# Math Help - Need help re - arraning this question

1. ## Need help re - arraning this question

Make x the subject:

y = (2/3)zx

And this one which I have already done but I need checking:

y = (6x - 7)/3

x = (3y + 7)/6

2. Originally Posted by BomberAF
Make x the subject:

y = (2/3)zx

And this one which I have already done but I need checking:

y = (6x - 7)/3

x = (3y + 7)/6
$y = \frac{6x-7}{3}$ gives $x = \frac{3y+7}{6}$

Similarly
$y = \frac{2zx}{3}$ gives $x = \frac{3y}{2z}$

3. Originally Posted by harish21
$y = \frac{6x-7}{3}$ gives $x = \frac{3y+7}{6}$

Similarly
$y = \frac{2zx}{3}$ gives $x = \frac{3y}{2z}$

Thanks for the answer to the first question but you have got the second one incorrect:

It is (2/3)yz, the y and z are not over the 3.

Could you also check this one for me:

y = (1/2)xz.

x = (2y)/z

4. Originally Posted by BomberAF
Thanks for the answer to the first question but you have got the second one incorrect:

It is (2/3)yz, the y and z are not over the 3.

Could you also check this one for me:

y = (1/2)xz.

x = (2y)/z
Okay. For the first question, i thought $zx$ was in the numerator.
If $zx$ is in the denominator, then your question becomes $y = \frac{2}{3zx}$ which gives:
$x= \frac{2}{3yz}$

For $y = \frac{1}{2xz}$ you get
$x = \frac{1}{2yz}$

hope that is what you are looking for!