B Confusion over transposing formulae question

[Buggsy GC]

Hello I'm working through transposing formulae of pearsons Anthony Croft and Robert Davidson's foundation maths 5th edition and I'm stuck on exercise 10.1 question 1)e)
We're it states that I must transpose the formula to make x the subject.
The formula is y = ¹/₂x and the answer x = ¹/₂y
I don't understand how they got this answer. As I did the working as
E)y = 1/2x
× 2
(2)(y) = ⁽¹⁾/₍₂₎x × ⁽²⁾/₍₁₎
2y = x
I base this way of a working off another question which I got correct
D)y = (¹)/(₂)(x - 7)
+ 7
y + 7 = ¹/₂x
× 2
(2)(y + 7) = ⁽¹⁾/₍₂₎x × ⁽²⁾/₍₁₎
2y + 14 = x
x = 2y + 14
If I am missing some rule that I'm meant to implement or if there is a error in the textbook your advise would be greatly appreciated.
Yours sincerely Buggsy

 

[mfb]

E)y = 1/2x
× 2
(2)(y) = ⁽¹⁾/₍₂₎x × ⁽²⁾/₍₁₎
2y = x

That is correct.

D)y = (¹)/(₂)(x - 7)
+ 7
y + 7 = ¹/₂x

That step is incorrect at the right side.

Try x=7 and see if you can find a value for y such that both the initial and the final formula are right.

The formula is y = ¹/₂x and the answer x = ¹/₂y

That does not make sense.

 

[PeroK]

If $y = \frac{1}{2x}$ then $x = \frac{1}{2y}$ is correct.

Try $x = 1$ for example.

 

[mfb]

I didn't expect x and y to be in the denominator.

 

[Buggsy GC]

I know but look

 

[Buggsy GC]

Pardon me Perok can you take a picture of your working and post it, for some reason your post shows up on my screen as a bunch of hashtags and letters.
Thank you

 

[PeroK]

Pardon me Perok can you take a picture of your working and post it, for some reason your post shows up on my screen as a bunch of hashtags and letters.
Thank you

I'm sorry to tell you I did it in my head!

 

[PeroK]

Hint: multiply by $x$ to start.

 

[Buggsy GC]

So it look like this https://www.physicsforums.com/attachments/108916
Ok second question how did you know to start with x not to just repeat the same action I used to get rid of the 1/2x in question d)
We're the answer was x =2y + 14 were the original formula was y = ¹/₂x - 7

 

[PeroK]

So it look like this https://www.physicsforums.com/attachments/108916
Ok second question how did you know to start with x not to just repeat the same action I used to get rid of the 1/2x in question d)
We're the answer was x =2y + 14 were the original formula was y = ¹/₂x - 7

That last equation is very different. That is

$y = \frac{x}{2} - 7$

 

[Buggsy GC]

I'm sorry this taking a while but can tell me is there a rule I should remember or an order of operation when rearranging formulas as I' m getting different answers depending on how I start to balance the equation even if I balance both sides all the way through

 

[Mark44]

I know but look
View attachment 108914View attachment 108915

But look at what? You posted lots of problems on what looks like two pages. Which problem are you asking about here?

 

[Buggsy GC]

Question e) and d) 10.1

 

[PeroK]

I'm sorry this taking a while but can tell me is there a rule I should remember or an order of operation when rearranging formulas as I' m getting different answers depending on how I start to balance the equation even if I balance both sides all the way through

That simply means you are making a mistake. There are no rules except you can: a) add the same thing to both sides of the equation; or, b) multiply both sides of the equation by the same thing. That same thing can be a number, such as $2$ a parameter, such as $a$, or a variable, such as $x$.

For example:

$y = ax + b$

$y-b = ax$ (added $-b$ to both sides, which is the same thing as subtracting $b$)

$\frac{y-b}{a} = x$ (multiplied both sides by $\frac{1}{a}$, which is the same as dividing by $a$)

Another example is:

$y = \frac{1}{x}$

$xy = 1$ (muliplied both sides by $x$)

$x = \frac{1}{y}$ (divided both sides by $y$)