What is the solution to the linear equation?
$6 k+10.5=3 k+12$
The problem asks us to solve the linear equation \(6k + 10.5 = 3k + 12\) for the variable \(k\). This is a linear equation in one variable, which means we can isolate the variable by performing algebraic operations on both sides to balance the equation. We are given four potential solutions to check: \(k = 0.5\), \(k = 2\), \(k = 7.3\), and \(k = 9\). Our strategy will be to use algebraic manipulation to find the exact value of \(k\).
To isolate \(k\), we start by collecting all terms containing \(k\) on one side of the equation. We do this by subtracting \(3k\) from both sides:
\[6k - 3k + 10.5 = 3k - 3k + 12\]
\[3k + 10.5 = 12\]
Next, we isolate the term with \(k\) by moving the constant term to the other side. We subtract \(10.5\) from both sides:
\[3k + 10.5 - 10.5 = 12 - 10.5\]
\[3k = 1.5\]
Finally, we solve for \(k\) by dividing both sides of the equation by the coefficient of \(k\), which is \(3\):
\[k = \frac{1.5}{3}\]
\[k = 0.5\]
Comparing this result to the given options, we confirm that \(k = 0.5\) is the correct solution.
Linear equations like this are foundational in economics and personal finance. For example, imagine two different subscription plans for a service: one costs \(\$10.50\) upfront plus \(\$6\) per month, and the other costs \(\$12\) upfront plus \(\$3\) per month. By setting the total cost expressions equal, like \(6k + 10.5 = 3k + 12\), you can determine the exact number of months, \(k\), where both plans end up costing the same amount. Finding this "break-even" point helps consumers make informed financial decisions.