Solving for An Algebraic Expression in Two Ways

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[curiousobserver]

There is a shorter version of second method. Multiply each term in second eq by 3 by using suitable part of first eq:
ax/x + by/y + cz/z = 3 * 5 =>
a+b+c = 15
(o:

[-basicmaths]

Good problem!
a=3/x, b=3/y,c=3/z
a+b+c =3/x+3/y+3/z
a+b+c=3*(1/x+1/y+1/z)=15

[ThAlEdison]

Slightly different second method:

Given
ax=ax=by=cz=3
1/x+1/y+1/z=5
Find a+b+c:
For the given equations to be true, none of a, b, c, x, y, z can equal 0. This means we are free to divide.
Notice that we know what "ax" is (3), so start by multiplying a+b+c by x
(a+b+c)x=ax+bx+cx=3+bx+cx
We also know what "by" is, so try multiplying by y

Finally, rounding things out, we know what "cz" is, so multiply by z

But we want to know a+b+c not (a+b+c)xyz, so divide both sides by xyz

[jmart]

I think that the method I used is similar to the second one.

3(1/x+1/y+1/z)=3*5=15

[paulortega]

Let U=1/X, V=1/Y, and W=1/Z. Then U+V+W = 5 and after substitution A/3 + B/3 + C/3 = 5 or (A+B+C+/3 = 5. Therefore (A+B+C)=15
Similar but I like to get away from denominators right away.

[Happy_Abe]

I enjoyed the 2nd method much more
Way more direct

[vuyyurubasant]

Instead, we can do simply as follows.
ax = 3 => a = 3/x
by = 3 => b = 3/y
cz = 3 => c = 3/z
Then
a+b+c = 3/x + 3/y + 3/z
= 3( 1/x + 1/y + 1/z)
Substitute the value of 1/x+1/y+1/z = 5 (which was given )
= 3 (5)
= 15 answer
Is it not simple 👆

[soumyadeepdas]

Take ax=by=cz=3 as eqn 1 and now divide the entire eqn. with xyz, on doin this you will get (a/yz) = (b/xz) = (c/xy) = (3/xyz). Also, yz = (xyz.a)/3, xz = (xyz.b)/3 and xy = (xyz.c)/3

Now take (1/x) + (1/y) + (1/z) = 5 as eqn 2, now on proceeding with this eqn. We get (yz + xz + xy)/xyz = 5. Now replace yz= (xyz.a)/3, xz = (xyz.b)/3 and xy = (xyz.c)/3 And now on substitution of tese values in eqn. 2 we will get (a/3) + (b/3) + (c/3) = 5. On solving we will get a+b+c =15. Just an alternate way of solving!! Thanks a lot @sybermath for bringing such fun questions every single day!!👍👍✊✊

[burlfromlondon]

I definitely preferred the 2nd method & it was the one I initially thought off at the start of the video

[raystinger]

So ax=by=cz=3
=> a = 3/x, b=3/y, c=3/z
Given 1/x + 1/y + 1/z = 5, we multiply this equation by 3:
=>3/x + 3/y + 3/z = 15
=> a + b + c = 15
I'm curious to know why this video is 6min long, it didn't take me that long to get the answer.

[miro.s]

the first method is valid if for at least 4 distinct points that are linearly independent ( = they create 3D simplex with non zero volume) with coordinates (x, y, z) the sum is 15, then 15 must be invariant regardless of points from domain.

[mdshamsulalammomin]

I solved this problem without paper and pen. a+b+c=15.
ax=by=cz=3 so x=3/a, y=3/b and z=3/c
Now 1/x=a/3, 1/y=b/3, 1/z=c/3
Again 1/x+1/y+1/z=5
Or, a/3+b/3+c/3=5
Hence a+b+c=3*5=15.

[kalyanbasak]

Good evening Sir, answer sharing a+b+c=15

[prabhudasmandal]

I thought of the 2nd method, the easy one and solved it.

[uhkfpov]

Multiply eq 2 by: abcxyz = (ax)(by)(cz) = 27
Then we get 9(a + b + c) = 5 * 27
So (a + b + c) = 15

[geraldillo]

this is easier;
a=3/x b=3/y c=3/z
a+b+c=3/x+3/y+3/z
=3(1/x+1/y+1/z)
=3*5
=15

[najimyadil]

a=3/X
b=3/y
C=3/z
1/X +1/y +1/z =5
3(1/X +1/y +1/z) =3*5
a+b+C = 15

[coolmangame]

nicee another question i was able to solve fairly quickly. I did exactly the 2nd method

[michaelempeigne]

I use the AM- HM inequality

( a + b + c ) / 3 > or = 3 / [ ( 1 / a ) + ( 1 / b ) + ( 1 / c ) ]

this means that [ ( 1 / a ) + ( 1 / b ) + ( 1 / c ) ] * ( a + b + c ) > or = 9


Now use ax = by = cz = 3 == >

[stvp]

The second method makes it more obvious why the answer is always 15–it might be more useful to younger learners