Solve for the value of the variable: $5x + 12 - 6x = 5x - 48$
The goal is to solve for the variable \(x\) in the equation \(5x + 12 - 6x = 5x - 48\). This is a linear equation in one variable, which means our objective is to isolate \(x\) by simplifying both sides of the equation and using algebraic operations to move terms involving \(x\) to one side and constant terms to the other.
First, we simplify the left-hand side by combining the like terms \(5x\) and \(-6x\). We have:
\[(5x - 6x) + 12 = 5x - 48\]
\[-x + 12 = 5x - 48\]
Next, we gather all the terms containing \(x\) on the right-hand side by adding \(x\) to both sides of the equation:
\[12 = 5x + x - 48\]
\[12 = 6x - 48\]
Now, we move the constant term \(-48\) to the left-hand side by adding \(48\) to both sides:
\[12 + 48 = 6x\]
\[60 = 6x\]
Finally, to solve for \(x\), we divide both sides by the coefficient of \(x\), which is \(6\):
\[x = \frac{60}{6}\]
\[x = 10\]
By substituting \(x = 10\) back into the original equation, we get \(5(10) + 12 - 6(10) = 50 + 12 - 60 = 2\) on the left side and \(5(10) - 48 = 50 - 48 = 2\) on the right side. Since both sides are equal, the solution is correct. The final value of \(x\) is \(\boxed{10}\).
Solving linear equations is a fundamental skill in algebra used in everyday scenarios. For example, imagine you are comparing two service plans: Plan A charges a base fee of \(\$12\) plus a rate, while Plan B is a different rate minus a one-time discount. By modeling these as an equation like \(5x + 12 - 6x = 5x - 48\), you can find the specific usage level \(x\) where both plans result in the same total cost, allowing you to make an informed financial decision.