If (x + 3) varies directly as y and x = 3 when y = 12, what is the value of x when y = 8?
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1
- \(\frac{1}{2}\)
- \(-\frac{}{}\)
- -1
The correct answer is: A
Explanation
\((x+3) \propto y\\∴x + 3 = ky\hspace{1mm}when\hspace{1mm}x = 3, y = 12\\
3+3 = 12k\\
∴ k = \frac{1}{2}\\
\Longrightarrow x + 3 = \frac{1}{2}y\) to find x when y = 8
\(x + 3 = \frac{1}{2}\times 8\\
x=4-3\\
x = 1\)