Addition of fractions with denominators that are also contain addition

[Concor]

I have been trying to solve this for a few days and i can't get anywhere. I've tried may different ways but they all end up with the equation growing out of control and me unable to see anyway to simplify after I've distributed some of the terms.
To be clear i have gone beyond the point shown in the image but then the equation becomes large and confusing so i left it there in this photo. If you think it would be helpful to see some of my previous attempts just ask.

 

[stapel]

View attachment 37142

I have been trying to solve this for a few days and i can't get anywhere. I've tried may different ways but they all end up with the equation growing out of control and me unable to see anyway to simplify after I've distributed some of the terms.
To be clear i have gone beyond the point shown in the image but then the equation becomes large and confusing so i left it there in this photo. If you think it would be helpful to see some of my previous attempts just ask.

You're on the right track, but don't forget that you're working with an equation. If you had been dealing only with an expression, then you'd be stuck with working with common denominators. But you're not! This means that you can multiply through to clear the denominators:

[imath]\qquad a(x - b)(x + c) + b(x + a)(x + c) = (a + b)(x + a)(x - b)[/imath]

It's still painful, but not quite so much.

[imath]\qquad \sout{ a(x^2 - bx + cx - bc) + b(x^2 +ax + cx + ac) = (a + b)(x^2 + ax - bc - ab) }[/imath]

[imath]\qquad a(x^2 - bx + cx - bc) + b(x^2 + ax + cx + ac) = (a + b)(x^2 + ax - bx - ab)[/imath]

[imath]\qquad \sout{ ax^2 - abc + acx - abc + bx^2 + abc + bcx + abc = ax^2 + bx^2 + a^2x -abc - a^2b + abx - b^2c - ab^2 }[/imath]

[imath]\qquad (ax^2 - abx + acx - abc) + (bx^2 + abx + bcx + abc) = ax^2 + bx^2 + a^2 x + abx - abx - b^2 x - a^2 b - ab^2[/imath]

[imath]\qquad ax^2 + bx^2 - abx + abx + acx + bcx - abc + abc = ax^2 + bx^2 + abx - abx + a^2x - b^2 x - a^2 b - ab^2[/imath]

[imath]\qquad acx + bcx = a^2x - b^2x - a^2b - ab^2[/imath]

Simplify where possible. Then gather everything onto one side of the equation, with zero on the other. You'll see that all this is, is a painful quadratic lumpy linear equation. And the Quadratic Formula can always fix that for you is entirely unnecessary.

 

[Concor]

You're on the right track, but don't forget that you're working with an equation. If you had been dealing only with an expression, then you'd be stuck with working with common denominators. But you're not! This means that you can multiply through to clear the denominators:

[imath]\qquad a(x - b)(x + c) + b(x + a)(x + c) = (a + b)(x + a)(x - b)[/imath]

It's still painful, but not quite so much.

[imath]\qquad a(x^2 - bx + cx - bc) + b(x^2 +ax + cx + ac) = (a + b)(x^2 + ax - bc - ab)[/imath]

[imath]\qquad ax^2 - abc + acx - abc + bx^2 + abc + bcx + abc = ax^2 + bx^2 + a^2x -abc - a^2b + abx - b^2c - ab^2[/imath]

Simplify where possible. Then gather everything onto one side of the equation, with zero on the other. You'll see that all this is, is a painful quadratic equation. And the Quadratic Formula can always fix that for you.

Thanks for the help. i tried again and got a bit further bit still wasn't able to solve it. When you said that the Quadratic formula could fix it for me what did you mean? also i included the working out so you could see where i am going wrong.

 

[The Highlander]

Thanks for the help. i tried again and got a bit further bit still wasn't able to solve it. When you said that the Quadratic formula could fix it for me what did you mean? also i included the working out so you could see where i am going wrong. View attachment 37166View attachment 37165

I'm afraid your attempt to move everything to the LHS and equate that to zero isn't correct.

I noticed that there were a couple of wee mistakes in the expansion of the brackets at Post #2, above, that I have corrected in red, below...

