Changing the subject of a formula

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Homework Help Overview

The discussion revolves around the problem of rearranging the formula p=(3-2t)/(4+t) to make t the subject. Participants explore the steps involved in this algebraic manipulation and question the correctness of their approaches.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants describe various steps taken to rearrange the formula, including multiplying both sides by (4+t), expanding brackets, and factorizing. There are questions about the notation used and whether the rearrangement is correct. Some participants suggest trying specific values for t or p to verify the results.

Discussion Status

The discussion is ongoing, with participants providing feedback on each other's notation and approaches. Some guidance has been offered regarding the use of parentheses in expressions, but there is no explicit consensus on the correctness of the final rearrangement.

Contextual Notes

There are concerns about the clarity of the notation used in the original expression and the implications of changing it after replies have been made. Participants also discuss the distinction between making t the subject and solving for t, indicating some confusion about terminology.

paulb203
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Homework Statement
Make t the subject of the formula

p=(3-2t)/(4+t)

The left hand side was given in the traditional fraction form, with numerator on top of denominator; this is just the same, yeah, with the slash?
Relevant Equations
N/A
Step 1

Mult.both sides by 4+t;

p(4+t)=3-2t

Step 2

Expand brackets;

4p+pt=3-2t

Step 3

Sub.4p from both sides;

pt=3-2t-4p

Step 4

Add 2t to both sides;

pt+2t=3-4p

Step 5

Factorise left side;

t(p+2)=3-4p

Step 6

Divide both sides by (p+2);

t=3-4p/p+2

Nb. I did try an online calculator but couldn't get it to work (probably me!)
Also, I puzzled over this for ages. Then watched a Maths Genie video (Rearranging Harder Formulae), where I learned that you could factorise, expand brackets, etc. I still don't know if my answer is correct though.
 
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Your notation for your initial and final expressions is wrong. You evaluate division before addition and subtraction, so p=3-2t/4+t means p=3+t/2. You either need to put in brackets and write p=(3-2t)/(4+t) or, better, use the LaTeX facility (see the LaTeX Guide linked below the reply box) and write $$p=\frac{3-2t}{4+t}$$

Your rearrangement seems to be correct, apart from that.
 
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Here is how to type in LaTeX:
https://www.physicsforums.com/help/latexhelp/

Whenever you write a division in a line by using ##/## or ##\div## you need definitely parentheses. It always has to be written as ##(\cdot)/(\cdot),## or even better ##((\cdot)/(\cdot)).## Any other notation is ambivalent and probably the reason why your online check went wrong. A function is usually written as ##y(x)= ...## Now you want to make ##t## the subject and ##p## the variable. So you should write
$$
x=\dfrac{3-2y(x)}{4+y(x)} \text{ or linearly } x=(3-2y(x))/(4+y(x))
$$
so that machines can distinguish between variable ##p=x## and function aka subject ##t=y(x)##.
 
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Ibix said:
Your notation for your initial and final expressions is wrong. You evaluate division before addition and subtraction, so p=3-2t/4+t means p=3+t/2. You either need to put in brackets and write p=(3-2t)/(4+t) or, better, use the LaTeX facility (see the LaTeX Guide linked below the reply box) and write $$p=\frac{3-2t}{4+t}$$

Your rearrangement seems to be correct, apart from that.
Thanks, Ibix. I've edited it accordingly.
 
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paulb203 said:
I still don't know if my answer is correct though.
Why don't you try some numbers for ##t## or ##p## and see if it works out?
 
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paulb203 said:
Homework Statement: Make t the subject of the formula

p=(3-2t)/(4+t)

The left hand side was given in the traditional fraction form, with numerator on top of denominator; this is just the same, yeah, with the slash?
Relevant Equations: N/A

Step 1

Mult.both sides by 4+t;

p(4+t)=3-2t

Step 2
. . .

t(p+2)=3-4p
Step 6

Divide both sides by (p+2);

t=3-4p/p+2

Nb. I did try an online calculator but couldn't get it to work (probably me!)
Also, I puzzled over this for ages. Then watched a Maths Genie video (Rearranging Harder Formulae), where I learned that you could factorise, expand brackets, etc. I still don't know if my answer is correct though.
I see that you changed (Edited) the original expression:

From: p=3-2t/4+t

To: p=(3-2t)/(4+t) .

Generally, you should avoid changing (Editing) a post, especially the OP, once the post has been replied to. If you do Edit, please include a notation which indicates that there was a change.

Notice that your final result also suffers from same mistake that was made originally in the initial expression. If/when you correct this, please put the correction in a new post.
 
Let me say a bit more. We start with the equation:

paulb203 said:
p=(3-2t)/(4+t)

What does this equation mean? Well, it means that whatever value ##t## takes, we can calculate ##p## using this equation.

Now, we come to the rearranged equation:
paulb203 said:
t=3-4p/p+2
This ought to say the same thing. Only this time, it's simpler to pick a value for ##p## and calculate the value for ##t##. But, of course, the pairs of values ##t, p## that satisfay the equation should be the same in both cases. If they are not, then the rearrangement must have a mistake in it.

Now, we cannot check all values of ##t, p##, but we could check a few. Such as:

From the first equation: ##t = 0 \ \Rightarrow \ p = \frac 3 4##.

And, from the second equation: ##\ p = \frac 3 4 \ \Rightarrow \ t = 0##.

You might want to try one or two more examples, just to check that everything looks okay.

The overriding point is that these are not just abstract manipulations, following some arbitrary mathematical rules. These manipulations involve essentially different forms of the same equation, that must be satisfied by the same pairs of value ##t, p## in each case. And, that can be confirmed or denied by some actual calculations.
 
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paulb203 said:
Homework Statement: Make t the subject of the formula
When you state the problem, you can call it "solving for t".
paulb203 said:
p=(3-2t)/(4+t)

The left hand side was given in the traditional fraction form, with numerator on top of denominator; this is just the same, yeah, with the slash?Step 6

Divide both sides by (p+2);

t=3-4p/p+2
Everything you did was basically right except for the last step. When you divided 3-4p by p+2, you should use parentheses to keep 3-4p together, (3-4p), and p+2 together, (p+2), That would give you t=(3-4p)/(p+2).
Some calculators make parentheses easy to type in, others do not.
 
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Paulb203, are you sure p was not intended to be defined as a function of t?
 
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  • #10
SammyS said:
I see that you changed (Edited) the original expression:

From: p=3-2t/4+t

To: p=(3-2t)/(4+t) .

Generally, you should avoid changing (Editing) a post, especially the OP, once the post has been replied to. If you do Edit, please include a notation which indicates that there was a change.

Notice that your final result also suffers from same mistake that was made originally in the initial expression. If/when you correct this, please put the correction in a new post.
Thanks, SammyS, I'll bear that in mind in future.
 
  • #11
WWGD said:
Paulb203, are you sure p was not intended to be defined as a function of t?
I don't know what 'defined as a function of t' means. I only know it literally said, "Make t the subject of the formula"
 
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  • #12
FactChecker said:
When you state the problem, you can call it "solving for t".

Everything you did was basically right except for the last step. When you divided 3-4p by p_2, you should use parentheses to keep 3-4p together, (3-4p), and p+2 together, (p+2), That would give you t=(3-4p)/(p+2).
Some calculators make parentheses easy to type in, others do not.
Thanks, FactChecker
 
  • #13
paulb203 said:
I don't know what 'defined as a function of t' means. I only know it literally said, "Make t the subject of the formula"
Ok, my bad, then it's more likely about solving for t.
 
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  • #14
PeroK said:
Why don't you try some numbers for ##t## or ##p## and see if it works out?
Thanks. I tried p=-1 and got t=7 [starting with t=(3-4p)/(2+p)]. Then plugged those into the original
p=(3-2t)/(4+t) and it seemed to work out, with -1=(-11)/11
 
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  • #15
WWGD said:
Ok, my bad, then it's more likely about solving for t.
No problem :) Although I'm not sure what the means either :)
I understand, "make t the subject of the formula" (I think). Initially p is the subject of the formula (p= etc, etc). Making t the subject just means rearranging it to, t= etc, etc, yeah?

I thought solving for t meant to come up with the value of t (?). In this case I can't work out the value of t without knowing the value of p, yeah?
 
  • #16
paulb203 said:
No problem :) Although I'm not sure what the means either :)
I understand, "make t the subject of the formula" (I think). Initially p is the subject of the formula (p= etc, etc). Making t the subject just means rearranging it to, t= etc, etc, yeah?

I thought solving for t meant to come up with the value of t (?). In this case I can't work out the value of t without knowing the value of p, yeah?
In this context, "make ##t## the subject of the formula"; "express ##t## as a function of ##p##"; and, "solve for ##t##" are different ways of saying the same thing.
 
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  • #17
WWGD said:
Paulb203, are you sure p was not intended to be defined as a function of t?
@WWGD, I think you are making an assumption here about the nature of the problem that wasn't in evidence.
WWGD said:
Ok, my bad, then it's more likely about solving for t.
"Make <something> the subject of the formula" is something of a Britishism, I believe.
paulb203 said:
I thought solving for t meant to come up with the value of t (?).
No. As @PeroK says below, it's just another way of saying "solve for a variable in terms of another."
PeroK said:
In this context, "make t the subject of the formula"; "express t as a function of p"; and, "solve for t" are different ways of saying the same thing.
Agree.
 
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