Help me prove easy algebra problem.

[Miike012]

Homework Statement

If x/a = y/b = z/c prove that
(x^2 +a^2)/(x+a) +(y^2+b^2)/(y+b) + (z^2+c^2)/(z+c) = ((x+y+z)^2 + (a+b+c)^2) / (x+y+z+a+b+c)

The Attempt at a Solution

Ive attempted and I have my work in paint document... I rather not recopy on here because I am to lazy to type.
I was hopeing if someone can help me I can email you the paint document... plus I have my questions highlighted in paint so it will be easier for who ever would like to help me.. I would appreciate it. thank you... Email me... my email is

englishhm@yahoo.com

 

[Delta2]

I think it would help if u set l=x/a=y/b=z/c hence x=al, y=bl, z=cl.

The right hand side will then become (l^2+1) (a+b+c)^2/(l+1)(a+b+c)=(a+b+c)*(l^2+1)/(l+1).

The left hand side will become a*(l^2+1)/(l+1) + b* (l^2+1)/(l+1) + c * (l^2+1)/(l+1). I think it is almost obvious now.

 

[Miike012]

I did that... but I get stuck at this one point...

 

[Delta2]

er mike, this point is obvious, allow me to doubt on what you have done so far and wether you stuck at this point.

 

[Miike012]

After adding them together then simplifying I came out with... ((k^2 + 1)(a+b+c))/(k+1)

Then the back of the book says that the next line is ...

((k^2(a+b+c)^2+(a+b+c)^2)/(k(a+b+c)+(a+b+c))

how did the additional (a+b+c)^2 pop up? and why did the original a+b+c in the numerator become squared ?
and why did k+1 in the denominator turn int(k(a+b+c)+(a+b+c))
?

 

[Miike012]

Sorry, I just picked up the book and started reading about 30 min ago... so I've never seen anything like this.

 

[Delta2]

Apparently your book doesn't process the right and left hand side separately (which i did and show that both sides are equal to (a+b+c)*(l^2+1)/(l+1)) but is doing a conversion from the left hand side to the right hand side. At that point it multiplies both the numerator and the denominator by (a+b+c) and continues its way to the right hand side.

 

[vish_al210]

Hi, they have just multiplied the numerator and the denominator by (a+b+c) to help simplify things.
It is lke multiplying and dividing a value or equation using a suitable value to help simplify. Since u multiply and divide using same value the actual equation does not change.

 

[Miike012]

Your right , it did not show that...
Thank you for your help...

 

[Miike012]

just curiouse... because I just finished algebra last semester... and I don't remeber doing any proofs in general... when does someone start learning proofs?

 

[vish_al210]

Well.. in India.. in our matriculation curriculam syllabus, we start doing/solving algebraic proofs.. or like the sum/problem u mentioned in about our 8th or 9th standard or in ur equivalent 9th grade I guess.. Well the age is roughly 14 - 16.
What are u referring to as proof..similar questions like the one u mentioned right??

 

[Miike012]

Yes, I am referring to the one that I showed. That is kind of sad that I have not learned any proofs, I am not 20, and you learned them when you were 15-16. education in america is not very good... I guess I will have to teach my self... which I do not mind doing.

 

[vish_al210]

I guess the human mind is capable of learning at any age..
As for the standard of education, I cannot comment on that..
for if you consider a comparison of 16yr olds, the inference may be different, but if you consider 24 yr olds then the comparison is different. Though we learn how to try and prove algebraic equations and trigonometry and analytical geometry very early, we start learning calculus at 15-16, but it is not very indepth, not all students learn indepth.. But yeah, it does expose all students (willingly//forcibly to all subjects.. presents sort of an overview.) whereas I guess u get to choose to study algebra//trigonometry//calculus when u need to and can choose to dwell more indepth into it. So don't know if you are being fair on evaluating ur system of education.. there are times when we envy ur curriculum...