1. ## Help!! Complex Fractions

can anybody help me?! I am really confused

1.) a-(5a/a+5) / a+(5a/a-5)

2.) (a/a-b) - (b/a+b) / (b/a-b) + (a/a+b)

3.) (a-b/a+b) + (b/a) / (a/b) - (a-b/a+b)

Thanks

2. Originally Posted by whoa
can anybody help me?! I am really confused

1.) a-(5a/a+5) / a+(5a/a-5)

2.) (a/a-b) - (b/a+b) / (b/a-b) + (a/a+b)

3.) (a-b/a+b) + (b/a) / (a/b) - (a-b/a+b)

Thanks
1) $\frac{{a - \frac{{5a}}{{a + 5}}}}{{a + \frac{{5a}}{{a - 5}}}} = \frac{{\frac{{a(a + 5) - 5a}}{{a + 5}}}}{{\frac{{a(a - 5) + 5a}}{{a - 5}}}} = \frac{{\frac{{a^2 + 5a - 5a}}{{a + 5}}}}{{\frac{{a^2 - 5a + 5a}}{{a - 5}}}} = \frac{{\frac{{a^2 }}{{a + 5}}}}{{\frac{{a^2 }}{{a - 5}}}} = \frac{{a^2 (a - 5)}}{{a^2 (a + 5)}} = \frac{{a - 5}}{{a + 5}}$

2) $\frac{{\frac{a}{{a - b}} - \frac{b}{{a + b}}}}{{\frac{b}{{a - b}} + \frac{a}{{a + b}}}} = \frac{{\frac{{a(a + b) - b(a - b)}}{{(a - b)(a + b)}}}}{{\frac{{b(a + b) + a(a - b)}}{{(a - b)(a + b)}}}} = \frac{{\frac{{a^2 + ab - ab + b^2 }}{{a^2 - b^2 }}}}{{\frac{{ab + b^2 + a^2 - ab}}{{a^2 - b^2 }}}} = \frac{{\frac{{a^2 + b^2 }}{{a^2 - b^2 }}}}{{\frac{{a^2 + b^2 }}{{a^2 - b^2 }}}} = \frac{{(a^2 + b^2 )(a^2 - b^2 )}}{{(a^2 - b^2 )(a^2 + b^2 )}} = 1$

3) $\frac{{\frac{{a - b}}{{a + b}} + \frac{b}{a}}}{{\frac{a}{b} - \frac{{a - b}}{{a + b}}}} = \frac{{\frac{{a(a - b) + b(a + b)}}{{a(a + b)}}}}{{\frac{{a(a + b) - b(a - b)}}{{b(a + b)}}}} = \frac{{\frac{{a^2 - ab + ab + b^2 }}{{a(a + b)}}}}{{\frac{{a^2 + ab - ab + b^2 }}{{b(a + b)}}}} = \frac{{\frac{{a^2 + b^2 }}{{a(a + b)}}}}{{\frac{{a^2 + b^2 }}{{b(a + b)}}}} = \frac{{(a^2 + b^2 )b(a + b)}}{{(a^2 + b^2 )a(a + b)}} = \frac{b}{a}$

3. thanks! by the way how did you get the final answer?

4. Hello, whoa!

The standard procedure for complex fractions is to multiply through by the LCD.
This allows us to cancel those annoying denominators.

$1)\;\frac{a - \frac{5a}{a+5}}{a + \frac{5a}{a-5}}$

The LCD is $(a+5)(a-5)$

. . $\frac{(a+5)(a-5)\cdot\left[a - \frac{5a}{a+5}\right]} {(a+5)(a-5)\cdot\left[a + \frac{5a}{a-5}\right] }\;= \;\frac{a(a+5)(a-5) - 5a(a-5)}{a(a+5)(a-5) + 5a(a+5)}$

Factor: . $\frac{a\cdot(a-5)\cdot[(a+5) - 5]}{a\cdot(a+5)\cdot[(a-5) + 5]} \;=\;\frac{a\cdot(a-5)\cdot a}{a\cdot(a+5)\cdot a} \;= \;\boxed{\frac{a-5}{a+5}}$

