Working with Ratios and Proportions

#Shorts #RatiosAndProportions via @YouTube @Apple @Desmos @NotabilityApp @googledocs @canva

[scottbadger]

a+b=cd, a+c=bd, b+c=ad
Add the lefts and rights.
a+b+a+c+b+c=cd+bd+ad
2(a+b+c)=d(c+b+a)
Divide out c+b+a
2=d

[Fantasyworld]

Owh, extremely good steep!!
Because that ratio.

a/b = a/b
(a+a)/(b+b) = 2a/2b = a/b
<that same equations>

a/b = c/d => c/d = 3a/3b
a/b = 3a/3b
(3+1)a/(3+1)b = a/b

:)

[mikehood]

I get that a, b & c are non- zero, but how can you just assert that a+b+c≠0?

[rorybrowne]

I never came across that property. Let's try it out.

So; if 3/6 = 5/10, we can deduce that 3+5 = 6+10.... which it does since it's 8/16, and each of the ratios are essentially 1/2.

So if a/b = c/d, then it can be shown that for some value k, c/d = ka/kb. To prove this for the c=ka, we calculate that c=(d/b)a, and then d=(d/b)b just cancels out so k=d/b.

therefore a+c = a + ka, and b+d is b+kb.
Therefore a+c/b+d = (a+ka) / (b+kb) = ((1+k)a) /((1+k)b) = a/b.

[trumpetbob]

I enjoy the shorts but is there a reason you are posting a bunch at a time rather than spreading them out? (I admit it could be a time zone thing, but of the many channels I'm subscribed to, your shorts appear in my feed back to back to back... and I would expect someone else to also be uploading at those times.)

[girolamocapita]

a+b=dc a+c=db b+c=da so 2(a+b+c)=d(a+b+c) so d=2

[leif]

Hello the answer is negative 1..didnt anyone else get the correct answer.?.if you cross multiply the first two equations you get (b+c)=-a and plug thst into the last equation and you get -a/a=d hence -1=d not two..surely someone else got this?