if y varies directly as x and inversely as z^2, and y=3 when x=2 and z=4, what is the value of x when y=6 and z=2?its algebra 2. please explain the steps so that i can understand what to do.
You need to notice that if y varies directly as x, then y is a multiple of x such that:
`y = n*x`
You should also notice that if y varies inversely as `z^2` , then y is a multiple of `1/z^2` such that:
`y = 1/z^2`
Hence, you may write y in terms of x and `z^2` such that:
`y = n(x/z^2)`
You need to substitute 2 for x, 3 for y and 4 for z in equation above to find n such that:
`3 = n*(2/4^2) => 16*3 = 2n => n = 8*3 => n = 24 `
Since you know the value of n, you may find x, when y = 6 and z = 2, such that:
`6 = 24*x/2^2 => 4*6 = 24*x => x = 24/24 => x = 1`
Hence, evaluating the value of x, under the given conditions yields `x = 1` .