Problem 63 - Algebra A Combined Function

Question

Solve. See Examples 1 through 7 $$ 3 x+\frac{5}{16}=\frac{3}{4}-\frac{1}{8} x-\frac{1}{2} $$

Step-by-Step Solution

Answer

The solution is \( x = -\frac{1}{50} \).

Step 1: Combine Like Terms on the Right-Hand Side

First, simplify the right-hand side of the equation. Combine \( \frac{3}{4} \) and \( -\frac{1}{2} \):\[ \frac{3}{4} - \frac{1}{2} = \frac{3}{4} - \frac{2}{4} = \frac{1}{4}. \] So the equation becomes: \[ 3x + \frac{5}{16} = \frac{1}{4} - \frac{1}{8}x. \]

Step 2: Eliminate Fractions

To eliminate fractions, multiply the entire equation by 16, the least common multiple of the denominators. \[ 16(3x + \frac{5}{16}) = 16(\frac{1}{4} - \frac{1}{8}x). \] Simplifying, we get: \[ 48x + 5 = 4 - 2x. \]

Step 3: Move Variables to One Side

Add \(2x\) to both sides to bring all the \(x\) terms to the left: \[ 48x + 2x + 5 = 4. \] Simplifying gives: \[ 50x + 5 = 4. \]

Step 4: Isolate the Variable Term

Subtract 5 from both sides to isolate the \(x\) term: \[ 50x = 4 - 5. \] This simplifies to: \[ 50x = -1. \]

Step 5: Solve for x

Divide both sides by 50 to solve for \(x\): \[ x = \frac{-1}{50}. \]

Key Concepts

Like Terms

In algebra, combining like terms is a fundamental step in simplifying expressions and solving equations. Like terms are terms in an equation that have identical variable parts raised to the same power. For example, in the expression \(3x\) and \(-\frac{1}{8}x\), both terms contain the variable \(x\), which makes them like terms. Here, our task was to simplify the right-hand side of the original equation by combining these fractions. We combined \(\frac{3}{4}\) and \(-\frac{1}{2}\) to get a single fraction, \(\frac{1}{4}\). This process helps reduce the complexity of the equation by consolidating terms, making the equation easier to work with.

• Always check if terms have the same variables and exponents before combining them.
• This step often involves adding or subtracting coefficients of like terms.

Identifying and combining like terms is crucial to simplifying equations and should be the first step in many algebraic problems that involve multiple terms.

Fraction Elimination

Eliminating fractions in an equation can greatly simplify the solving process. Fractions tend to complicate calculations, so using the least common multiple (LCM) to eliminate them is helpful. For the equation \(3x + \frac{5}{16} = \frac{1}{4} - \frac{1}{8}x\), we determined that 16 is the LCM of all denominators. We multiplied every term in the equation by 16 to eliminate fractions across the equation.
This method transforms the equation into one that only involves whole numbers: \(48x + 5 = 4 - 2x\).

• Identify the LCM of denominators in the equation.
• Multiply every term by this LCM to clear out fractional terms.

Eliminating fractions makes the arithmetic easier and reduces possibilities for errors later on.

Variable Isolation

Isolating variables is a critical step to streamline solutions and solve for unknowns effectively. After eliminating fractions in the previous step, our goal is to get all the variable terms on one side of the equation. In the equation \(48x + 5 = 4 - 2x\), we moved the \(-2x\) on the right over to the left by adding \(2x\) to both sides. This process leads to the equation \(50x + 5 = 4\).
This action consolidates the variable terms on one side, setting the stage for further simplification.

• The objective is to isolate the terms containing the unknown variable on one side.
• Perform inverse operations to move terms across the equation.

Successful isolation of variables simplifies the path to solving the equation and understanding the value that makes the equation true.

Solve for x

Finally, solving for \(x\) means finding the value of the variable that satisfies the equation. Once we had \(50x + 5 = 4\), we aimed to isolate \(x\) by removing the constant term on its side of the equation. Subtracting 5 from both sides led to \(50x = -1\). The last step was dividing both sides by 50 to give \(x = \frac{-1}{50}\).
In this stage, we:

• Removed any remaining constant terms from the side with the variable by using subtraction or addition.
• Divided by the coefficient of the variable to completely solve for the variable's value.

Solving for \(x\) provides the exact value for which both sides of the original equation are equal. Understanding this process is fundamental for solving any linear equation efficiently.