2}+\frac{8}{x}=\frac{x}{x-2}\), the common denominator would be \x(x-2)\. Multiplying each term by this common denominator allows you to clear the fractions:
\[x(x-2) \left( \frac{2}{x-2}+\frac{8}{x} \right)=x(x-2)\left( \frac{x}{x-2}\right)\]
After distributing, the equation simplifies to: \2x + 8x - 16 = x^2\. The elimination of fractions makes solving the equation easier.

quadratic equations

When you solve rational expressions, sometimes the equation simplifies into a quadratic equation. A quadratic equation is any equation that can be written in the form \[ax^2 + bx + c = 0\]. In the example provided, the rational equation turned into the quadratic equation \[x^2 - 10x + 16 = 0\]. Solving a quadratic equation typically involves factoring, using the quadratic formula, or completing the square.
For this exercise, we can factor the quadratic equation:
\[0 = (x-8)(x-2)\]
Setting each factor to zero gives the solutions:
\[x-8=0 \Rightarrow x=8\]
\[x-2=0 \Rightarrow x=2\]

extraneous solutions

An extraneous solution is a result that emerges from solving an equation but does not satisfy the original equation. When dealing with rational expressions, always check your solutions in the original equation to ensure they are not extraneous. For instance, in this problem, when you substitute \x = 2\ back into the original equation \(\frac{2}{x-2}+\frac{8}{x}=\frac{x}{x-2}\), the denominator becomes zero. This means \x = 2\ is an extraneous solution. Thus, the valid solution is only \x=8\.