Problem 18 - Algebra 1: Concepts and Skills

Question

Solve the equation. \(\frac{2}{3}(x+8)=8\)

Step-by-Step Solution

Answer

The solution to the equation is \(x=4\).

Step 1: Start with the Original Equation

The original equation is \(\frac{2}{3}(x+8)=8\).

Step 2: Distribute

We need to distribute the \(\frac{2}{3}\) into the parentheses. This gives us \(\frac{2}{3}x+\frac{16}{3}=8\).

Step 3: Subtract \(\frac{16}{3}\) from both sides

To isolate the term with \(x\), subtract \(\frac{16}{3}\) from both sides of the equation. This gives us \(\frac{2}{3}x=8-\frac{16}{3}\). This simplifies to \(\frac{2}{3}x=\frac{8}{1}-\frac{16}{3}\).

Step 4: Unify the fractions on the right side

Before subtracting the fractions, find a common denominator, which is 3 in this case. This gives \(\frac{2}{3}x=\frac{24}{3}-\frac{16}{3}\). Now you can subtract the fractions and simplify. This results in \(\frac{2}{3}x=\frac{8}{3}\).

Step 5: Solve for \(x\)

To get \(x\) by itself, multiply each side of the equation by the reciprocal of \(\frac{2}{3}\), which is \(\frac{3}{2}\). This gives \(x=\frac{3}{2}\cdot\frac{8}{3}\) which simplifies to \(x=4\).

Key Concepts

Fraction Distribution

In mathematics, when dealing with equations involving fractions, the process of fraction distribution may come into play. This involves applying a fraction to each item inside a bracket or parentheses. Consider the equation \( \frac{2}{3}(x+8) = 8 \).

• The fraction \( \frac{2}{3} \) must be distributed across the terms inside the parentheses, meaning both \( x \) and \( 8 \) will be multiplied by \( \frac{2}{3} \).
• This results in \( \frac{2}{3} \times x + \frac{2}{3} \times 8 \), which simplifies to \( \frac{2}{3}x + \frac{16}{3} \).

Bringing fractions into a distribution helps in effectively breaking down and simplifying equations. Always ensure each term within the parentheses is accounted for when applying distribution.

Isolating Variables

Isolating the variable is a critical step in solving equations, as it helps to find its value by itself on one side of the equation. When you have an equation like \( \frac{2}{3}x + \frac{16}{3} = 8 \), the goal is to get \( x \) alone on one side.

• The first task is to eliminate any constant numbers attached to the variable.
• Subtract \( \frac{16}{3} \) from both sides so that you're left with the term \( \frac{2}{3}x \).
• Now the equation becomes \( \frac{2}{3}x = 8 - \frac{16}{3} \).

Isolating the variable often involves moving terms across the equation and using inverse operations, like addition and subtraction, until the variable stands alone.

Fraction Subtraction

When two fractions need to be subtracted, like \( 8 - \frac{16}{3} \), it's crucial to convert them to have a common denominator. Fraction subtraction is all about ensuring both fractions align properly.

• If one number is a whole number, like 8, it can be presented as a fraction with 1 as its denominator, becoming \( \frac{8}{1} \).
• Next, determine a common denominator, which in this case is 3.
• Convert \( \frac{8}{1} \) to \( \frac{24}{3} \), matching the denominator of \( \frac{16}{3} \).
• Now subtract: \( \frac{24}{3} - \frac{16}{3} = \frac{8}{3} \).

The result, \( \frac{8}{3} \), is a simplified fraction obtained by subtracting two like fractions.

Reciprocal Multiplication

Finally, solving for the variable often requires the use of reciprocal multiplication, especially when dealing with fractional coefficients. Once the equation is simplified to \( \frac{2}{3}x = \frac{8}{3} \), the next step is to clear the fraction by using its reciprocal.

• The reciprocal of \( \frac{2}{3} \) is \( \frac{3}{2} \). This means you will multiply both sides of the equation by \( \frac{3}{2} \).
• This transforms the equation: \( x = \frac{3}{2} \times \frac{8}{3} \).
• On simplifying, the fractions \( \frac{3}{2} \) and \( \frac{2}{3} \) cancel each other out on the left side, leaving \( x \) by itself.
• On the right side, multiply the remaining fractions to get: \( x = \frac{24}{6} \), which simplifies to \( x = 4 \).

Reciprocal multiplication is essential for eliminating fractional coefficients, allowing you to solve for the variable effectively.