Why Do We Divide Both Sides in Formula Transposition?

  • Thread starter wewelsburg
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The "x"s cancel on the right: y+ 5= y+ x. Then you can "cancel" the y's by subtracting y from both sides: 5= x. Of course, 5= x also implies that y+ x= 5+ x= 5+ 5= 10.
  • #1
wewelsburg
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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
 
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  • #2
wewelsburg said:
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??
I don't see any division by two-- you multiply both sides by two to eliminate the factor of 1/2 premultiplying a.
Also I think I got eg2 wrong, my sense of multiplication and division is off.
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:
[tex]l=n^2ua[/tex]
[tex]\frac{l}{ua}=n^2[/tex]
[tex]n=\sqrt{\frac{l}{ua}}[/tex]
 
  • #3
Thanks Cristo.
 
  • #4
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 separate 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
wewelsburg said:
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 separate x value.
NO! x(y+ x)= x*y+ x*x= x^2+ xy, not x^2+ y.

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.
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.

Is this because there are like characters on both sides of the equation (x and y)?
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
 

Related to Why Do We Divide Both Sides in Formula Transposition?

What is the basic transposition formula problem?

The basic transposition formula problem involves solving equations by rearranging the variables to isolate the desired variable. This is often used in mathematics and science to solve for unknown values in equations.

How do I solve a basic transposition formula problem?

To solve a basic transposition formula problem, you must first identify which variable you want to isolate. Then, use the inverse operations to move all other variables to the other side of the equation. Finally, solve for the desired variable.

What are the inverse operations used in a basic transposition formula problem?

The inverse operations used in a basic transposition formula problem include addition and subtraction, multiplication and division, and exponentiation and logarithms. These operations are used to move variables to the other side of the equation.

Are there any rules or guidelines to follow when solving a basic transposition formula problem?

Yes, there are a few rules and guidelines to follow when solving a basic transposition formula problem. These include keeping the equation balanced by performing the same operations on both sides, being careful when dealing with negative numbers and fractions, and checking your work to ensure you have solved correctly.

Can basic transposition formula problems be used in real-world applications?

Absolutely! Basic transposition formula problems are commonly used in fields such as physics, chemistry, and engineering to solve for unknown values in equations. They can also be applied in everyday situations, such as calculating the cost of a sale item with a discount.

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