Question

If Y varies directly as x and inversely as the square of z. y=20 when x=50 and z =5. How do you find y when x=3 and z=6?

Answer 1

`y=c/z^2" "->" "y=500/6^2" "->" "y=13 8/9 -> 125/9`

`y=kx " "->" " y=2/5xx3" "->" "y=6/5`

Explanation:

As both `x` and `z` are related to `y` we can express them as follows:

Given:`" "color(red)( y)" "color(blue)(alpha)" "color(red)( x)" "color(blue)(alpha)" "color(red)( 1/z^2)`

Where `alpha` means proportional to.

Let `k" and "c` be constants of variation. Then we have:

`y=k x=c/z^2`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Determine the values of "k" and "c)`

Base condition is that at `y=20 ; x=50 ; z=5`

`color(brown)("To determine "k)`
`=>y=kx" "->" "20=k(50)`

Divide both sides by 50

`20/50=kxx50/50" "`

but `50/50=1`

`color(brown)(k=20/50=2/5`

'...........................................................
`color(brown)("To determine "c)`

`=>y=c/z^2" "->" "20=c/(5^2)`

Multiply both sides by 25

`color(brown)(c=20xx25= 500)`

'.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Determine the value of "y" at "x=3)`

`y=kx " "->" " y=2/5xx3`

`color(blue)( y=6/5)`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Determine the value of "y" at "z=6)`

`y=c/z^2" "->" "y=500/6^2`

`y=13 8/9 -> 125/9`