Problem 29 - Elementary and Intermediate Algebra

Question

Solve each equation and check the result. If an equation has no solution, so indicate. $$ \frac{4}{5}-\frac{1}{10 x}=\frac{7}{15} $$

Step-by-Step Solution

Answer

The solution is \(x = \frac{3}{10}\).

Step 1: Eliminate Fractions

To eliminate the fractions, multiply every term in the equation by the least common multiple (LCM) of the denominators. Here, the denominators are 5, 10x, and 15. The LCM is 30x. Multiply the entire equation by 30x:\[ 30x \left( \frac{4}{5} \right) - 30x \left( \frac{1}{10x} \right) = 30x \left( \frac{7}{15} \right) \] This simplifies to:\[ 24x - 3 = 14x \]

Step 2: Rearrange the Equation

Rearrange the equation by moving the \'x\' terms to one side and constants to the other. Subtract \(14x\) from both sides:\[ 24x - 14x - 3 = 0 \]Simplified:\[ 10x - 3 = 0 \]

Step 3: Solve for x

Now solve for \(x\). Add 3 to both sides:\[ 10x = 3 \]Divide both sides by 10:\[ x = \frac{3}{10} \]

Step 4: Verify the Solution

Substitute \( x = \frac{3}{10} \) back into the original equation to verify the solution:Original equation:\[ \frac{4}{5} - \frac{1}{10 \times \frac{3}{10}} = \frac{7}{15} \]This simplifies to:\[ \frac{4}{5} - \frac{1}{3} = \frac{7}{15} \]Convert fractions to have a common denominator (15):\[ \frac{12}{15} - \frac{5}{15} = \frac{7}{15} \]This holds true:\[ \frac{7}{15} = \frac{7}{15} \] Thus, the solution is verified.

Key Concepts

Eliminating Fractions

When you encounter an equation with fractions, it often helps to eliminate them entirely. This makes the equation easier to solve. To do this, multiply every term by a number that will cancel out each denominator.
For example, in the equation \( \frac{4}{5} - \frac{1}{10x} = \frac{7}{15} \), the goal is to get rid of the fractions. Find the least common multiple (LCM) of all the denominators to decide the number to multiply by.
Multiplying through by this LCM will eliminate all fractions from the equation, and make it simpler to handle. This step is crucial because working without fractions is generally more straightforward, especially when dealing with linear equations.
By multiplying each term by the LCM, you turn the fractions into whole numbers, which then allows you to proceed with conventional algebraic methods like moving terms to one side. This is why eliminating fractions is a valuable technique in solving equations.

Least Common Multiple

The Least Common Multiple (LCM) is a critical concept when dealing with fractions, especially in equations. The LCM of a set of numbers is the smallest positive integer that is evenly divisible by all the numbers in the set.
For our equation, with denominators 5, 10x, and 15, we identify the LCM to help eliminate fractions.
To find the LCM, you might have to consider the numerical values and any variable components. Here, the LCM is 30x, because:

• 30 is the smallest number divisible by 5, 10, and 15.
• x is included because it is a part of one denominator.

By using the LCM to multiply the entire equation, you ensure all fractions are eliminated. Using the LCM is an efficient way to prepare the equation for an easier and quicker resolution.

Verifying Solutions

Once you find a solution to an equation, verification is an essential step. It involves substituting the solution back into the original equation to check if it results in a true statement.
For the equation \( \frac{4}{5} - \frac{1}{10x} = \frac{7}{15} \), let's verify the solution \( x = \frac{3}{10} \).
Insert \( x = \frac{3}{10} \) back into each term:

• Calculate \( \frac{1}{10 \times \frac{3}{10}} \).
• Simplify the new fractions, and compare both sides.

The left side needs to equal the right side of the initial equation. As calculated,\( \frac{7}{15} = \frac{7}{15} \), confirming that the solution \( x = \frac{3}{10} \) is correct.
Verification not only boosts confidence in your solution but also ensures there were no errors in the multiplication, division, or arithmetic while solving the equation.