Rearranging Equations
What is the subject?
Consider the following SUVAT formula. The following formula is used to find the velocity of an object over time. In the formula V (velocity) is known as the subject of the formula. That is; VELOCITY is the subject we’re trying to solve or find.
In the formula v stands for final velocity, u stands for initial velocity, a stands for acceleration, and t stands for time.
The formula may be used used to find the velocity of an object given that you know the initial velocity of the object, its acceleration, and the time (how long it has moved for). Suppose we wanted to find time (t) in the formula, we would have to know what the value of v, u and a is.
We rearrange an equation to change the subject by making another variable the subject to be found. To find time we would have to rearrange the equation to make t the subject so that time can be found. There a few basic rules that you have to remember.
To make t the subject we have to make sure that t is the only letter left standing on one side of the equation.
We have to move u to the other side so that the term with a is left on its own. To do this we have to subtract u from both sides of the equation.
This operation should provide the following equation.
We must divide both sides by a to make sure that t is left on its own.
This operation should provide the following equation.
Lastly we rewrite the equation in a more easy readable form.
Rearranging equations with brackets
In this section we shall explore how to rearrange equations which involve brackets.
Make y the subject in the equation x=k(y+z)
There are a few basic steps that we must follow in rearranging this equation. In this example we must rearrange the equation;
There is a multiplication on one side of the equation. Since y is on the other side we have to find a way to leave it on its own. To get rid of k on the other side we must divide both sides by k.
This operation provdes the following results;
We have y one the other side but with another term or letter which we have to get rid of. To get rid of this we must subtract z from both sides.
This operation should provide the following results.
We rewrite it in an easy to approach form;
We have managed to make y the subject
Rearranging equations with fractions
This section explores how to rearrange equations with fractions.
Note that;
It is a good idea to get rid of the fractions first. We do this by multiplying both sides by x..
This shoulf provide the following results;
Since we’re trying to make x the subject we have to get rid of 5b, we do this by dividing both sides by 5b.
This operation should provide the following result.
This can be rewritten as;
Rearranging equations with quadratics
The following shows an equation which involves quadratics;
This example contains a sqaure which might make the rearranging seem complicated. But all we have to do is undo the square.
Getting rid of the fraction should simplify the equation. We can do this by multiplying both sides by e
This operation should provide the following result.
We must leave x2 on its own since we’re trying to make x the subject. To do this we must get rid of f on the left hand of the equation..
The letter x is not yet alone. We still have to get rid of a from the left hand side. To get rid of a we must divide both sides by a.
This operation should provide the following result.
To make x2 x we must square root it, this means we must square root both sides if both sides of the equal are to remain equal.
This operation should provide the following answer.
First we multiple each side with (x – b)2 to get rid of the fractions.
This operation should provide the following results.
Divide both sides with c so that you leave the term with x alone.
This operation should provide the following results.
To get rid of the square in (x – b)2 we have to square root it. What we do to one side we must also do to the other side.
This operation should provide the following result.
The letter should be the only letter left standing on one side of the equation. There is still the letter b on the right hand side of the equation. To get rid of it we must add b to both sides of the equation.
This operation should provide the following result.
Thankyou for all the help on this but im still slightly confused on what you do if you have a more basic one like this x/2=a and you have to rearrange to make x the subject. please help my teacher didnt explain it very well.
This is a good method to use when rearranging fractions. I assume the question is arranged as below;
a is the same as a/1, So you rewrite it like this. I hope you realise why.
You then “cross multiply”. You multiply x with 1 and 2 with a.
This leaves;
It does not matter if 2a comes first or x comes first. It is the same.
Is it always possible to rearrange an equation to get one term? In physics we have the equation of motion s=ut+1/2at^2. I can’t see how to rearrange that to describe t in terms of u, a and s.
break it down on how to solve for Q
5,967=[Q(3-0.3)-40,000](.85)+(4,000x.15)
Thanks you =)
what if you have y=a/(x+b)2 making x the subject
to type squared, di it like this x^2
multiply both sides by (x+b)^2
y*(x+b)^2=a
then divide both sides by y
(x+b)^2=a/y
square root both sides to get rid of the squared
x+b= root(a/y)
subtract b from both sides
x=root(a/y)-b
I still find it VERY confusing! 🙁
HELP :
B = U(N)(I) / L
Need to find U
how do u rearrange the equation y=4(3+x) to make x the subject??
Hello Lucy,
I will answer that for you. First you will need to expand the expressions so you get;
\[ y = 12 + 4x \]
Move 12 to the left hand side to get;
\[ y – 12 = 4x \]
The divide both sides by 4 to get;
\[ \frac{y – 12}{4} = x \]
It does not matter if x comes first because it is the same;
\[ x = \frac{y – 12}{4} \]
p= sqaureroot(1-m2)/m
m=?
