Problem 104 - Intermediate Algebra

Question

Solve each equation. $$ \frac{2}{3}(3 m-2)=\frac{3}{4} m+\frac{11}{12} $$

Step-by-Step Solution

Answer

The solution is \( m = \frac{9}{5} \).

Step 1: Distribute and Simplify

First, distribute the fraction \( \frac{2}{3} \) on the left side of the equation: \[ \frac{2}{3}(3m - 2) = \frac{2}{3} \times 3m - \frac{2}{3} \times 2 = 2m - \frac{4}{3}. \] The equation becomes: \[ 2m - \frac{4}{3} = \frac{3}{4}m + \frac{11}{12}. \]

Step 2: Eliminate Fractions

To eliminate fractions, find the least common multiple of the denominators (3, 4, and 12). This is 12. Multiply every term by 12: \[ 12(2m) - 12\left(\frac{4}{3}\right) = 12\left(\frac{3}{4}m\right) + 12\left(\frac{11}{12}\right). \] This simplifies to: \[ 24m - 16 = 9m + 11. \]

Step 3: Collect Like Terms

Subtract \(9m\) from both sides to collect like terms: \[ 24m - 9m - 16 = 11. \] Which simplifies to:\[ 15m - 16 = 11. \]

Step 4: Solve for \(m\)

Add 16 to both sides to isolate the term with \(m\): \[ 15m = 27. \] Then, divide both sides by 15 to solve for \(m\): \[ m = \frac{27}{15}. \] Simplify the fraction:\[ m = \frac{9}{5}. \]

Key Concepts

Distributive Property

The distributive property is a fundamental aspect of algebra that helps simplify expressions by distributing a single term across terms within parentheses. In this problem, the distributive property is applied to the expression \( \frac{2}{3}(3m - 2) \). To apply the distributive property, you multiply \( \frac{2}{3} \) by each term inside the parentheses individually. This means you take \( \frac{2}{3} \times 3m \) and \( \frac{2}{3} \times (-2) \).

• \( \frac{2}{3} \times 3m = 2m \)
• \( \frac{2}{3} \times (-2) = -\frac{4}{3} \)

After distribution, the expression simplifies to \( 2m - \frac{4}{3} \). By distributing the factor across each term inside the brackets, we can effectively work with a simpler linear equation. The step of distribution is crucial as it sets up the equation for further simplification and solution.

Eliminating Fractions

Fractions in equations can make calculations tricky. One of the strategies to simplify dealing with fractions is to eliminate them by finding a common multiple. In this exercise, the denominators are 3, 4, and 12.
To eliminate fractions:

• Identify the least common multiple (LCM) of 3, 4, and 12, which is 12 in this case.
• Multiply every term in the equation by 12.

This step ensures that each term is transformed into whole numbers, making the equation easier to solve. After multiplication, each part of the equation transforms as follows:

• \( 12 \times 2m - 12 \times \frac{4}{3} = 24m - 16 \)
• \( 12 \times \frac{3}{4}m + 12 \times \frac{11}{12} = 9m + 11 \)

The equation becomes \( 24m - 16 = 9m + 11 \), allowing for straightforward algebraic manipulation.

Collecting Like Terms

Once fractions are eliminated, we can focus on organizing and simplifying the equation further by collecting like terms. Collecting like terms involves combining terms on both sides of the equation that have the same variables or are constants.
In the equation \( 24m - 16 = 9m + 11 \), the terms involving \( m \) are on both sides, so we perform the following steps:

• Subtract \( 9m \) from both sides: \( 24m - 9m - 16 = 11 \)
• Simplify to \( 15m - 16 = 11 \)

By organizing terms on one side, we can handle the equation better. This operation is essential for isolating the variable on one side, which then leads us to solving for the unknown.

Simplifying Fractions

The final step after solving the equation is often simplifying the fraction obtained as the solution. Simplification makes the solution cleaner and easier to understand.
The equation we have solved comes down to \( 15m = 27 \). To find \( m \), we divide both sides by 15:

• \( m = \frac{27}{15} \)

We simplify \( \frac{27}{15} \) by finding the greatest common divisor (GCD) of 27 and 15, which is 3. Divide both numerator and denominator by 3 to get:

• \( \frac{27}{15} = \frac{9}{5} \)

Thus, the value of \( m \) is \( \frac{9}{5} \), a simplified form presenting the solution in its clearest form.