Hello JadeKiara Originally Posted by
JadeKiara Do you realise that the equation was $\displaystyle y=2x-5/3$?
I'm sorry, I am new to LaTex, and it was supposed to be:
1. y=2x-5/3
2. y=/sqrt{4+2x}
3. y=1/{x^3+8}
4. y={k+ax}/{k-ax}
mathaddict's answer to #1 was based on the equation $\displaystyle y = 2x -\frac53$, and was correct.
However, #3 was (correctly) based on your first post. But, following your alteration, it should be: $\displaystyle y = \frac{1}{x^3+8}$
$\displaystyle \Rightarrow x^3+8 = \frac{1}{y}$ (Do you understand that step?)
$\displaystyle \Rightarrow x^3 = \frac{1}{y}-8$
$\displaystyle \Rightarrow x = \sqrt[3]{\frac{1}{y}-8}$
which could be written instead as$\displaystyle \Rightarrow x = \sqrt[3]{\frac{1-8y}{y}}$
For #2: $\displaystyle y = \sqrt{4+2x}$
$\displaystyle \Rightarrow y^2 = 4+2x$
Can you complete it from here?
And #4 is a bit trickier: $\displaystyle y = \frac{k+ax}{k-ax}$
$\displaystyle \Rightarrow y(k-ax)=k+ax$
$\displaystyle \Rightarrow ky-ayx=k+ax$
$\displaystyle \Rightarrow ky-k = ax+ayx$
$\displaystyle \Rightarrow k(y-1)=a(1+y)x$ (Do you see how I've factorised each side?)
$\displaystyle \Rightarrow x = \frac{k(y-1)}{a(1+y)}$
Grandad