Problem 34 - Introductory Algebra for College Students

Question

Solve each rational equation. $$\frac{2}{x-2}+\frac{4}{x}=2$$

Step-by-Step Solution

Answer

The solution to the rational equation \(\frac{2}{x-2}+\frac{4}{x}=2\) is \(x = 0\).

Step 1 - Find the Common Denominator

In the equation \(\frac{2}{x-2}+\frac{4}{x}=2\), the denominators are \(x-2\) and \(x\). The common denominator of these two denominations is found by multiplying the two of them. Therefore, the least common denominator is \(x(x-2)\).

Step 2 - Convert to Polynomial Equation

To clear the fractions, multiply every term of the equation by the least common denominator \(x(x-2)\). This yields: \[2x(x)+4x(x-2)=2x(x-2)x.\] Simplify the equation to obtain: \(2x^2+4x^2-8x=2x^2\).

Step 3 - Solve for x

Condense the equation to a form we can solve, that is \(4x^2 - 8x = 0\). Factor this equation, we can think of it as \(4x(x - 2) = 0\). Set each factor that consists of a variable to zero and solve for x. Doing this we find the solutions to the equation \(x = 0\) and \(x = 2\).

Step 4 - Check the Solution

You should check that these solutions for \(x\) don't make the denominator of any fraction in the original equation equal zero. If you substitute \(x = 0\) into the original equation, none of the denominators are equal to zero, so \(x = 0\) is a valid solution. But, if you substitute \(x = 2\) into the original equation, the denominator of the first fraction would be equal to zero, which is not allowable. Therefore, \(x = 2\) must be excluded.

Key Concepts

Common Denominator

When working with rational equations like \(\frac{2}{x-2} + \frac{4}{x} = 2\), we want to combine these fractions to solve for \(x\). However, before doing that, it's essential to find a common denominator—a value that both denominators can divide into without a remainder.

Imagine the problem like different people running on separate loops in a park. To bring them together at the same point, they need to follow a path that's a multiple of their individual loops. Similarly, we combine the denominators \(x-2\) and \(x\) by finding the least number that both can divide into—like a shared path for our runners. This number is called the least common denominator (LCD).

Calculating the LCD by multiplying the two denominators gives us \(x(x-2)\). It's the shared path that makes working with the equation simpler, just as it would be easier to manage a group of runners if they all followed one track.

Polynomial Equation

After finding the common denominator, the next step is creating a polynomial equation. By multiplying every term in the rational equation by the LCD, we essentially clear the fractions. This is akin to transforming a group of mixed fruits into a single type of fruit, such as apples, making them easier to count.

For our equation, by multiplying every term by \(x(x-2)\), the denominators cancel out, leaving us with a simpler polynomial equation to solve: \(4x^2 - 8x = 0\). Polynomial equations like this are far easier to work with. Think of them as a tray where all apples are now visible and ready to be arranged for easy counting—solving for \(x\) is just like counting how many apples are on the tray.

Factoring Polynomials

To solve for \(x\) in our polynomial equation \(4x^2 - 8x = 0\), we need to factor the polynomial. Factoring is similar to breaking down a complex recipe into its basic ingredients, making it clearer to understand what's in it.

In this equation, we can factor out \(4x\), simplifying the equation to \(4x(x - 2) = 0\).

Why Factoring Matters

Factoring makes it evident that there are potential values of \(x\) that could turn the entire expression into zero (like removing an ingredient from a recipe to see if the dish still works). These values are important because they're the possible solutions for our equation.

Solution Checking

Finding potential solutions for an equation isn't enough—verification is equally crucial. This is the solution checking phase, which involves substituting the possible solutions back into the original rational equation to confirm their validity. Think of it as tasting a dish to make sure the seasoning is just right.

For our equation, one of the proposed solutions was \(x = 2\), but when we substitute it back into the original rational equation, we hit a snag—the denominator in the first term becomes zero, which is like a recipe that requires water, but there's none. This discrepancy invalidates \(x = 2\) as a solution. Only after we ensure that none of the denominators turn into zero can we confirm the valid solutions—just like double-checking each ingredient to ensure the dish turns out perfect.