Problem 40 - Elementary Algebra

Question

\(\frac{h}{4}+\frac{h}{5}-\frac{h}{6}=1\)

Step-by-Step Solution

Answer

The value of \(h\) is \(\frac{60}{17}\).

Step 1: Find a Common Denominator

First, identify the common denominator for the fractions in the equation. The denominators are 4, 5, and 6. The least common multiple of these numbers is 60. We will rewrite each fraction with 60 as the denominator.

Step 2: Rewrite Fractions

Convert each term to have the common denominator of 60. This means:- \(\frac{h}{4} = \frac{15h}{60}\)- \(\frac{h}{5} = \frac{12h}{60}\)- \(\frac{h}{6} = \frac{10h}{60}\)

Step 3: Combine Fractions

Combine the fractions:\[ \frac{15h}{60} + \frac{12h}{60} - \frac{10h}{60} = \frac{(15h + 12h - 10h)}{60} = \frac{17h}{60} \]

Step 4: Solve the Equation

Set the combined fraction equal to 1:\[ \frac{17h}{60} = 1 \]Multiply both sides by 60 to clear the fraction:\[ 17h = 60 \]

Step 5: Solve for h

Divide both sides by 17 to solve for \(h\):\[ h = \frac{60}{17} \]

Key Concepts

Common Denominator

Working with fractions often means dealing with different denominators, which can make calculations cumbersome. This is where the concept of a common denominator becomes essential. A **common denominator** is a shared denominator between various fractions that allows you to seamlessly perform operations like addition or subtraction.
To find a common denominator, you need the least common multiple (LCM) of the denominators involved. In the equation \( \frac{h}{4} + \frac{h}{5} - \frac{h}{6} = 1 \), the denominators are 4, 5, and 6. Here is how to find their LCM:

• List the multiples of each number.
• 4: 4, 8, 12, 16, 20, 24, ...
• 5: 5, 10, 15, 20, 25, ...
• 6: 6, 12, 18, 24, 30, ...

The smallest common multiple you can spot between these lists is 60. So, 60 becomes the common denominator. By converting all fractions to have this common denominator, calculations become much simpler and more straightforward.

Fraction Operations

Once you have a common denominator, you can easily perform operations with fractions, such as addition and subtraction. In our equation, after converting each term to have a denominator of 60, the fractions become:

• \( \frac{h}{4} = \frac{15h}{60} \)
• \( \frac{h}{5} = \frac{12h}{60} \)
• \( \frac{h}{6} = \frac{10h}{60} \)

By performing **fraction operations**, you simply bring the numerators together as if they were integers, while maintaining the common denominator:
\[ \frac{15h}{60} + \frac{12h}{60} - \frac{10h}{60} = \frac{17h}{60} \]
The process involves adding the numerators 15h, 12h, and subtracting 10h to end up with a single fraction. It's crucial to remember that only the numerators change during these operations, as the common denominator remains fixed. This keeps the overall equation balanced.

Solving Linear Equations

The goal of solving linear equations is to find an unknown variable, often represented by symbols like \( x \) or \( h \). In our exercise, after the fraction operations and rewriting the equation, we were left with \( \frac{17h}{60} = 1 \).
To solve for \( h \) in this linear equation, we need to eliminate the fraction by taking the following steps:

• Multiply both sides of the equation by the common denominator, here it's 60, resulting in:
\[ 17h = 60 \]
• This step ensures that \( h \) is isolated with a coefficient of 17.
• Finally, divide both sides by 17 to solve for \( h \):
\[ h = \frac{60}{17} \]

And there you have it! By using basic algebraic manipulation and fraction operations, you arrive at the solution for \( h \). Remember, these steps can apply to any linear equation involving fractions, making them a handy tool in solving many algebra problems.