rearranging formula - why is it?

Question

need clarification on this point - when rearranging formula

make s subject of
r = 4(3-s) I will resolve this so that the "r" comes at the end of the equation on the RHS :
ie

s=12 - r/4 or

but with an x and y equation the y goes to the beginning of the RHS equation ie

make x the subject

y = 3x +5

becomes

x = y-5/3

why is it not

x = 5-y/3?


glad for any clarification

Answer 1

y = 3x +5
subtract 5 from both sides
y-5 = 3x + 5 - 5 {to isolate 3x}
y-5 = 3x + 0 = 3x
now multiply both sides by 1/3 {to get x}
1/3(y-5) = 1/3(3x) = 1*x = x
you have to be careful with what operations you are performing

Answer 2

You seem to have memorised a rule of some kind for switching formulae round. I feel you should work from first principles in questions of this type.

You need to be clear in your mind about expressions such as a - b/c. When written on a single line without brackets, this means a - (b/c), as multiplication and division take precedence over addition and subtraction.

It is not equivalent to a fraction written normally with numerator a - b and denominator b. To write that on a single line, you must write (a - b)/c.

When a fraction is written in the normal 'double-deck' arrangement, the long bar line in the middle brackets together terms in the numerator and those in the denominator.

A fraction with a - b on top and c + d underneath, written on a single line is:
(a - b) / (c + d)
and not
a - b / c + d
which really means
a - (b/c) + d.

You will see many questions written wrongly on this site. Answerers often have to guess where the brackets should be, based on familiarity with questions of the type being asked. Once in a while, they interpret the question wrongly, but not through any fault of their own.

This is how I suggest you switch round your formulae:

r = 4(3 - s)
Divide each side by 4:
r / 4 = 3 - s
Add s to each side:
s + r/4 = 3
Subtract r/4 from each side:
s = 3 - r/4
or
s = (12 - 4r) / 4 if you want the result in a style similar to that of the next question.

y = 3x + 5
Subtract 5 from each side:
y - 5 = 3x
Divide by 3:
x = (y - 5) / 3

Answer 3

It does not matter is the variable goes first and constant go last. It is a matter of taste.
so, if y = 3x + 5
or 3x = y - 5
x = y/3 - 5/3

It is the same as x = -5/3 + y/3.

In this case, s = 12 - r/4
it is the same equation as s = -r/4 + 12.

Answer 4

there is one simple rule for rearranging ur variables i.e. ur values and expression should remain the same
let me show u steps for ur question
r=4(3-s)
now to solve it either u can div. both the RHS and LHS by 4 or
u can time the whole bracket with 4
let times it
r=12-4s
now as u want s by itself
then take 4s to the other side by adding both side by 4s
r+4s=12-4s+4s
now same with the r sub. r both the sides
4s=12-r
divide both the side by 4
s=(12-r)/4

now with ur second example
y=3x+5
sub 5 both the side
y-5=3x
and now divide both the side with 3
(y-5)/3=x
which is same as
x=(y-5)/3

hope this helped u

Answer 5

There are some good and bad answers here. The answers - in there simplest terms - are

s=3-(r/4)

and

x=(y-5)/3

Be careful with your brackets because y-5/3 is not the same as (y-5)/3. It reminds you which operations are to be done first.

If you need any more help email me!

Answer 6

y = 3x + 5
3x = y - 5

x = y/3 - 5/3 or (y - 5) / 3 is what you get.

In your first one, r = 4(3-s); r = 12 - 4s

-4s = r - 12

s = -r/4 + 3