Rearrange Vs = U/P so that U is the subject
Given `Vs=U/P` :
`P(Vs)=U` using the multiplication property of equality
`U=P(Vs)` using the symmetric property of equality.
`(Vs)^2=U/P` square both sides
I just noticed your comment about the square root. Follow my same pattern, except that when you perform the opposite operation of the square root of U/P you must raise this to the second power.
Anytime you have roots, you must perform an exponent to cancel the root, and work your way through solving the variable.
*A square root means that you square it to cancel the root.
*A third root means that you cube it to cancel the root, and so on.
And when you square U/P to cancel the square root, don't forget to square the other side of the equation in order to balance the equation.
When you square both sides, you end up with `(Vs)^2=U/P`
Now you may focus on the P, following my same instructions
down to your answer which is `P(Vs)^2=U`
Your problem begins as Vs = U/P and you are faced with the task of making U the subject. Let's begin by rewording the problem to make it more clear. In algebra, we call this "solving" for U, which just means that we must get 'U' on one side of the equation and by itself (either side is fine, but it must be alone).
Let's look at a similar problem, just without as many variables (letters) before we begin:
2(5) = A/10 - Solve for A (make A the subject)
In order to solve for A, you must first multiply 2 and 5 ("simplifying the problem") which gives you
10 = A/10
Then to finish the problem, you simply work on getting the A by itself. This is called 'isolating the variable' (getting the letter by itself). In this case your variable - A - is being divided by 10, so you must perform the opposite of division to undue the division. Because the opposite of division is multiplication, you will multiply A/10 X 10 to leave you with 10A/10 in which the 10 will divide by 10 (cancel each other out) leaving you with 1A or simply A.
But what you do to one side of the equation, you MUST do to the other side. This is called "balancing the equation". Many students get caught on this one rule.
So in order to balance the equation, we will then multiply 10 X 10 which gives us 100.
So now we have 100 = A, which is a true statement because when you plug the value of 100 in for A you get
2(5) = 100/10 which is 10 = 100/10 which ends at 10 = 10 and is therefore a true statement, making the problem solved.
Now that we have related the question you asked down to a simpler problem, let's apply this logic to your particular problem.
You have Vs = U/P
Isolate the variable 'U'.
Because U is being divided by P and we want U by itself, we will perform the opposite operation of division by multiplying U/P X P.
This gives us Vs = U/P X P which gives us Vs = P(U)/P
Because variables follow the same rules as integers (numbers) your P's will cancel as they divide out leaving you with
Vs = U BUT you CANNOT forget to perform the same operation to both sides. Balance that equation.
When you multiply U/P X P, you must also multiply 'P' to the other side of the equation, leaving you with
P(V)(s) = U or if you want to write it the other way for more digestible reading, you may also write the answer as
U = P(V)(s)
Either way states the correct answer.
I did actually put a square root symbol before the division but the system deleted it as I posted it