# Thread: Transposition of formulae question

1. ## Transposition of formulae question

Hi guys I'm having difficulties approaching some of these problems, pointing me toward the right direction and checking my answers would be much appreciated

Question:

Rearrange each of the following equations making x the subject:

1.) w= y/ x + c
= x = yc-wc/w-c

2.) y^3= (xz)^3
= x= sqroot y^3-z^3

3.) zx= yx + w
= x= w/z-y

In the likely event that I've got all/any wrong I can provide the steps which got me into this mess on request.....thanks in advance guys

2. Originally Posted by komodo
Hi guys I'm having difficulties approaching some of these problems, pointing me toward the right direction and checking my answers would be much appreciated

Question:

Rearrange each of the following equations making x the subject:

1.) w= y/ x + c
= x = yc-wc/w-c

2.) y^3= (xz)^3
= x= sqroot y^3-z^3

3.) zx= yx + w
= x= w/z-y

In the likely event that I've got all/any wrong I can provide the steps which got me into this mess on request.....thanks in advance guys
This is very unclear. Firstly, 4) cannot be rearranged as it is - there is no $\displaystyle x$ - did you leave a typo? Or did you just want to solve 8^3/2?

As for the others:

$\displaystyle 1)w=\frac{y}{x}+c$
Subtract $\displaystyle c$ from both sides:
$\displaystyle w-c=\frac{y}{x}$
Multiply both sides by $\displaystyle x$:
$\displaystyle x(w-c)=y$
Divide both sides by the factored term:
$\displaystyle x=\frac{y}{w-c}$

2)$\displaystyle y^3=(xz)^3$
Take the cube root of both sides:
$\displaystyle y=xz$
Divide both sides by z:
$\displaystyle x=\frac{y}{z}$

3) You have the correct solution (I believe), but your notation is incorrect. Think about brackets, and the order of operations.

3. Sorry 4.) should be v=8x^3/2

I'm guessing for 3 there should be a brackets around z-y?

4. Originally Posted by komodo
Sorry 4.) should be v=8x^3/2

I'm guessing for 3 there should be a brackets around z-y?
Yeah. Do you think you can have a go at 4) by yourself, using the above concepts? What is the inverse of dividing by two, multiplying by eight and cubing something? Take it one step at a time.

5. Originally Posted by Quacky
Yeah. Do you think you can have a go at 4) by yourself, using the above concepts? What is the inverse of dividing by two, multiplying by eight and cubing something? Take it one step at a time.
Going step by step this is what I got:

v=8x^3/2

= v/8 = x^3/2

= 2.v/8 = x^3

x= cuberoot 2.v/8

Close enough???

6. Originally Posted by komodo
Going step by step this is what I got:

v=8x^3/2

= v/8 = x^3/2

= 2.v/8 = x^3

x= Squareroot 2.v/8

Close enough???
Yeah, except you want to take the cube root at the last stage, not the square root. And remember that $\displaystyle \frac{2v}{8}=\frac{v}{4}$

Edit: I see that you've corrected this yourself while I was replying.

7. Originally Posted by Quacky
Yeah, except you want to take the cube root at the last stage, not the square root. and remember that $\displaystyle \frac{2v}{8}=\frac{v}{4}$
Thats great thanks mate

By the way the 1st question should have been w= y/ (x+b) so in other words y divided by both of them not just y/x +c.....my fault I should have written it down more clearly.

8. Originally Posted by komodo
Thats great thanks mate

By the way the 1st question should have been w= y/ (x+b) so in other words y divided by both of them not just y/x +c.....my fault I should have written it down more clearly.
Ok, start by multiplying both sides of the equation by $\displaystyle (x+b)$ to give:
$\displaystyle w(x+b)=y$

Can you take it from here? Expand the brackets, isolate the terms involving $\displaystyle x$, and from there, finishing should be quite simple.

9. Originally Posted by Quacky
Ok, start by multiplying both sides of the equation by $\displaystyle (x+b)$ to give:
$\displaystyle w(x+b)=y$

Can you take it from here? Expand the brackets, isolate the terms involving $\displaystyle x$, and from there, finishing should be quite simple.
w(x+b)=y

wx+wb=y

wx= y-wb

x= y-wb / w

Is this correct?

10. Originally Posted by komodo
w(x+b)=y

wx+wb=y

wx= y-wb

x= y-wb / w

Is this correct?
Yep.

11. Originally Posted by Quacky
Yep.
Thanks for your help once again much appreciated

12. Originally Posted by komodo
w(x+b)=y

wx+wb=y

wx= y-wb

x= y-wb / w

Is this correct?
No it's not. You MUST use the brackets when you intend for subtraction to happen before division.

It should be (y - wb)/w

13. Hi komodo,
solve for x v =8x^3/2
rearrange x^3/2 =v/8
raise each equality to the 2/3 power ((x)^3/2)^2/3= (v^2/3 )/8^2/3
simplify x=( v^2/3)/4

14. This is the perfect example of why you need to use brackets where they're needed.

We can't tell if it's $\displaystyle \displaystyle v = \frac{8x^3}{2}$ or $\displaystyle \displaystyle v = 8x^{\frac{3}{2}}$

15. Originally Posted by Prove It
This is the perfect example of why you need to use brackets where they're needed.

We can't tell if it's $\displaystyle \displaystyle v = \frac{8x^3}{2}$ or $\displaystyle \displaystyle v = 8x^{\frac{3}{2}}$
It should have been $\displaystyle \displaystyle v = 8x^{\frac{3}{2}}$

Originally Posted by Prove It
No it's not. You MUST use the brackets when you intend for subtraction to happen before division.

It should be (y - wb)/w
Thanks

Originally Posted by bjhopper
Hi komodo,
solve for x v =8x^3/2
rearrange x^3/2 =v/8
raise each equality to the 2/3 power ((x)^3/2)^2/3= (v^2/3 )/8^2/3
simplify x=( v^2/3)/4
Hi bjhopper

Is x true for both x=(v^2/3) /4 and x=cuberoot v / 4 or have I confused everyone by not writting it down correctly?

Thanks guys

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