# Homework Help: Basic transpostion formula problem

1. Jul 4, 2007

### wewelsburg

Hi,
formula transposition is basic but I have trouble in doing it sometimes. eg1 is the correct answer but why do we divide both sides by two??
Also I think I got eg2 wrong, my sense of multiplication and division is off.

S=UT+1/2at^2 make a the subject
* Deduct UT from both sides S-UT=1/2AT^2
*multiply both sides by 2 2(S-UT)=AT^2
*Divide both sides by t^2 A=2(S-UT)/T^2

EG2
L=N^2 X UA/ 1 make N the subject of the formula.
*divide both sides by UA N^2/1=L/UA
*multiply 1 N^2=L/UA X 1
*square the equation N=square root of L/UA X 1

2. Jul 4, 2007

### cristo

Staff Emeritus
I don't see any division by two-- you multiply both sides by two to eliminate the factor of 1/2 premultiplying a.
2. looks fine to me. However, you have factors of 1 floating around which really don't need to be written. So, I would do 2. as follows:
$$l=n^2ua$$
$$\frac{l}{ua}=n^2$$
$$n=\sqrt{\frac{l}{ua}}$$

3. Jul 4, 2007

### wewelsburg

Thanks Cristo.

4. Jul 6, 2007

### wewelsburg

Also, I have another transposed formulae with the right working and answer, but I've misunderstood some details.

Find Y
y/(y + x)+ 5=x
a) Multiply each side by y+x to get rid of the fraction
y+5(y+x)=xy+x^2
b) Multiply out of brackets
y+5y+5x=xy+x^2
c) Get all y terms on the left hand side of equation by subtracting xy
6y-xy+5x=x^2
d) Subtract 5x
6y-xy=x^2-5x
e) Divide 6-x to isolate y for final answer
y=x^2-5x/6-x

I don't understand step b) where we get a value of x^2. Because x(y+x) would give us x^2 + y, not another seperate x value.
Also in all the tanspositions I have done so far if you use a figure on one side of the equation it cancels out on that sideof the equation (eg V=IR x R would cancel R out RHS) but in step a this doesn't happen. Is this because there are like characters on both sides of the equation (x and y)???

5. Jul 6, 2007

### HallsofIvy

NO! x(y+ x)= x*y+ x*x= x^2+ xy, not x^2+ y.

I don't understand what you mean by "use a figure on one side of the equation". If you want to solve V= IR for I, you must divide both sides of the equation by R: V/R= (IR)/R and now the two R's cancel:V/R= I.

It certainly did "happen". Multiplying both sides of the equation by x+ y "cancels" the denominator on the left: (y/y+x)(y+x)= y