Question

try again
your answer is incorrect.

• part (a): your answer is incorrect.

suppose that y varies directly with x, and y = 12 when x = 15.
(a) write a direct variation equation that relates x and y.
equation: \\(\frac{4}{5}x\\)
(b) find y when x = 2.
\\(y = \frac{8}{5}\\)

Explanation:

Part (a)

Step1: Recall direct variation formula

The formula for direct variation is \( y = kx \), where \( k \) is the constant of variation.

Step2: Find the constant \( k \)

We know that \( y = 12 \) when \( x = 15 \). Substitute these values into the formula: \( 12 = k \times 15 \).
To solve for \( k \), divide both sides by 15: \( k=\frac{12}{15}=\frac{4}{5} \).

Step3: Write the direct variation equation

Substitute \( k = \frac{4}{5} \) back into the direct variation formula: \( y=\frac{4}{5}x \). Wait, but the system said the previous answer was incorrect? Wait, no, maybe the input's previous answer was missing the \( y=\) part. Wait, the user's previous answer for part (a) was \( \frac{4}{5}x \), but the equation should be \( y = \frac{4}{5}x \). Oh, maybe that's the mistake. So the correct equation is \( y=\frac{4}{5}x \).

Part (b)

Step1: Use the direct variation equation

We have the equation \( y=\frac{4}{5}x \) from part (a).

Step2: Substitute \( x = 2 \) into the equation

Substitute \( x = 2 \) into \( y=\frac{4}{5}x \): \( y=\frac{4}{5}\times2=\frac{8}{5} \). Wait, but let's check again. Wait, part (a) correct equation: \( y = kx \), \( k = 12/15 = 4/5 \), so \( y = (4/5)x \). Then for \( x = 2 \), \( y = (4/5)*2 = 8/5 = 1.6 \). But maybe the user's mistake was in part (a) not writing the full equation \( y = \frac{4}{5}x \) (if the previous answer was just \( \frac{4}{5}x \) without \( y = \)).

Part (a) Final Answer

Step1: Use the equation from part (a)

We have \( y=\frac{4}{5}x \).

Step2: Substitute \( x = 2 \)

\( y=\frac{4}{5}\times2=\frac{8}{5} \) (or \( 1.6 \)).

Answer:

\( y=\frac{4}{5}x \)

Part (b) Final Answer