Suppose $y$ varies directly with $x$. If $y=6$ when $x=-2$, find $x$ when $y=15$.
The problem states that \(y\) varies directly with \(x\). This means that there exists a constant of variation \(k\) such that the relationship between the two variables can be expressed as \(y = kx\). We are given a specific case: when \(x = -2\), \(y = 6\). To proceed, we first need to identify the constant \(k\) using this information.
To find the value of the constant \(k\), we substitute the given values \(x = -2\) and \(y = 6\) into the equation \(y = kx\):
\[6 = k \cdot (-2)\]
Now, we solve for \(k\) by dividing both sides by \(-2\):
\[k = \frac{6}{-2} = -3\]
Thus, the specific relationship between \(x\) and \(y\) is \(y = -3x\).
With the relationship \(y = -3x\) established, we can find the value of \(x\) when \(y = 15\). Substitute \(y = 15\) into the equation:
\[15 = -3x\]
Now, solve for \(x\) by dividing both sides by \(-3\):
\[x = \frac{15}{-3} = -5\]
Therefore, when \(y = 15\), \(x\) must be \(-5\). This matches the first option provided.
Direct variation is a fundamental concept in modeling real-world relationships where one quantity changes in proportion to another. For example, if you are filling a pool with a garden hose, the total volume of water in the pool (y) varies directly with the time (x) the hose is left running. If you know that it takes 6 minutes to fill a certain amount, you can determine how much longer or shorter it will take to fill a different target volume, assuming the flow rate remains constant. Understanding this helps in planning tasks that depend on time-based or rate-based inputs.