How to solve for x quadratic equation?

Question

Thank you for the feedback. One place where I messed up and should have been more clear is that the 2 was a variable and is open to change so I should have denoted that differently. I've chosen to denote it as $n$.

I’ve updated the post taking into account the kind user edits that helped make the post more readable.

You can calculate the average rate of two rates given both rates and the percentage of total for one rate. Average rate = rate2 * (rate2 % of total) + rate1 * (1-(% of total rate2))

My attempt to solve this was to add variables denoted as the following and solve for x: $z = (y * nx) + ((1-y)*x) $

Where $z$ = average rate $x$ = rate 1 $y$ = % of total of rate 2 $n$ = ratio of rate2/rate1

I expanded to get: $z = y(1-y) + xy + nx(1-y) + nx^2$.

I'm lost after this though. Any help on next steps would be greatly appreciated

I know the answer is: $x = \dfrac{z}{ (y * n) + (1-y)}$.

The question came about by trying to solve for rate1 of two rates given:

• Ratio of rate2/rate1
• Average rate
• % of the total that rate2 makes up

I'd like to understand the steps and logic to properly solve for rate1 though.

Answer 1

Hint: $$z = 2xy + x(1 - y)$$ is not quadratic, but only linear in $x$. (Why?)

Can you finish?

Answer 2

Thank you to Blue/Arnie/dxiv !

That was just a massive mess up on my end. I know it does not work that way. I was doing it on paper and must have just written it poorly and then let the error flow through.

$z=(y∗nx)+((1−y)∗x)$
$z=ynx+x−xy z=x(yn+1−y)$
Answer: $x = \dfrac{z}{ (yn) + (1-y)}$