Question

The variables x=9 and y=15 varies directly. How do you write an equation that relates the variables and find x when y=-5?

Answer 1

`x=3` when `y=5`

Explanation:

When y increases, x also increases in the same proportion and when y decreases, x also decreases in the same proportion.
So `x:y=x1:y1`
Or `x/y=(x1)/(y1)`
We know, y=15, x=9 and y1=5
`:.x1=((x/y)(y1)=(9/15)5=45/15=3`
`x1=3`

Answer 2

The equation is `y=5/3x`
and
`y=-5color(white)("xx")rarrcolor(white)("xx")x=-3`

Explanation:

If `x` and `y` vary directly, then
`color(white)("XXX")y=color(blue)c * x` for some constant `color(blue)c`

Given that `x=9` and `y=15` is one solution to this equation:
`color(white)("XXX")15=color(blue)c * 9`

`color(white)("XXX")rarr color(blue)c=15/9=color(blue)(5/3)`

So the relating equation is `y=color(blue)(5/3)x`

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

When `y=-5`, the equation becomes:
`color(white)("XXX")-5=color(blue)(5/3)color(green)x`

`color(white)("XXX")rarr color(green)x=(-5)xx3/5)=color(green)(-3)`