[imath]\qquad     a(x - b)(x + c) + b(x + a)(x + c) = (a + b)(x + a)(x - b)[/imath]

[imath]\qquad\implies a(x^2 - bx + cx - bc) + b(x^2 +ax + cx + ac) = (a + b)(x^2 + ax - b{\color{red}x} - ab)[/imath]

[imath]\qquad\implies ax^2 - ab{\color{red}x} + acx - abc + bx^2 + ab{\color{red}x} + bcx + abc = ax^2 + bx^2 + a^2x -abc - a^2b + abx - b^2{\color{red}x} - ab^2[/imath]

[imath]\qquad\implies (a+b)x^2 + (2ab+ac+bc)x = (a+b)x^2 + (a^2+ab-b^2)x -ab(a+b+c)[/imath]

[imath]\qquad\implies (2ab+ac+bc)x = (a^2+ab-b^2)x -ab(a+b+c)[/imath]

However (in the fourth & fifth lines I've added above), when I gather the x² terms, the x¹ terms and the x⁰ terms (which is what was suggested you do before moving everything to the LHS), I find that the x² term, \(\displaystyle (a+b)x^2\), appears on both sides (thereby cancelling each other out) and so I don't get the expected(?) quadratic in x.

I battled for some time to try and resolve the remaining terms into an expression for x in terms of \(\displaystyle a, b\) & \(\displaystyle c\) but since I had both \(\displaystyle a^2\) and  \(\displaystyle b^2\) appearing in the coefficient(s) of x, I too struggled to find any way to make x the subject of the equation with anything remotely elegant (& comprehensible?) on the other side of the equals sign!

I've probably missed some simple rationalisation (or rearrangement) somewhere but I'm sure someone else can demonstrate how to better resolve this.

 

[Dr.Peterson]

I battled for some time to try and resolve the remaining terms into an expression for x in terms of a,b & c but since I had both a² and  b² appearing in the coefficient(s) of x, I too struggled to find any way to make x the subject of the equation with anything remotely elegant (& comprehensible?) on the other side of the equals sign!

I, too find that the equation is linear in x; that gives a very simple result: x is a simple fraction involving a, b, and c. ( A key step in simplifying is to cancel (a+b); but I see no reason to be bothered that a^2 and b^2 are both present as coefficients.)

All that is needed is a fresh and careful start. I wouldn't try to find an error in the work shown, but just start over, checking every line you write. And I would definitely start as @stapel did.

 

[The Highlander]

All that is needed is a fresh and careful start. I wouldn't try to find an error in the work shown, but just start over, checking every line you write. And I would definitely start as @stapel did.

I did start with @stapel's approach but there were several errors in her expansion which I had to correct before I could proceed.

However, unfortunately, I missed one of her little slips which carried through into my subsequent algebraic manipulation and, I'm afraid the mistake I missed was compounded by me also making similar errors in my own work which meant (as you have succinctly pointed out) that it was impossible for me to reach the correct resolution of x in terms of \(\displaystyle a, b\) & \(\displaystyle c\).

I have now fixed the final mistake in the first three lines and reworked the remaining two (that were fully my responsibility) to the point where progression to a final resolution should be fairly obvious/simple...

[imath]\qquad     a(x - b)(x + c) + b(x + a)(x + c) = (a + b)(x + a)(x - b)[/imath]

[imath]\qquad\implies a(x^2 - bx + cx - bc) + b(x^2 +ax + cx + ac) = (a + b)(x^2 + ax - b{\color{red}x} - ab)[/imath]

[imath]\qquad\implies ax^2 - ab{\color{red}x} + acx - abc + bx^2 + ab{\color{red}x} + bcx + abc = ax^2 + bx^2 + a^2x -ab{\color{red}x} - a^2b + abx - b^2{\color{red}x} - ab^2[/imath]

[imath]\qquad\implies (a+b)x^2 + (a+b)cx = (a+b)x^2 + (a^2 - b^2)x -a^2b-ab^2[/imath]

[imath]\qquad\implies (a+b)cx = (a^2 - b^2)x -(a+b)ab[/imath]

I was able (from the last statement above) to go on to find the "simple fraction involving a, b, and c" that @Dr.Peterson refers to above but I won't demonstrate that here (and now) so that the OP has the chance to do at least some work of his/her own to reach the answer (unless s/he never shows up again).

@Concor: Please now come back and show us that you are able to complete the problem as stated (using the assistance that has been afforded in the forum). Thank you.


Hope that helps.

 

[Dr.Peterson]

I took another look at this and found that the work is easier if you first simplify each side separately, as fractions.

[imath]\displaystyle\frac{a}{x+a}+\frac{b}{x-b}[/imath] simplifies nicely, and then you can factor out [imath](a+b)[/imath] from both sides and eliminate it.

(This assumes it isn't zero, which wasn't stated in the problem.)

What's left is pretty easy, after you cross-multiply or clear fractions.