Question

`y` varies inversely with `x`, and `x=4.5` when `y=2.4`. What is the value of `x` when the value of `y=4.32`?

Answer 1

`color(blue)(x=2.5)`

Explanation:

Inverse variation is given by:

`y prop k/x^n`

Where `bbk` is the constant of variation.

To find `bbk` we substitute `x=4.5` and `y=2.4`

`2.4 = k/4.5`

`k=2.4*4.5=10.8`

When `y=4.32`

`4.32=10.8/x`

`x=10.8/4.32=2.5`

Answer 2

`x=2.5`

Explanation:

Direct variation uses the equation `y=kx`

Inverse variation uses the equation `y = k/x`

Where `k` represents the constant of variation.

In order to solve this problem we need to use the numbers provided for the first scenario to solve for the constant of
variation `k` and then use `k` to solve for the second set of numbers.

`x_1=4.5`
`y_1=2.4`
`k=?`

`y=k/x` Equation of Inverse Variation

`2.4 = k/4.5`
Use the multiplicative inverse to isolate `k`

`4.5*2.4 = k/cancel(4.5) * cancel4.5`

`k = 10.8`

`x_2=?`
`y_2=4.32`
`k=10.8`

`y=k/x` Equation of Inverse Variation

`4.32 = 10.8/x`
Use the multiplicative inverse to bring `x` out of the denominator

`x*4.32 = 10.8/cancel(x) * cancelx`

`4.32x = 10.8`

Divide both sides by `4.32` to isolate `x`

`(cancel(4.32)x)/cancel4.32 = 10.8/4.32`

`x=2.5`

Answer 3

`x = 2.5`

Explanation:

`y prop 1/x -> "Inverse variation"`

`y = k/x`, where `k` is constant

When;

`x = 4.5 and y = 2.4`

Substituting the values of `x and y` into the equation..

`2.4 = k/4.5`

`2.4/1 = k/4.5`

Cross multiplying;

`2.4 xx 4.5 = k xx 1`

`10.8 = k`

Therefore;

`k = 10.8`

Now the relationship between the two unknowns becomes;

`y = 10.8/x`

What is `x` when `y = 4.32`

Substituting the value of `y` into the relationship equation..

`4.32 = 10.8/x`

`4.32/1 = 10.8/x`

Cross multiplying;

`4.32 xx x = 10.8 xx 1`

`4.32x = 10.8`

`x = 10.8/4.32`

`x = 2.5`