Question

Identify the linear function(s).

• A. $\frac{x}{y}=\sqrt{5}$
• B. $\frac{1}{x}=\frac{1}{y}$
• C. $x=y^0$
• D. $y=x^{\frac{5}{3}-\frac{2}{3}}+\frac{1}{3}$

Answer 1

Answer: (A), (B), (D)

$y=x^{\frac{5}{3}-\frac{2}{3}}+\frac{1}{3}$

For a linear function, $y^b=c{x}^d+e$
$b=1,d=1$ or $0.$

Let's analyze each function.
$\frac{x}{y}=\sqrt{5}$
$\Rightarrow y=\frac{1}{\sqrt{5}}x$
Here, the exponent of $x$ and $y$ is 1.
Hence, it is a linear function.

$\frac{1}{x}=\frac{1}{y}$
(by cross-multiplying)
$\Rightarrow y=x$
Again, the exponent of $x$ and $y$ is 1.
Hence, it is a linear function.

$x=y^0$
$\Rightarrow x=1$
Here, the exponent of $y$ is 0.
Hence, it is not a linear function.
$\because$ at $x=1$, $y$ can have any value, which does not follow definition of a function.

$y=x^{\frac{5}{3}-\frac{2}{3}}+\frac{1}{3}$
$\Rightarrow y=x^{\frac{5-2}{3}}+\frac{1}{3}$
$\Rightarrow y=x^{\frac{3}{3}}+\frac{1}{3}$
$\Rightarrow y=x^1+\frac{1}{3}$
Here, the exponent of $x$ and $y$ is 1.
Hence, it is a linear function.