Solve the following equation for $a$.
$5 y=c(6+a)$
To solve the equation \(5y = c(6 + a)\) for the variable \(a\), we need to isolate \(a\) on one side of the equation. We assume that \(c eq 0\) to perform the division. The equation relates variables \(a, c,\) and \(y\), and our goal is to express \(a\) in terms of the other two variables.
Start with the original equation:
\[5y = c(6 + a)\]
Divide both sides of the equation by \(c\) to isolate the term containing \(a\):
\[\frac{5y}{c} = 6 + a\]
Next, subtract 6 from both sides of the equation to isolate \(a\):
\[a = \frac{5y}{c} - 6\]
This gives us the variable \(a\) expressed in terms of \(y\) and \(c\).
This algebraic manipulation is essential in many fields, such as physics or finance, where you often need to rearrange a formula to solve for a specific component. For example, if the formula for the total cost of a project is given by \(C = n(f + v)\), where \(n\) is the number of units, \(f\) is a fixed cost, and \(v\) is the variable cost per unit, you could rearrange it to find the variable cost \(v = \frac{C}{n} - f\) if you know the total cost and the number of units. Practicing these rearrangements helps build a strong foundation for solving more complex equations in science and economics.