- Distribute the \(2\) to get \(6x + 8 + 5 = 23\).
- Combine like terms to get \(6x + 13 = 23\).
- Subtract \(13\) from both sides to get \(6x = 10\).
- Divide by \(6\) and simplify the fraction to find \(x = \boxed{\frac{5}{3}}\).
Explanation
- Analysis of the Equation
We are tasked with solving the linear equation \(2(3x + 4) + 5 = 23\) for the variable \(x\). To do this, we will apply standard algebraic techniques such as distributing terms across parentheses, combining constants, and performing inverse operations to isolate \(x\).
- Distribution and Combining Like Terms
First, we distribute the coefficient \(2\) into the expression \((3x + 4)\):
\[2 \times 3x + 2 \times 4 + 5 = 23\]
\[6x + 8 + 5 = 23\]
Next, we combine the constant terms \(8\) and \(5\) on the left side:
\[6x + 13 = 23\]
- Isolating and Solving for x
To isolate the term with \(x\), we subtract \(13\) from both sides of the equation:
\[6x = 23 - 13\]
\[6x = 10\]
Finally, we divide both sides by \(6\) to solve for \(x\):
\[x = \frac{10}{6}\]
Simplifying the fraction by dividing both the numerator and the denominator by their greatest common divisor, \(2\), we obtain:
\[x = \frac{5}{3}\]
Examples
In many practical situations, such as calculating budget adjustments or scaling ingredients in a recipe, we encounter linear equations. For example, if a company has a fixed daily operating cost of \(\$5\) and an additional cost of \(\$2\) for each unit produced, and the total daily cost is \(\$23\), where the number of units produced is given by \((3x+4)\), we use this exact algebraic process to determine how many units were processed. Solving for \(x\) helps us understand the variables involved in such systems.