Question

Use the following information. In a direct variation, the ratio \(\frac{y}{x}\) is constant. If \(\left(x_{1}, y_{1}\right)\) and \(\left(x_{2}, y_{2}\right)\) are solutions of the equation \(\frac{y}{x}=k,\) then \(\frac{y_{1}}{x_{1}}=k\) and \(\frac{y_{2}}{x_{2}}=k .\) Use the proportion \(\frac{y_{1}}{x_{1}}=\frac{y_{2}}{x_{2}}\) to find the missing value. Find \(y_{2}\) when \(x_{1}=-4, y_{1}=8,\) and \(x_{2}=-1\)

Short answer

The value of \(y_2\) equals to 2

Step-by-step solution

Understand the given variables

Given in the exercise are the following values: \(x_1 = -4, y_1 = 8,\) and \(x_2 = -1\). The goal is to solve for \(y_2\)

Apply direct variation

According to direct variations principle, the ratios of \(y_1\) to \(x_1\) and \(y_2\) to \(x_2\) should be equal. Therefore, we can set up the following equation: \(\frac{y_1}{x_1} = \frac{y_2}{x_2}\)

Plug in given values and solve for \(y_2\)

Plugging in the given values we get: \(\frac{8}{-4} = \frac{y_2}{-1}\) Solving for \(y_2\), we get \(y_2 = 2\)