Problem 43 - Algebra A Combined Function

Question

Solve each equation for \(x\). $$ \frac{x-3}{2}=\frac{x-5}{6} $$

Step-by-Step Solution

Answer

The solution is \( x = 2 \).

Step 1: Understand the Equation

The given equation is \( \frac{x-3}{2}=\frac{x-5}{6} \). This is an equation involving fractions with \( x \) in both numerator places.

Step 2: Eliminate the Fractions

To eliminate the fractions, we can multiply every term by the least common multiple (LCM) of the denominators, which is 6. Thus, we multiply the entire equation by 6: \[ 6 \cdot \frac{x-3}{2} = 6 \cdot \frac{x-5}{6} \] Simplifying gives: \[ 3(x-3) = x-5 \]

Step 3: Distribute and Simplify

Now, distribute the 3 on the left side:\[ 3(x-3) = 3x - 9 \] So the equation becomes:\[ 3x - 9 = x - 5 \]

Step 4: Isolate the Variable Term

Subtract \( x \) from both sides to bring all the \( x \)-terms to one side:\[ 3x - x - 9 = x - x - 5 \] This simplifies to:\[ 2x - 9 = -5 \]

Step 5: Solve for x

Add 9 to both sides to isolate the term with \( x \):\[ 2x - 9 + 9 = -5 + 9 \] Simplify to get:\[ 2x = 4 \] Now, divide both sides by 2:\[ x = 2 \]

Step 6: Verify the Solution

Replace \( x \) with 2 in the original equation to ensure it satisfies the equation:Substitute into the original equation:\[ \frac{2-3}{2} = \frac{2-5}{6} \] Simplifying both sides gives:\[ \frac{-1}{2} = \frac{-3}{6} \] The right side also simplifies to \( \frac{-1}{2} \), thus verifying our solution is correct.

Key Concepts

Fractions in Equations

Fractions can sometimes make solving equations seem more complicated than they are. However, understanding the role they play is essential. In an equation like \(\frac{x-3}{2} = \frac{x-5}{6}\), the fractions indicate division by their respective denominators. When encountering fractions in equations, aim to simplify the problem by eliminating the fractions or making their terms easier to handle. This can often be done by finding a common multiple for the denominators, setting the stage for solving the equation more straightforwardly. Don't be intimidated by fractions; with the right techniques, they can be easily managed.

Least Common Multiple in Equations

The least common multiple (LCM) is a valuable tool in solving equations involving fractions. It helps in eliminating the fractions by scaling the terms up to integers, thus simplifying the equation. For the equation \(\frac{x-3}{2} = \frac{x-5}{6}\), you find the LCM of 2 and 6, which is 6. By multiplying every term by this LCM:

• Everything becomes a clear integer without fractions.
• You can handle the equation as a simple linear equation.

This makes the equation easier to solve and reduces potential arithmetic errors.

Distributive Property

The distributive property is an algebra rule that allows us to multiply and distribute a number across terms inside a parenthesis. From our equation \(3(x-3) = x-5\), by using the distributive property, multiply 3 by each term inside the parenthesis:

• 3 multiplied by \(x\) gives \(3x\).
• 3 multiplied by -3 gives -9.

This results in \(3x - 9 = x - 5\). Utilizing the distributive property here simplifies and provides a way to handle such equations by breaking down complex expressions into manageable parts.

Variable Isolation Techniques

Isolating the variable in an equation involves manipulating the equation so that the variable stands alone on one side. This allows us to determine its value.For \(3x - 9 = x - 5\), the goal is to move all terms with \(x\) to one side and constants to the other:

• Subtract \(x\) from both sides to get \(2x - 9 = -5\).
• Add 9 to both sides to arrive at \(2x = 4\).
• Finally, divide by 2 to find \(x = 2\).

Each step, whether adding, subtracting, or dividing, brings you closer to isolating the variable, making it crucial to maintaining the balance of the equation.