Solve.
$2(x-3)=19-5(4 x+1)$
The given linear equation is \(2(x - 3) = 19 - 5(4x + 1)\). Our objective is to find the value of \(x\) that satisfies this equation by expanding the expressions, collecting like terms, and isolating the variable.
First, we expand both sides of the equation. On the left side, we distribute the \(2\):
\[2(x - 3) = 2x - 6\]
On the right side, we distribute the \(-5\):
\[19 - 5(4x + 1) = 19 - 20x - 5 = 14 - 20x\]
Equating these, we have:
\[2x - 6 = 14 - 20x\]
Next, we collect all terms involving \(x\) on one side and constant terms on the other. Adding \(20x\) to both sides gives:
\[2x + 20x - 6 = 14\]
\[22x - 6 = 14\]
Adding \(6\) to both sides results in:
\[22x = 20\]
Finally, we solve for \(x\) by dividing both sides by \(22\):
\[x = \frac{20}{22}\]
Simplifying the fraction by dividing the numerator and denominator by their greatest common divisor, \(2\), we get:
\[x = \frac{10}{11}\]
Thus, the solution to the equation is \(x = \boxed{\frac{10}{11}}\).
Linear equations are used frequently in budgeting. Suppose you are planning an event where you have a set budget and need to account for both fixed costs and costs per person. If you represent the number of guests as \(x\), the relationship between your expenses and the total budget can be modeled as a linear equation, allowing you to solve for the maximum number of guests you can invite while staying within your financial limits. This simple algebraic process helps ensure that your resources are allocated efficiently.