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 there exists a constant \(k
eq 0\) such that the relationship between \(y\) and \(x\) is given by the equation:
\[y = kx\]
We are given that \(y = 6\) when \(x = -2\). We can use this information to determine the value of the constant \(k\).
To find \(k\), substitute the given values \(y = 6\) and \(x = -2\) into the equation \(y = kx\):
\[6 = k \times (-2)\]
Now, solve for \(k\) by dividing both sides by \(-2\):
\[k = \frac{6}{-2} = -3\]
Thus, the specific relationship between \(y\) and \(x\) is given by:
\[y = -3x\]
Now that we have the equation \(y = -3x\), we need to find the value of \(x\) when \(y = 15\). Substitute \(y = 15\) into the equation:
\[15 = -3x\]
Solve for \(x\) by dividing both sides by \(-3\):
\[x = \frac{15}{-3} = -5\]
Therefore, when \(y = 15\), the value of \(x\) is \(-5\).
Direct variation is a fundamental concept used in many real-world applications. For example, if you are buying items at a fixed price, the total cost \(C\) varies directly with the number of items \(n\) purchased, following the formula \(C = pn\), where \(p\) is the price per item. If you know the price of one item, you can use this relationship to quickly calculate the cost for any number of items. This concept helps in budgeting, estimating expenses, and understanding proportional changes in daily tasks.