**As
most will know by now, I am a reader of Quora, the Question and
Answer platform but not a contributor. It is rather interesting in the banality that shows
up at times. Having studie****d****
mathematics in my past life I smile at some of the questions that are
asked. Mostly I ruefully smile and think to myself ‘What the *ell’?
You continually get silly formulae type questions such as – From an
unknown number 5 is deducted, multiplied by 10 and the result is 82,
what is the unknown number? Well, if you work it out which will take
all of two seconds you will know the answer is 13.2 as (x-5)10=82.
So, opening brackets to multiply the constituents 10x-50=82.
Transferring the negative to its positive side 10x= 82+50 or 10x=132.
****So
x=132/10= 13.2**

**
In
order to check you can put the answer into the formula like so –
(13.2-5)10=82 deducting and multiplying 8.2x10= 82. ****So,
correct.**

**One
of the rules that lots of people do not get is the use of the ‘equal’
sign, the =. This is where a positive number, if you move it across,
becomes negative. And of course, a negative becomes positive.
Similarly for multiplications and divisions, as one can see above.
This is how you can ‘solve’ many of these type of formulae, even
if they are more complicated. **

**There
are literally hundreds of these types of questions. But why, if
questioners do not know these most basic arithmetic questions why not
just b****u****y
a basic arithmetic book? Surely, Quora is not really the place to keep
clogging up valuable space? Ah well, let’s not be difficult about
it and let these people have their fun. At least I am not paying the
bill. **

**One
of the uses is solving simultaneous equations, which are basically
two separate formula with two or sometimes more unknowns. Like this: **

**3x+
7y= 27 and 5x+ 2y=16. So you have two unknowns x and y and two equations or formula. To solve this
we have to work on x first (not always as you may be able to use the
values of y). And the way to do this is to make the values of x
the same. So, the coefficients of x are 3 and 5 so we can multiply 3
by 5 to get 15 in both cases. Then 3x+7y becomes 15x+ 35y= 135. We
must multiply both parts by the same value and so 5x+ 2y= 15x+6y= 87.
W****e****
can ‘ignore the 15x ****for
now ****
and look at the y values. 35y and 6y. We now subtract ****6****
from ****35****=
29 and 29y= 87 ****(as
135-48 = 87)****,
y= 87/29 = 3. All that is left then is to substitute the found y
value in either of the ****formula
or as they can be called equations to find x. So, 3x+ 7y is then 3x+7x3 – 3x+
21=27, 3x= 6 and therefore x has to be 2. As 6 /3 =2. Remember a
multiplication that is moved over, the 3x, the number 3 becomes part
of the division as shown.**

**Just
a simple working and hopefully it shows how that works. If you feel
confident now you could try the simultaneous equations below:**

**5(x+2y)-(3x+11y)=14
and 7x-9y-3(x-4y)=38. Good luck, no need to tell me. But remember it
is the practice to clear brackets first! ****So
I will give you the first bracket value. You note the multiply value
is 5 therefore 5x+10y-(etc etc). Values inside the brackets are
multiplied by the value outside the brackets. Have a practice, have a
go and don’t watch that interesting TV soap opera!**

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