Problem 28 - Algebra for College Students

Question

Solve each equation. \(\frac{3 a-1}{4}+\frac{a-2}{3}-\frac{a-1}{5}=\frac{21}{20}\)

Step-by-Step Solution

Answer

The solution to the equation is \(a = 2\).

Step 1: Identify a Common Denominator

To solve the equation, we'll first need to tackle the fractions. The denominators we have to consider are 4, 3, 5, and 20. To add or subtract fractions, we need a common denominator. The least common multiple (LCM) of these numbers is 60.

Step 2: Express Each Fraction with the Common Denominator

Rewrite each term so that each fraction has the denominator 60. \(\frac{3a-1}{4} = \frac{15(3a-1)}{60}, \frac{a-2}{3} = \frac{20(a-2)}{60}, \frac{a-1}{5} = \frac{12(a-1)}{60}, \text{and } \frac{21}{20} = \frac{63}{60}.\)

Step 3: Rewrite the Equation

With each term converted, the equation is now: \(\frac{15(3a-1)}{60} + \frac{20(a-2)}{60} - \frac{12(a-1)}{60} = \frac{63}{60}\).

Step 4: Combine the Fractions

Since all the fractions have the common denominator of 60, we can combine the numerators: \(15(3a-1) + 20(a-2) - 12(a-1) = 63.\)

Step 5: Simplify the Numerators

Simplify the expression by distributing and combining like terms:\(45a - 15 + 20a - 40 - 12a + 12 = 63.\)Combine the terms: \(53a - 43 = 63.\)

Step 6: Solve for a

Isolate variable \(a\) by adding 43 to both sides of the equation: \(53a = 106.\)Divide both sides by 53: \(a = 2.\)

Key Concepts

Common Denominator

When working with fractions, particularly in algebraic equations, a key step is finding a common denominator. This allows you to compare, add, or subtract the fractions without changing their values. In our exercise, the denominators are 4, 3, 5, and 20. To handle these fractions efficiently, a shared denominator is essential. By using a common denominator, you ensure each fraction is expressed in equivalent terms. This simplifies calculations significantly, as it permits straightforward addition and subtraction by focusing on numerators only. Remember, the common denominator must be a multiple of each of the original denominators, encompassing all fractions involved in the equation. Thus, it represents a standard term for fraction calculation and balances the equation for further processing.

Least Common Multiple

The least common multiple (LCM) is a crucial concept when determining a common denominator. It's the smallest positive number that is divisible by each of the denominators you're working with. In our original problem, to find the LCM of 4, 3, 5, and 20, we list the multiples or use prime factorization. For example:

• The multiples of 4 are 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60...
• The multiples of 3 are 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60...
• And so on, with similar lists for 5 and 20.

The LCM of these numbers is 60, as it's the first number that appears in all lists. Using the LCM as a common denominator simplifies the addition and subtraction of fractions, making calculations seamless and the equation simple to solve.

Fraction Addition and Subtraction

Once a common denominator has been established, adding and subtracting fractions becomes a straightforward process. In our equation, the fractions are expressed with the common denominator of 60. To operate on these fractions, we adjust the numerators accordingly:

• For the first term: \(\frac{3a-1}{4} = \frac{15(3a-1)}{60}\)
• For the second term:\(\frac{a-2}{3} = \frac{20(a-2)}{60}\)
• For the third term:\(\frac{a-1}{5} = \frac{12(a-1)}{60}\)

With all fractions sharing the denominator of 60, you can straightforwardly add or subtract by working solely with the numerators:o After combining the numerators:\(15(3a-1) + 20(a-2) - 12(a-1) = 63\)From here, solving the equation involves simplifying this to isolate the variable, by managing only the numerators while maintaining equality.

Equation Solving Steps

Solving an algebraic equation involves a systematic process that allows us to find the value of the unknown variable. In this exercise:

• Start by finding a common denominator and rewriting each fraction accordingly, as described in previous sections.
• Next, combine the fractions by adding or subtracting their numerators since they all share the same denominator.
• Simplify the equation by distributing through the numerators and combining like terms, which makes it easier to manage. In our example, the result is \(53a-43=63\).
• Finally, isolate the variable by performing inverse operations. Begin by adding 43 to both sides to get \(53a=106\), and then divide by 53 to solve for \(a\), resulting in \(a=2\).

By adhering to a step-by-step approach, each part of the equation is simplified and solved in a logical sequence, revealing the solution seamlessly. This process enhances understanding and accuracy in solving algebraic problems.