Equation: 1/a + 1/b + 1/x = 1/(a + b + x) for a, b not equal to zero

[Adderall]

How can i solve this ecuation ?

. . .\(\displaystyle \dfrac{1}{a}\, +\, \dfrac{1}{b}\, +\, \dfrac{1}{x}\, =\, \dfrac{1}{a\, +\, b\, +\, x},\, a,\, b\, \neq\, 0\)

a,b are not 0 and the ecuation has a finite number of solutions. Thank you very much !

 

[ksdhart2]

What are your thoughts? What have you tried? For instance, you first found a common denominator for the left hand side, added the fractions... and then what? Please comply with the Read Before Postinghttps://www.freemathhelp.com/forum/threads/54004-Read-Before-Posting thread that's stickied at the top of every sub-forum, and share with us all of the work you've done on this problem, even the parts you know for sure are wrong. Thank you.

 

[Adderall]

I tried but.. I have no idea to continue

S=Sum of solutions, I must to find S.

Thank you very much, I hope it's ok now.

 

[mmm4444bot]

I tried...

You made a good start.

Instead of combining the three abx terms into 3abx, subtract one abx from each side.

Then you will have 2abx on the left and zero on the right.

Next, consider this part on the left:

a^2x + b^2x + 2abx

If we factor out x, we get:

x(a^2 + 2ab + b^2)

Do you recognize a^2 + 2ab + b^2 as belonging to one of the special factoring patterns?

Likewise, you can group and factor the two x^2 terms.

The goal is to get a quadratic form because that can be solved by completing the square. :cool:

 

[ksdhart2]

Well, I'm not quite sure what you mean by "sum of solutions," nor am I sure why you think there are a finite number of solutions. Unless there's restrictions placed on these variables (which you didn't include in the problem statement), there are, in fact, infinitely many solutions. That aside, you're on the right track, but the way forward may not be necessarily obvious. There's some tricky factoring that can be done here. Starting from where you left off:

\(\displaystyle 2abx + a^2(b+x) + b^2(a+x) + x^2(a+b)=0\)

As I'm sure you've noticed, this particular bit of factoring didn't really help, so let's "undo" it and expand everything back out:

\(\displaystyle 2abx + a^2b + a^2x + b^2a + b^2x + x^2a + x^2b=0\)

Now if we rearrange the terms a bit, we can maybe more clearly see where to go next:

\(\displaystyle (ba^2 + ab^2) + (b^2x + ax^2) + (2abx + a^2x + x^2b)=0\)

Essentially, you'll want to use factoring by groupinghttp://www.mesacc.edu/~scotz47781/mat120/notes/factoring/grouping/grouping.html. What do you get when you continue on from here?

 

[Adderall]

Thank you, now I must to know the solutions. In the problem ennounce say that ecuation has a finite number of solutions. What this mean ?

 

[mmm4444bot]

Thank you, now I must to know the solutions.

In the last line, the factorization is not correct.

Fix this part:

(a+b)(x^2 + x + 1 + ab)

In the problem [statement, it says] that [the equation] has a finite number of solutions. What this mean ?

It means that a does not equal -b.

Otherwise, the given equation would simplify to:

1/x = 1/x

and that has an infinite number of solutions. :cool:

You have a quadratic polynomial above, so there are two solutions for x (in terms of a and b).

 

[Adderall]

I don't understand. The sum of soltions must be S=-a-b but why ? ( This is the answer from book ).

If I solve x+ab+x^2=0 x1+x2=-1.

 

[mmm4444bot]

I don't understand. The sum of soltions must be S=-a-b but why ?

Because the solutions are x=-a or x=-b.

If I solve x+ab+x^2=0 x1+x2=-1.

That quadratic polynomial is still not correct.

Check your factorization, by multiplying it out. That ought to help you find the error. :cool: