# Changing the subject of a formula

• paulb203
paulb203
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

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.

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.

paulb203
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)##.

e_jane and paulb203
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.

Ibix
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?

paulb203 and Ibix
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.

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.

paulb203
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.

Last edited:
paulb203
Paulb203, are you sure p was not intended to be defined as a function of t?

paulb203
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.

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"

WWGD
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

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"

paulb203
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

PeroK
WWGD 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?

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.

paulb203
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:
"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.

paulb203, e_jane and PeroK

## What does "changing the subject of a formula" mean?

Changing the subject of a formula means rearranging the formula so that a different variable is isolated on one side of the equation. Essentially, it involves solving the equation for a different variable than the one originally isolated.

## Why is changing the subject of a formula important?

Changing the subject of a formula is important because it allows you to express relationships between variables in different ways, making it easier to solve problems and understand the underlying relationships in mathematical and scientific contexts.

## What are the basic steps to change the subject of a formula?

The basic steps to change the subject of a formula include: identifying the variable you want to isolate, using inverse operations to move other terms to the opposite side of the equation, and simplifying the equation as needed. This process may involve addition, subtraction, multiplication, division, and other algebraic manipulations.

## Can you provide an example of changing the subject of a formula?

Sure! Consider the formula for the area of a rectangle, A = l * w, where A is the area, l is the length, and w is the width. To change the subject to solve for the width (w), you would divide both sides by the length (l): w = A / l.

## Are there any common mistakes to avoid when changing the subject of a formula?

Common mistakes include not applying inverse operations correctly, forgetting to apply operations to all terms in the equation, and making arithmetic errors. It's also important to maintain the balance of the equation by performing the same operation on both sides.

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