Given the equation $-15a = 75$, solve for $a$.
We are given the linear equation \(-15a = 75\), and our goal is to find the value of the variable \(a\). This is a standard algebraic problem where we aim to isolate the variable on one side of the equation. To do this, we need to understand the relationship between the coefficient \(-15\) and the variable \(a\). Since \(a\) is multiplied by \(-15\), we can reverse this operation by performing the inverse operation, which is division.
To solve for \(a\), we divide both sides of the equation by the coefficient of \(a\), which is \(-15\):
\[\frac{-15a}{-15} = \frac{75}{-15}\]
By simplifying the left side, we get \(a\). Now, we calculate the quotient on the right side:
\[a = \frac{75}{-15}\]
Dividing \(75\) by \(15\) gives \(5\), and because the numerator is positive and the denominator is negative, the resulting value must be negative. Thus, \(a = -5\).
We have found that \(a = -5\). Comparing this result to the given options (\(-60\), \(-5\), \(5\), and \(90\)), we see that our calculated value corresponds to the second option. The value \(-5\) satisfies the original equation since \(-15 \times (-5) = 75\). Therefore, the correct value for \(a\) is \(-5\).
Understanding how to solve for a variable in a linear equation is a fundamental skill used in many practical situations, such as calculating unit prices or managing budgets. For example, if you know that a total cost for 15 identical items is \$-\$75\$ (representing a credit or a reduction in debt), you can use this method to determine the cost per item. By solving the equation \(15x = -75\), you find that each item has a value of \(x = -5\), which is useful for tracking your financial balances or distributing costs evenly across multiple accounts.