I feel so dumb, since I've been doing this for like years, but was recently told in a case like this: -5 = 11/8(10) + b if you were to multiply every term by 8 to get rid of the fraction, that same rule wouldn't apply to the 10 being multiplied within the bracket. I have two questions;
a) Why is this the case?
b) Are there any similar scenarios like this? Does it apply to all multiplication, or just some?
#General Math Question
lime ravine
copper irisBOT
lime ravine
This is why I never reduce any equation!
icy scroll
Well
If you have ten halves of chocolate bars
And you multiply what you have by 2
How many chocolate bars do you have ?
If you want to write this mathematically, you have:
$10 \times \frac{1}{2}$ chocolate bars which can also be written (in a less natural way...)
$\frac{1}{2} (10)$
round cedarBOT
Daddy_314
icy scroll
So if you multiply all that you have by 2, how many chocolates do you have ?
If you do this logically (without using fractions) you would find you must have 10 chocolates in the end
Mathematically this translates to
$2 \times (\frac{1}{2} \times 10)
= \frac{2}{2} \times 10
= 1 \times 10
= 10$
round cedarBOT
Daddy_314
icy scroll
As you can see the "2" doesnt multiply both the fraction AND the ten, it wouldnt make sense
Another way to do it would be:
$2 \times ( \frac{1}{2} (10))
= 2 \times \frac{1}{2} \times 10
= 2 \times 10 \times \frac{1}{2}
= 20 \times \frac{1}{2}
= \frac{20}{2}
= 10$
round cedarBOT
Daddy_314
icy scroll
So the result stays the same if the 2 and 10 are first multiplied together
Or if the 2 and 1 are
This is called the commutativity of multiplication: you can do multiplication in any order you like
lime ravine
ok so what I got from this is that your conclusion is this you can work around this rule? sorry if i didn't understand
lime ravine
so it holds true for all multiplication
icy scroll
But maybe you didnt understand my example?
icy scroll
but if you have a multiplication you cant get rid of the fraction?
I dont understand what you mean by this
lime ravine
what i mean is like you cant multiply the multiplication to. geet rid of all the other fractions no matter what?
icy scroll
I only see one fraction in your example
lime ravine
in general
lime ravine
yes
icy scroll
$\frac{x}{3} = \frac{y}{2} + 1$
round cedarBOT
Daddy_314
lime ravine
yes
icy scroll
If you multiply both sides by 3, what do you get ?
lime ravine
dont you have to multiply by 6 because its the lcd
icy scroll
Lcm
lime ravine
yeah
lime ravine
if there was multiplication beside that y/2 for example like y/2(8) no matteer what you wouldnt multtiply it by 6 also?
icy scroll
If we had
$\frac{x}{3} = \frac{y}{2}(8) + 1$
round cedarBOT
Daddy_314
lime ravine
yes
icy scroll
Then we can rewrite it as
$\frac{x}{3} = \frac{y}{2} \times 8 + 1$
round cedarBOT
Daddy_314
lime ravine
yes
icy scroll
$\frac{x}{3} = \frac{y \times 8}{2} + 1$
round cedarBOT
Daddy_314
lime ravine
so it stays the same
icy scroll
No wait a bit
The magic will now happen
$\frac{x}{3} = \frac{y \times 4 \times 2}{2} + 1$
round cedarBOT
Daddy_314
round cedarBOT
Daddy_314
icy scroll
So we only need to multiply by 3
Of course you can multiply by 6 from the beginning if you didnt notice this
It wont change the result
But it will just make your computations a bit longer
icy scroll
Don't you mean y(48) ?
icy scroll
It can become y(48) if you want it to !
$6 \times\frac{x}{3} = 6 \times \frac{y}{2}(8) + 6$
$6 \times\frac{x}{3} = \frac{y}{2}(48) + 6$
round cedarBOT
Daddy_314
lime ravine
yes in that case y/2 would remain a fraction
icy scroll
Yeah but
This new form
Is uglier
Because 48/2 is an integer
And 6/3 also
So it isnt very nice
To keep it this way
It is correct
But it's not the best
It can be simplified more
lime ravine
but this is not correct right : 5 = 2/5(10) + b. so you do 25 = 2(50) + 5b
icy scroll
No, it is not
You multiplied by 5 two times
I think you are confusing these two rules:
$5(a+b) = 5a +5b$
Versus
$5ab = (5a) \times b = a \times (5b) \neq (5a)(5b)$
round cedarBOT
Daddy_314
lime ravine
oh yeah i see what you mean
because the multiplication is basically connected
so if you were to divide the multiplication and the # before it by 5 each it would be 2 times multiplied by 5 because they're connected
lime ravine
you said in that case you multiplied by 5 two times because you multipled the multiplier by 5 and the number before it by 5, which would be 2 times.
icy scroll
Yes
You can consider this example:
If you have 3 bags
Each bag has 6 candies
How many candies do you have ?
lime ravine
and that cant happen because like you demonstrated before they're basically connected so you only have to multiply by 5 once
18
icy scroll
$3 \times 6 = 18$
round cedarBOT
Daddy_314
icy scroll
Imagine I tell you: I will multiply all the candies you have by 5
So you will have $5 \times 18$
round cedarBOT
Daddy_314
icy scroll
Which is 90
This is the same as:
$5 \times ( 3 \times 6)
= (5 \times 3) \times 6
= 15 \times 6 = 90$
round cedarBOT
Daddy_314
icy scroll
OR
This is the same as:
$5 \times ( 3 \times 6)
= (5 \times 6) \times 3
= 30\times 3 = 90$
round cedarBOT
Daddy_314
icy scroll
So it is the same if:
Case 1:
Each bag has 30 candies
(You have three bags)
Case 2:
Each bag has 6 candies
But you get 15 bags instead of 3
So you either have 5 times more candies in each bag (but only get 3 bags)
Or you get 5 times more bags (each has 6 candies)
This is why we multiply by 5 either the number of candies in a bag
Or we multiply by 5 the number of bags
But not both
icy scroll
😁