rearranging equation

[Guest]

Hi forum,

I am trying to rearrange the following equation in terms of d as I then want to interpret the result when Y=90, but I am having some trouble, as wouldn't d^h just cancel out in the equation?

Can anyone assist with this?

y=(100xd^h)/(d^h+k^h)

[Guest]

No, in a/(a+ b) the two "a"s do not cancel out. If a= 2 and b= 5, for example, 2/(2+ 5)= 2/7, not 1/5. Also I am going to interpret the "x" in 100xd^h as multiplication, not another variable. Algebraically that would be better written 100d^h.

I would start by replacing d^h by "u" and write it as y= 100u/(u+ k^h).
Multiply on both sides by u+ k^h: y(u+ k^h)= yu+ yk^h= 100u.
Subtract yu from both sides: yk^h= 100u- yu= (100- y)u.
Divide both sides by 100- y: yk^h/(100- y)= u.

Now put d^h back for u: d^h= yk^h/(100- y).
And finally take the kth root of both sides: d= [k^hy/(100-y)]^{1/h}= k[(100- y)]{1/h}.

[Guest]

oh wow! I really love this explanation- just also want to point you to some great new high school solutions. Also would appreciate if ya'll can point me to more such resources- i just began homeschooling my kids in this pandemic.