$2)\;\frac{\frac{a}{a-b} - \frac{b}{a+b}}{\frac{b}{a-b} + \frac{a}{a+b}}$

The LCD is $(a- b)(a+b)$

. . $\frac{(a-b)(a+b)\cdot\left[\frac{a}{a-b} - \frac{b}{a+b}\right]}{(a-b)(a+b)\cdot\left[\frac{b}{a-b} + \frac{a}{a+b}\right]} \;= \;\frac{a(a+b) - b(a-b)}{b(a+b) + a(a-b)}$

. . . $= \;\frac{a^2+ab - ab + b^2}{ab + b^2 + a^2 - ab} \;=\;\frac{a^2+b^2}{a^2+b^2} \;= \;\boxed{1}$

$3)\;\frac{\frac{a-b}{a+b} + \frac{b}{a}}{\frac{a}{b} - \frac{a-b}{a+b}}$

The LCD is $ab(a+b)$

. . $\frac{ab(a+b)\cdot\left[\frac{a-b}{a+b} + \frac{b}{a}\right]}{ab(a+b)\cdot\left[\frac{a}{b} - \frac{a-b}{a+b}\right]}
\;= \;\frac{ab(a-b) + b^2(a + b)}{a^2(a+b) - ab(a-b)}$
$= \;\frac{a^2b - ab^2 + ab^2 + b^3}{a^3 + a^2b - a^2b + ab^2}$

. . $= \;\frac{a^2b + b^3}{a^3 + ab^2} \;=\;\frac{b(a^2+b^2)}{a(a^2+b^2)} \;=\;\boxed{\frac{b}{a}}$

5. Ah ok. thanks............ but I still have a problem. what if the numbers are raised by negative one? here are some examples:

1.) (3x^-1 - 2y^-1) / (2y^-1 - 3x^-1)

2.) {(3a)^-1 - (2b)^-1} / {(2b)^-1 + (3a)^-1}

3.) (ab^-1 + ba^-1) / {(ab)^-1}

6. Originally Posted by whoa
Ah ok. thanks............ but I still have a problem. what if the numbers are raised by negative one? here are some examples:

1.) (3x^-1 - 2y^-1) / (2y^-1 - 3x^-1)

2.) {(3a)^-1 - (2b)^-1} / {(2b)^-1 + (3a)^-1}

3.) (ab^-1 + ba^-1) / {(ab)^-1}

General rule when having number raised by negative number is:

$\frac{1}{{a^m }} = a^{ - m} ,a \in R\backslash \{ 0\} ,m \in N$

I will show you examples from your equations for such numbers but you should try to do all this by yourself since you have been given help from Soroban and me for similar examples.

So we have $3x^{ - 1} = 3\frac{1}{x} = \frac{3}{x}$
Notice that we only raised x by -1 not 3x.

You have also $(3a)^{ - 1} = \frac{1}{{3a}}$.
Notice that we raised 3a by -1 not just a.

7. uhm... i'm not sure here's my answer:

1.) (3/x -2/y)/(2/y-3/x) = {3(y)-2(x)/xy} = (3y-2x/xy)/(2x-3y/xy)= (3y-2x/xy) * (xy/2x-3y)=////////// (3y-2x)/(2x-3y)final answer

2.) (1/3a-1/2b)/(1/2b+1/3a) = {2b-3a/(3a)(2b)}/{(3a+2b/(2b)(3a)}={2b-3a/(3a)(2b)}*{3a+2b/(2b)(3a)}= ////////-3a/3a,,, final answer

3.) (1/ab+1/ba)/(1/ab)= (1/ab)/(1/ab)= (1/ab*ab/1)=///////////1,,, final answer

hehe i'm not sure

8. Originally Posted by whoa
uhm... i'm not sure here's my answer:

1.) (3/x -2/y)/(2/y-3/x) = {3(y)-2(x)/xy} = (3y-2x/xy)/(2x-3y/xy)= (3y-2x/xy) * (xy/2x-3y)=////////// (3y-2x)/(2x-3y)final answer

2.) (1/3a-1/2b)/(1/2b+1/3a) = {2b-3a/(3a)(2b)}/{(3a+2b/(2b)(3a)}={2b-3a/(3a)(2b)}*{3a+2b/(2b)(3a)}= ////////-3a/3a,,, final answer

3.) (1/ab+1/ba)/(1/ab)= (1/ab)/(1/ab)= (1/ab*ab/1)=///////////1,,, final answer

hehe i'm not sure
1) Correct.

