Problem 52 - Algebra 1

Question

Solve \(\frac{1}{3}(7 x+5)=3 x-5\) A 15 B -5 C 10 D \(\frac{-5}{2}\)

Step-by-Step Solution

Answer

C 10

Step 1: Distribute the 1/3 to terms inside the parentheses

Multiply each term inside the parentheses on the left side of the equation by 1/3. This results in \(\frac{1}{3} * 7x + \frac{1}{3} * 5 = 3x - 5\), which simplifies to \(\frac{7}{3}x + \frac{5}{3} = 3x - 5\).

Step 2: Simplify the equation

In order to isolate \(x\), you must first move the \(x\) terms to one side and the constant terms to the other side. To do so, subtract \(\frac{7}{3}x\) from both sides of the equation, which results in \(\frac{5}{3} = 3x - \frac{7}{3}x - 5\). Secondly, add 5 to both sides of the equation, resulting in \(\frac{5}{3} + 5 = 3x - \frac{7}{3}x\). This simplifies to \(\frac{20}{3} = \frac{2}{3}x\).

Step 3: Solve for x

Finally, to solve for \(x\), divide both sides of the equation by \(\frac{2}{3}\). This gives \(x = \frac{20}{3} / \frac{2}{3}\), which simplifies to \(x = 10\).

Key Concepts

Distributive Property

When solving linear equations that include parentheses, the distributive property is a vital tool. The distributive property states that a term outside the parentheses can be multiplied by each term within the parentheses. This is written as \( a(b + c) = ab + ac \). In our exercise, we have a fraction outside the parentheses, \( \frac{1}{3} \), which we will distribute over the terms inside the parentheses, \( 7x + 5 \). This means:

• Multiply \( \frac{1}{3} \) by \( 7x \), resulting in \( \frac{7}{3}x \).
• Multiply \( \frac{1}{3} \) by \( 5 \), resulting in \( \frac{5}{3} \).

This step transforms the original equation \( \frac{1}{3}(7x + 5) = 3x - 5 \) into \( \frac{7}{3}x + \frac{5}{3} = 3x - 5 \). The distributive property plays a crucial role in simplifying equations at this stage.

Isolating Variable

The process of isolating the variable in an equation is key to finding its solution. The goal is to have our variable, \( x \), on one side of the equation, and all constant terms on the other. Start by addressing the equation:

• Subtract \( \frac{7}{3}x \) from both sides to remove it from the left side, resulting in \( 3x - \frac{7}{3}x \) on the right side.
• Add 5 to both sides to eliminate the constant term on the right, transforming it into \( \frac{5}{3} + 5 \).

This manipulation results in \( \frac{20}{3} = \frac{2}{3}x \). By performing these operations, we have successfully grouped like terms, readying the equation for the final step to solve for \( x \). Effective rearrangement is critical in finding solutions to algebraic equations.

Division of Fractions

Once the variable is isolated to one side, solving for it often involves dealing with fractions, especially in equations like these. The result \( \frac{20}{3} = \frac{2}{3}x \) requires us to undo the multiplication of \( x \) by \( \frac{2}{3} \). To solve for \( x \), divide both sides of the equation by \( \frac{2}{3} \).

• Remember, dividing by a fraction is equivalent to multiplying by its reciprocal.
• The reciprocal of \( \frac{2}{3} \) is \( \frac{3}{2} \).

Thus, multiply both sides by \( \frac{3}{2} \):
\[ x = \frac{20}{3} \times \frac{3}{2} \]
Simplifying, the result is \( x = 10 \). Understanding division of fractions is essential for tackling many algebraic problems where fractions appear.