How can I do these?

Make x the subject:

1. $\displaystyle y= 2x-5/3 $

2. $\displaystyle y= /sqrt{4+2x}$

3. $\displaystyle y= 1/x^3+8 $

4. $\displaystyle y= k+ax/k-ax $

Thanks in advance!

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- Nov 27th 2009, 12:03 AMJadeKiaraMaking X The Subject
How can I do these?

Make x the subject:

1. $\displaystyle y= 2x-5/3 $

2. $\displaystyle y= /sqrt{4+2x}$

3. $\displaystyle y= 1/x^3+8 $

4. $\displaystyle y= k+ax/k-ax $

Thanks in advance! - Nov 27th 2009, 12:10 AMmathaddict
HI

the idea here is to throw everything so that x is the only term at one side .

(1) y=2x-5/3

Multiply the whole thing by 3 ie 3y=6x-5

6x=3y+5

x=(3y+5)/6

$\displaystyle y=\frac{1}{x^3}+8$

$\displaystyle y-8=\frac{1}{x^3}$

$\displaystyle

x^3=\frac{1}{y-8}

$

$\displaystyle x=\frac{1}{(y-8)^{\frac{1}{3}}}$

Try out the rest . - Nov 27th 2009, 12:14 AMJadeKiara
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} - Nov 27th 2009, 12:44 AMGrandad
Hello JadeKiaramathaddict'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:$\displaystyle y = \frac{1}{x^3+8}$which could be written instead as

$\displaystyle \Rightarrow x^3+8 = \frac{1}{y}$ (Do you understand that step?)

$\displaystyle \Rightarrow x^3 = \frac{1}{y}-8$

$\displaystyle \Rightarrow x = \sqrt[3]{\frac{1}{y}-8}$

$\displaystyle \Rightarrow x = \sqrt[3]{\frac{1-8y}{y}}$For #2:

$\displaystyle y = \sqrt{4+2x}$Can you complete it from here?

$\displaystyle \Rightarrow y^2 = 4+2x$

And #4 is a bit trickier:$\displaystyle y = \frac{k+ax}{k-ax}$Grandad

$\displaystyle \Rightarrow y(k-ax)=k+ax$

$\displaystyle \Rightarrow ky-ayx=k+ax$

$\displaystyle \Rightarrow ky-k = ax+ayx$

$\displaystyle \Rightarrow k(y-1)=a(1+y)x$ (Do you see how I've factorised each side?)

$\displaystyle \Rightarrow x = \frac{k(y-1)}{a(1+y)}$