MHB Rearranging Formula, making x the subject from y=5(x+3)

[Subliminal1]

Hi, we were rearranging formula to make $$x$$ the subject and for one equation our answer was different to our lecturers, but they failed to explain why they were right and we were wrong- if someone could that would be great.

The equation with our working:

$$y=5\left(x+3\right)$$

$$\frac{y}{5}=x+3$$

$$\frac{y}{5}-3=x$$

We're told to do "the opposite" for this, so:
We isolated the variable ($$x$$) by dividing each side by factors that don't contain the variable ($$5$$).
Then it's simple by subtracting $$3$$ from $$x$$.
Replacing $$x$$ with 2 as an example to test our result:

$$y=5\left(2+3\right)$$ (or) $$y=5$$x$$5$$ (or) $$y=25$$

$$\frac{25}{5}=5$$, $$−3=2$$, $$x=2$$

Even Mathway gives the same working, so I'm a little lost. I can't remember exactly what our lecturer said the answer was, I believe it was $$y=1.2$$, but as I said they wouldn't tell us how they calculated that.
I have posted this question on a couple of other sites as it's still sat in the moderation queue on the first one we tried.

Thanks

 

[Greg]

What was the complete problem you were given, exactly as stated?

 

[Subliminal1]

Hi, the problem was literally "make $$x$$ the subject:" with a list of equations such as:

Q:$$y=x+3$$
A:$$x=y-3$$

Q:$$y=x$$x$$5$$
A:$$x=\frac{y}{5}$$

 

[I like Serena]

Hi Subliminal, welcome to MHB! ;)

Your working is all correct.
If desired, we can substitute $y=1.2$ to find the corresponding value for $x$.

 

[Subliminal1]

Hi, thanks for confirming this for me. Now to ask our lecturer point blank to explain their workings. Any tips on how to do this respectfully? ha, just joking.

Hi Subliminal, welcome to MHB! ;)

Your working is all correct.
If desired, we can substitute $y=1.2$ to find the corresponding value for $x$.

Funnily enough I was here a few years ago https://mathhelpboards.com/basic-probability-statistics-23/joint-probability-independent-events-16280-post75844.html#post75844!

I'm not actually sure what you mean here, it's been a while since I've studied maths!

Thanks again.

 

[I like Serena]

Hi, thanks for confirming this for me. Now to ask our lecturer point blank to explain their workings. Any tips on how to do this respectfully? ha, just joking.

Funnily enough I was here a few years ago https://mathhelpboards.com/basic-probability-statistics-23/joint-probability-independent-events-16280-post75844.html#post75844!

I'm not actually sure what you mean here, it's been a while since I've studied maths!

Thanks again.

Ah, my mistake. I saw someone with 4 posts that joined in September.
I just assumed that it was a couple of days ago without actually looking at the year. (Blush)

 

[Subliminal1]

It's no problem, I appreciate the hospitality here, it helps this forum stand out amongst the others!

 

[I like Serena]

Anyway, you found $\frac{y}{5}-3=x$.
This is 'just' a relation. We cannot solve it any further and find unique numbers to satisfy it.
However, for a given $y$ we can find the corresponding value of $x$ now.

 

[Subliminal1]

Sorry, I'm still not following. With my working I've replaced $$x$$ with $$1.2$$, as per our lecturers solution and it does make sense, but I'm unable to figure how the equation was simplified to get it:

$$x=1.2$$

$$y=5\left(1.2+3\right)$$

$$y=21$$

$$\frac{21}{5}-3=1.2=x$$

So, I can see it's "right", I just don't see how to work it to get that result.

Edit: I'm just confusing myself now >.<

 

[I like Serena]

Sorry, I'm still not following. With my working I've replaced $$x$$ with $$1.2$$, as per our lecturers solution and it does make sense, but I'm unable to figure how the equation was simplified to get it:

$$x=1.2$$

$$y=5\left(1.2+3\right)$$

$$y=21$$

$$\frac{21}{5}-3=1.2=x$$

So, I can see it's "right", I just don't see how to work it to get that result.

You already gave an arbitrary example with $x=2$ to verify the result.
Your lecturer picked a different example with $x=1.2$ to verify it.
Both are correct.

 

[Subliminal1]

Ok, so as we have to show our working, how would they have come to the conclusion that $$x=1.2$$?
Thanks

 

[I like Serena]

Ok, so as we have to show our working, how would they have come to the conclusion that $$x=1.2$$?
Thanks

They didn't. They just picked $x=1.2$ arbitrarily as an example. Any other value would be fine as well.

 

[Subliminal1]

Oh, it just clicked...

The entire point here is that $$x$$ can be anything, it's the method we use to find $$x$$ that is being assessed.

Sorry, I appreciate your patience!