Question

If x varies inversely as the square of y and if x=3 when y=-1, what is the value of x when y =3

Answer 1

Answer:

$x$ would be $(1/3)$ when $y = 3$.

Step-by-step explanation:

The question states that $x$ varies "inversely" as $y^2$. In other words, there is a non-zero number $a$ (a constant) such that:

$\displaystyle x = \frac{a}{y^2}$.

The challenging part is to find the value of $a$ in this equation.

Given that $x = 3$ when $y = -1$, replace the $x$ in this equation with $3$ and $y$ with $(-1)$. This equation should still be valid:

$\displaystyle 3 = \frac{a}{(-1)^{2}}$.

Solve for $a$:

$a = 3 \times (-1)^2 = 3$.

Hence, the relation between $x$ and $y$ becomes:

$\displaystyle x = \frac{3}{y^2}$.

Find the value of $x$ when $y = 3$ by replacing the $y$ in this equation with $3$.

$\displaystyle x = \frac{3}{3^2} = \frac{1}{3}$.

In other words, the value of $x$ would be $\displaystyle \frac{1}{3}$ when $y = 3$.