Simplifying An Algebraic Expression | Two Ways

[@seanfraser3125]

Let S be the value of the second expression; that is, the value we want to find.
Add 2 times the first equation to the second equation. With some rearranging we have
1/(1+2x) + 2x/(1+2x) + 1/(1+2y) +
2y/(1+2y) + 1/(1+2z) + 2z/(1+2z)
= 2 + S
Combine like terms:
(1+2x)/(1+2x) + (1+2y)/(1+2y) + (1+2z)/(1+2z) = 2 + S
All three terms on the left-hand side are 1, so we have 3 = 2 + S. Thus S = 1.

[@musicsubicandcebu1774]

The best method so far!

[@lifeisajourney4340]

We can set 1+2x=a or x = (a-1)/2
1+2y=b or y = (b-1)/2
1+2z=c or z = (c-1)/2
substituting x, y, z in the eqn we get
x/a + y/b + z/c =1
(a-1)/2a + (b-1)/2b + (c-1)/2c = 1 (Multiply both sides by 2)
(a-1)/a + (b-1)/b + (c-1)/c = 2
Now (a-1)/a = 1 - 1/a and so on
so we get 3 - 1/a - 1/b - 1/c = 2
and 1 = 1/a +1/b +1/c

[@thorinpalladino2826]

1 is obviously an answer. The tricky part is showing that it is the only answer.

[@falknfurter]

Eliminating the fractions in both equations also works nicely: (1) Multiplying the first equation with 2(1+2x)(1+2y)(1+2z) yields 2x(1+2y)(1+2z) + 2y(1+2x)(1+2z) + 2z(1+2x)(1+2y) = 2(1+2x)(1+2y)(2+2z). Although this looks like a monster, it can be tamed after you (2) multiply the second equation with (1+2x)(1+2y)(1+2z). The second equation becomes (1+2y)(1+2z)+(1+2x)(1+2z)+(1+2x)(1+2y) = A(1+2x)(1+2y)(1+2z). (3) Taming these monster expressions is done by adding both equations together. The result is 3(1+2x)(1+2y)(1+2z) = (2+A)(1+2x)(1+2y)(1+2z). The parenthesis with x,y and z cancel out and we have the result 3 = 2+A and finally A=1.

[@quark67000]

Directly used your second method (before seeing the video), which is very direct and elegant.

[@mohitg196]

I tried a guess and check method and I putted the values for x,y,z=1 and same for the second equation and I got the 2nd equation equal to 1.

[@shoshosalah3447]

That's same I made ❤

[@mehulpunia6174]

please try this :
If aα^2+bα+c=5α^2+3α
aβ^2+bβ+c=5β^2+3β
& aγ^2+bγ+c=5γ^2+3γ
then find the value of a,b,&c

[@diwakar500]

a=5,b=3,c=0

[@razin4419]

3th method:
I added -1-1-1=-3 to each terms and I found 1 also.

[@ManjulaMathew-wb3zn]

Or you can multiply the original equation by 2 then subtract 1 from each fraction. Result would be
-A=2-3
A=1

[@ManjulaMathew-wb3zn]

Third method here.
Let A= the required function and subtract each fraction from 1
3-A=1-1/(2x+1) + 1-1/(2y+1) + 1-1/(2z+[)
3-A=2(LHS of the first equation =1)
3-A=2*1
A=1

[@gdmathguy]

I did the second method but I MISCALCULATED IT and it became -0.5 instead as a solution so that's annoying

[@rakenzarnsworld2]

Let's put x = 1, y = 1, z = 1
1/3 + 1/3 + 1/3 = 1
Answer: 1

[@broytingaravsol]

crazy, i've ever been to the answer but skipped

[@yakupbuyankara5903]

1.

[@mathswan1607]

1

[@mega_mango]

ez

[@stevemonkey6666]

1