How do i rearrange 4x = 8x-2/3 to make x the subhect??
a bit late to the party, but it’s very simple – what you do to one side, you have to do to the other. In this case, 4x = 8x-2/3
1. multiply both sides by 3 to remove the fraction
12x = 8x-2
2. you can now minus 8x from both sides so it will become:
4x = -2
I’ll let you figure out the rest 😉
0.17 = x + 2.4 (0.1 – x)
Hi Rebecca,
I am not sure if you have received the answer to your problem, but this is how I would have done it. 0.17=x+2.4(0.1-x)
1) take the last part 2.4(0.1-x) and multiply everything in the bracket with 2.4 you have;
\[ 2.4 × 0.1- 2.4x = 0.24 -2.4x \]
2) Then you put it back in your formula;
So you have;
\[ 0.17 = x + 0.24 – 2.4x \]
3) Then what I did, you collect all the terms with X;
so you have
\[ x – 2.4x =- 1.4x \]
4) Then I had \[ 0.17 = 0.24 – 1.4x \]
5) Then you move those nu without X from right to left but remember to change the sign to a negative as 0.24 is on the right positive, you will have ;
\[ 0.17 – 0.24 = – 1.4x \]
6) Then you divide both sides by 1.4 and you are left with the fraction as 1.4 ÷ 1.4 X will be cancelled and you are left with X. So you are left with X = 0.17 -0.25 ÷ 1.4. Remember this is a fraction after the equal. I am having trouble with the keyboard, doesn’t have fraction on my tab. Hopefully it is correct but that was my logic in solving that . What is yours? Maybe I am wrong, maybe you get a better answer,let me know. Bye,good luck.
Rearrange the following equation to find x in terms of a and b, in
simplest form (assume that a ‘= 4b):
a(a − x) + 4b(x − 4b) = 0.
You have to try otherwise you may not get any reply. I also think you have written the question wrong. I will start you off.
If we have to write the new expression in terms of a and b then we won’t need this “Assume that a = 4b” because we have to include a and b in the final expression when x is the subject. So I will ignore that. We’re starting with;
\[ a(a – x) + 4b(x – 4b) = 0 \]
First will need to get rid of the factors ( expand the expression ).
\[ a^2 – ax + 4bx – 16b^2 = 0 \]
We collect like terms and make sure that all terms that contain x are on one side.
\[ -ax + 4bx = 16b^2 – a^2 \]
You should be able to finish off from here.
h=bx-a/c
what is ch=bx-a
What are you trying to find?
Hello. 22/m = 9 + m. how would I go about finding m?
Hello Sean,
You have to show that you tried, so that we know where you’re stuck. I assume you have the expression as;
\[ \frac{22}{m} = 9 + m \]
I will help you get started. We must have m on its own on one side. To get rid of m on one side we must multiply both sides by m. Notice that if you do so you’ll have to divide the ms in the fraction side. Or you could use the cross multiply technique described above. When you do this you will get;
\[ 22 = m(9 + m) \]
It is very easy to continue from here.
Hi how would i go about rearranging this?To find x “0=5x^4+3x^-4”
Hello Peter,
This is not a rearranging question. You have to solve the value of x.
Hello Great Site… This helped me alot. But i still don’t understand the rules of like ‘negative or positive’. I have this sum “F=mg-kv^2” make v the subject… I keep getting answers like this… sqrt(60*250/-0.1*10)… I know the answer but have no clue how to properly do this question.
Thankyou so much
Charlie
Hello Great Site… This helped me alot. But i still don’t understand the rules of like ‘negative or positive’. I have this sum “F=mg-kv^2″ make v the subject… I keep getting answers like this… sqrt(60*250/-0.1*10)… I know the answer but have no clue how to properly do this question.
1) Add kv^2 to both sides to get F+kv^2=mg
2)Take F for kv^2 = mg-F
3) Divide by k for v^2 = (mg-F)/k
4) Square root to get:
v = sqrt((mg-F)/k)
1) Add kv^2 to both sides to get F+kv^2=mg
2)Take F for kv^2 = mg-F
3) Divide by k for v^2 = (mg-F)/k
4) Square root to get:
v = sqrt((mg-F)/k)
Helpful site.. though how do you rearrange A=(a+b/2)h so that a is the subject? Any help would be so so appreciated. Thanks!
how would you rearrange this
vo = (v2-v1)rf/r1
to make v2 the subject
Thanks very much for sharing your knowledge!! it has made maths simpler!!
Hey, this has been really helpful but I’m a bit confused; how come in the question in the comments y=4(3+x) do you expand the brackets but not in the example x=k(y+z)
hey, I’ve had mixed information on how to rearrange this equation
a=b/4-c
To find ‘b’ or find ‘c’? And is that ‘(b/4)-c’ or ‘b/(4-c)’?
BIDMAS
So (b/4)-c
If you want to rearrange for b:
Add c to both sides to get a+c=b/4
Times both sides by 4 to get either b = 4(a+c) or b = 4a + 4c
For c:
Add c to both sides to get a+c = b/4
Take away a from both sides to get c = b/4-a
Grammar error.
In the third line.
Well done.
When simplifying indicies how would you do (mn^-2)4?
And how would u make a and then d the subject of this formula: s=n/2(2a+d(n-1)?
Many thanks!
What about a(x+1)=b(x+2)
Hi,
I am hoping someone can help me rearrange the following formula so that ‘i’ is the subject. I’m struggling to figure it out!
n = (ln(FV/PV))/(ln(1+i)
Thanks!
Hello,I was wondering as to how you would solve a problem if the variable you were solving for was a numerator and not a denominator