Rearranging a formula to make subject - Can someone please check my working?

[matheus]

Hey guys,

I have made an attempt at rearranging the formula below and I am looking for someone to just have a look over it and see if I got it right?

Really appreciate any pointers or tips

Thank you! :-)

 

[earboth]

Hey guys,

I have made an attempt at rearranging the formula below and I am looking for someone to just have a look over it and see if I got it right?Really appreciate any pointers or tips

Thank you! :-)

Hallo,

all your calculations are OK. Nevertheless you can factor out (-3) in the numerator and the denominator and afterwards cancel.

 

[Jameson]

Hi there. Actually I think I see an error going from line 1 to line 2. You also missed a double negative. We start with:
$$2a = b-c \Big(\frac{b}{3}-4\Big) $$
Now you multiply every term by 3 and you should get the following:
$$6a = 3b-3c\Big(\frac{b}{3}-4\Big) $$
You multiplied by 3 on the outside and inside of the parentheses and we just need to do one or the other, since the whole expression is just one term. From here you can distribute the $3c$ to the terms in parentheses and note that $(-3c)(-4)={\color{red}+}12c$.
\begin{align*}6a &= 3b-bc+12c \\
6a-12c &= 3b-bc \\
6a-12c &= b(3-c)
\end{align*}
I'll let you take it from here. You have to play around with the fraction in the end to get the final form you need. Make sense? :)

 

[matheus]

Hi guys,

Many thanks for the replies! :-D

I am currently having another look at where I went wrong, I do struggle with these sorts of questions :-(

 

[matheus]

Hi guys,

So after taking a break from this and then going back to it, here is my attempt:

How does it look? :-/

Thanks guys :-)

 

[earboth]

How does it look? :-/

Thanks guys :-)

Hello,

there are no mistakes except in the very last line: Why do you change \(\displaystyle (3-c)\) into \(\displaystyle (c-3)\) ?

The correct answer is: \(\displaystyle b = \frac{6(a-2c)}{3-c}\)

In fact you changed the sign of the term because: \(\displaystyle (3-c)=(-1) \cdot (c-3)\)

 

[matheus]

Thank you so much earboth, for checking out my work and also pointing that last line out.

I will blame user error :-D

Corrected:

Hopefully that should do the trick :-)

 

[earboth]

...

Hopefully that should do the trick :-)

Hello,

now everything is OK. (Yes)

 

[matheus]

Hi earboth,

Thank you and the other posters for your help with my formula. I am very very grateful!

Have a great day :D