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

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In summary, our lecturers were rearranging formulas 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.
  • #1
Subliminal1
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Hi, we were rearranging formula to make \(\displaystyle x\) 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:

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

\(\displaystyle \frac{y}{5}=x+3\)

\(\displaystyle \frac{y}{5}-3=x\)

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

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

\(\displaystyle \frac{25}{5}=5\), \(\displaystyle −3=2\), \(\displaystyle x=2\)

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 \(\displaystyle y=1.2\), 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
 
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  • #2
What was the complete problem you were given, exactly as stated?
 
  • #3
Hi, the problem was literally "make \(\displaystyle x\) the subject:" with a list of equations such as:

Q:\(\displaystyle y=x+3\)
A:\(\displaystyle x=y-3\)

Q:\(\displaystyle y=x\)x\(\displaystyle 5\)
A:\(\displaystyle x=\frac{y}{5}\)
 
  • #4
Hi Subliminal, welcome to MHB! ;)

Your working is all correct.
If desired, we can substitute $y=1.2$ to find the corresponding value for $x$.
 
  • #5
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.

I like Serena said:
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.
 
  • #6
Subliminal said:
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)
 
  • #7
It's no problem, I appreciate the hospitality here, it helps this forum stand out amongst the others!
 
  • #8
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.
 
  • #9
Sorry, I'm still not following. With my working I've replaced \(\displaystyle x\) with \(\displaystyle 1.2\), as per our lecturers solution and it does make sense, but I'm unable to figure how the equation was simplified to get it:

\(\displaystyle x=1.2\)

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

\(\displaystyle y=21\)

\(\displaystyle \frac{21}{5}-3=1.2=x\)

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 >.<
 
  • #10
Subliminal said:
Sorry, I'm still not following. With my working I've replaced \(\displaystyle x\) with \(\displaystyle 1.2\), as per our lecturers solution and it does make sense, but I'm unable to figure how the equation was simplified to get it:

\(\displaystyle x=1.2\)

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

\(\displaystyle y=21\)

\(\displaystyle \frac{21}{5}-3=1.2=x\)

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.
 
  • #11
Ok, so as we have to show our working, how would they have come to the conclusion that \(\displaystyle x=1.2\)?
Thanks
 
  • #12
Subliminal said:
Ok, so as we have to show our working, how would they have come to the conclusion that \(\displaystyle x=1.2\)?
Thanks

They didn't. They just picked $x=1.2$ arbitrarily as an example. Any other value would be fine as well.
 
  • #13
Oh, it just clicked...

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

Sorry, I appreciate your patience!
 

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

What is the purpose of rearranging a formula?

The purpose of rearranging a formula is to solve for a specific variable or make it the subject of the formula. This can help in solving equations and making predictions in scientific experiments.

How do you make x the subject in a formula?

To make x the subject in a formula, you need to isolate the variable x on one side of the equation. This can be done by using inverse operations, such as adding, subtracting, multiplying, or dividing, to both sides of the equation until x is alone on one side.

What is the first step in rearranging a formula to make x the subject?

The first step in rearranging a formula to make x the subject is to identify which operations are being performed on x. Then, use inverse operations to move those operations to the other side of the equation, isolating x on one side.

Can I rearrange a formula if there are parentheses around x?

Yes, you can still rearrange a formula if there are parentheses around x. The parentheses indicate that the operations inside should be performed first, so you would need to use the distributive property to remove the parentheses before continuing to rearrange the formula.

Are there any limitations to rearranging a formula?

There are some limitations to rearranging a formula, such as when the formula involves complex functions or multiple variables. In these cases, it may not be possible to isolate x and make it the subject of the formula. Additionally, some formulas may have multiple solutions for x, so it is important to check your work and verify that your solution is valid.

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