Solve the inequality $-25 z \leq 2$, and write the solution in interval notation.
The objective is to solve the linear inequality \(-25z \leq 2\) for the real variable \(z\). The inequality involves a coefficient of \(-25\) attached to \(z\). To isolate \(z\), we must perform operations that maintain the equality relationship while adhering to the fundamental rules of inequality manipulation. Specifically, we must recall that multiplying or dividing both sides of an inequality by a negative number reverses the direction of the inequality sign.
To isolate \(z\), we divide both sides of the inequality \(-25z \leq 2\) by \(-25\). As stated in the analysis, because we are dividing by a negative number, the inequality sign \(\leq\) must be flipped to \(\geq\). This yields the following steps:
\[-25z \leq 2\]
\[\frac{-25z}{-25} \geq \frac{2}{-25}\]
\[z \geq -\frac{2}{25}\]
Thus, the condition for the inequality to hold is that \(z\) must be greater than or equal to \(-\frac{2}{25}\).
Having found that \(z \geq -\frac{2}{25}\), we represent this set of all possible values for \(z\) in interval notation. Since the inequality is "greater than or equal to", we include the endpoint \(-\frac{2}{25}\) in our interval using a square bracket, and the values extend to positive infinity. Therefore, the solution set is \(\left[-\frac{2}{25}, \infty\right)\).
Inequalities are fundamental in real-world scenarios where constraints exist. For example, if you are planning a budget for a project and the cost \(C\) is modeled by \(C = 25z\) where \(z\) represents the amount of material purchased, and you have a maximum budget of \(\$2\) available in a specific currency-adjusted account where the balance has dropped, you might be solving for how much you can effectively 'avoid' spending or account for in a negative balance scenario. Knowing that \(z \geq -0.08\) tells you the minimum constraint required to keep your balance within acceptable limits.