Problem 10 - Algebra for College Students

Question

For Problems \(1-44\), solve each equation. $$ \frac{5}{7 x}-\frac{5}{6}=\frac{1}{6 x} $$

Step-by-Step Solution

Answer

The solution is \( x = \frac{23}{35} \).

Step 1: Identify a Common Denominator

First, identify the least common denominator (LCD) for the fractions in the equation. The terms involve denominators that include constants and a variable 'x': 7x, 6, and 6x. The LCD here would be 42x.

Step 2: Multiply Each Term by the LCD

Multiply each term of the equation \[ \frac{5}{7x} - \frac{5}{6} = \frac{1}{6x} \] by the common denominator, 42x, to eliminate the fractions. This yields: \[ 42x \times \frac{5}{7x} - 42x \times \frac{5}{6} = 42x \times \frac{1}{6x}. \]

Step 3: Simplify Each Term

Simplify the equation: - For the first term: \( 42x \times \frac{5}{7x} = 30 \). - For the second term: \( 42x \times \frac{5}{6} = 35x \). - For the third term: \( 42x \times \frac{1}{6x} = 7 \). Thus, the equation simplifies to: \[ 30 - 35x = 7. \]

Step 4: Solve for x

Rearrange the equation to solve for \( x \): - Start by isolating the \( 35x \) term: \[ 30 - 7 = 35x \] \[ 23 = 35x. \] - Finally, solve for \( x \) by dividing both sides by 35: \[ x = \frac{23}{35}. \]

Key Concepts

Least Common Denominator

The least common denominator (LCD) is a crucial concept when solving equations involving algebraic fractions. To understand it, imagine you have various fractions, each with different denominators—in this case, 7x, 6, and 6x. The LCD is the smallest number that all these denominators can divide into without leaving a remainder.

For the equation \( \frac{5}{7x}-\frac{5}{6}=\frac{1}{6x} \), finding the LCD helps us manage the fractions more easily by consolidating them into a single form. Here, the LCD is 42x because:

• 7x (composed of 7 and x, with 7 going into 42 six times, hence it fits perfectly).
• 6 fits into 42 seven times.
• 6x, already having x, fits directly into 42x.

This step ensures that when we multiply through by the LCD, the fractions disappear, simplifying the equation to integers or simpler expressions.

Equation Solving

Equation solving is the process of finding the value of the unknown variable that makes the equation true. In this instance, we are simplifying the equation using the LCD to move forward with solving.

After identifying the LCD as 42x, each term in the equation \( \frac{5}{7x} - \frac{5}{6} = \frac{1}{6x} \) is multiplied by this common denominator. Why do this? Essentially, multiplying through by 42x, like magic, clears out the fractions, simplifying the terms so we can see clearly what we are dealing with. The transformed terms turn into:

• \( 42x \times \frac{5}{7x} \) simplifies to 30 because the x's and some constants cancel out nicely.
• \( 42x \times \frac{5}{6} \) becomes 35x once calculated.
• \( 42x \times \frac{1}{6x} \) becomes 7 upon cancellation.

Now, the equation looks cleaner: \( 30 - 35x = 7 \), making solving for x straightforward.

Simplification

Simplification in algebraic expressions involves reducing them to their most readable and straightforward form. This skill allows us to make sense of complex equations by carefully transforming them into simpler ones.

After multiplying each term by the LCD (42x), we performed simplification that led us to having no fractions in the equation. For the given example:

• First, calculate the results of each multiplication to convert the equation \( 42x \times \frac{5}{7x} - 42x \times \frac{5}{6} = 42x \times \frac{1}{6x} \) into the simplified form: 30 - 35x = 7.
• Combine and eliminate terms by balancing both sides of the equation.

Each of the original terms has been simplified by multiplication and reduction, transforming our view of the equation, primarily dealing now with simple arithmetic rather than fractions.

Variable Isolation

Variable isolation is the technique of rearranging an equation to have the variable of interest by itself on one side of the equation. It's the last step to solve for the variable explicitly.

Here, the simplified equation \( 30 - 35x = 7 \) pivots around making the variable x all by itself. Follow these steps for isolation:

• Start by moving numbers to the opposite side of the equation: Subtract 7 from 30 to find the remainder to balance the equation, leading to \( 23 = 35x \).
• Then, solve for x by dividing both sides of the equation by 35 to clear any coefficient in front of x: \( x = \frac{23}{35} \).

This results in x being explicitly solved for, showcasing the power of effectively rearranging and simplifying to isolate variables.