Problem 20 - College Algebra

Question

Exercises \(17-30\) contain equations with constants in denominators. Solve each equation. $$ \frac{x}{5}-\frac{1}{2}=\frac{x}{6} $$

Step-by-Step Solution

Answer

The solution of the given equation is \(x=15\).

Step 1: Eliminate the Denominators

First, eliminate the denominators in the equation by multiplying every term by 30. This will give the equation: \( 30 * \frac{x}{5} - 30 * \frac{1}{2} = 30 * \frac{x}{6} \). Simplify this equation to obtain: \( 6x - 15 = 5x \)

Step 2: Rearrange the Equation

Next, rearrange this equation to isolate x. Subtract 5x from both sides to get: \(6x - 5x = 15\). This simplifies to \(x=15\).

Key Concepts

Algebraic Fractions

Algebraic fractions are similar to regular fractions, but they contain algebraic expressions in the numerator and/or the denominator. Just like their numerical counterparts, the goal when working with algebraic fractions is to perform operations such as addition, subtraction, multiplication, division or solving equations. For example, in the given exercise, we have the algebraic equation \(\frac{x}{5}-\frac{1}{2}=\frac{x}{6}\). These fractions may appear daunting at first due to the presence of variables, but the process to solve them usually involves finding a common denominator, eliminating the fractions, and then solving the resulting linear equation..

To work effectively with algebraic fractions, one must understand operations with regular fractions, how to find a least common denominator (LCD), and how to simplify expressions. It's important to remember not to cross multiply immediately if there are more than two fractions or if they are part of a more complex equation. Instead, we first find a common factor that can eliminate all denominators when multiplied with each term.

Common Denominator Method

The common denominator method is a foundational technique in algebra that is used to solve equations involving algebraic fractions. The key idea is to find a least common denominator (LCD) that can be used to eliminate the fractions. In our exercise, each fraction has a different denominator. To find an LCD we look for the least common multiple (LCM) of the denominators, which in this case are 5, 2, and 6. The LCM of these numbers is 30, so multiplying each term by 30 will clear the fractions and give us a simpler equation to solve.

When using this method, you multiply every term on both sides of the equation by the LCD. This step needs to be done carefully to ensure each term is treated equally to avoid mistakes. It's critical that the entire term, including its sign, is multiplied by the LCD. This approach transforms an equation with algebraic fractions into a linear equation which is often easier to solve.

Linear Equations

Linear equations are equations of the first degree, meaning they contain variables that are raised only to the power of one. When the variables are isolated on one side, and the constants on the other, linear equations have the general form \(ax + b = 0\), where \(a\) and \(b\) are constants. Solving linear equations involves isolating the variable to find its value.

In our exercise, after eliminating the denominators and simplifying, we obtained a linear equation: \(6x - 15 = 5x\). To solve this, we follow a set of steps, such as combining like terms and isolating the variable. By subtracting \(5x\) from both sides, we simplify down to \(x = 15\), which is our solution. Remember, checking your answer by substituting back into the original equation is essential to ensure the solution is correct. To fully understand linear equations, practice solving them with different coefficients and constants, and soon the process will become more intuitive.