2) Answer is not correct. You have done it almost ok but at the end you made an error.
Solution is:
$\frac{{\frac{1}{{3a}} - \frac{1}{{2b}}}}{{\frac{1}{{2b}} + \frac{1}{{3a}}}} = \frac{{\frac{{2b - 3a}}{{6ab}}}}{{\frac{{3a + 2b}}{{6ab}}}} = \frac{{6ab(2b - 3a)}}{{6ab(3a + 2b)}} = \frac{{2b - 3a}}{{3a + 2b}}$

$\frac{{2b - 3a}}{{3a + 2b}}$ can't be further simplifed as you did to $- \frac{{3a}}{{3a}}$

3) Not correct. You made a mistake at start.

It's not $ab^{ - 1} = \frac{1}{{ab}}$ because only b is raised by -1. If it were $(ab)^{ - 1}$ then it would be $(ab)^{ - 1} = \frac{1}{{ab}}$.
Correct is $ab^{ - 1} = a\frac{1}{b} = \frac{a}{b}$.

Now, try to do 3) again.

9. wow! thanks

here's no. 3 i'm not sure again hehe

(a/b+b/a) / (1/ab)= [a(b)+b(a)/ab] / 1/ab= (ab+ba/ab) / 1/ab= (ab+ba/ab)*ab/1=///////// (ab+ba/1),,,, final answer

10. Hello, whoa!

Why don't you try it my way? . . . Or don't you understand it?

$1)\;\;\frac{3x^{-1} - 2y^{-1}}{2y^{-1} - 3x^{-1}}$

We have: . $\frac{\frac{3}{x} - \frac{2}{y}}{\frac{2}{y} - \frac{3}{x}}$ . . . . The LCD is $xy$

. . $\frac{xy\,\left(\frac{3}{x} - \frac{2}{y}\right)}{xy\,\left(\frac{2}{y} - \frac{3}{x}\right)} \;=\;\frac{\not{x }y\cdot\frac{3}{\not{x }} - x\!\!\not{y }\cdot\frac{2}{\not{y }}}{x\!\!\not{y}\cdot\frac{2}{\not{y}} - \not{x}y\cdot\frac{3}{\not{x}}} \;=\;\frac{3y - 2x}{2x - 3y} \:=\:\frac{-(2x - 3y)}{2x-3y} \;=\;-1$

$2)\;\;\frac{(3a)^{-1} - (2b)^{-1}}{2b)^{-1} + (3a)^{-1}}$

We have: . $\frac{\frac{1}{3a} - \frac{1}{2b}}{\frac{1}{2b} + \frac{1}{3a}}$ . . . . . The LCD is $6ab$

. . $\frac{6ab\left(\frac{1}{3a} - \frac{1}{2b}\right)}{6ab\left(\frac{1}{2b} + \frac{1}{3a}\right)} \;= \;\frac{\not{6 }^2\!\!\not{a}b\cdot\frac{1}{\not{3}\not{a}} - \not{6 }^3a\!\!\not{b}\cdot\frac{1}{\not{2}\not{b}}}{\not {6}^3a\!\!\not{b}\cdot\frac{1}{\not{2}\not{b}} - \not{6}^2\!\!\not{a}b\cdot\frac{1}{\not{3}\not{a}} } \;=\;\frac{2b -3a}{3a+2b}$ . . .
cannot be reduced

$3)\;\;\frac{ab^{-1} + ba^{-1}}{(ab)^{-1}}$

We have: . $\frac{\frac{a}{b} + \frac{b}{a}}{\frac{1}{ab}}$ . . . .The LCD is $ab$

. . $\frac{ab\left(\frac{a}{b} + \frac{b}{a}\right)}{ab\left(\frac{1}{ab}\right)} \;= \;\frac{a\!\!\not{b}\cdot\frac{a}{\not{b}} + \not{a }b\cdot\frac{b}{\not{a }}}{\not{a }\!\!\not{b}\cdot\frac{1}{\not{a}\not{b}}} \;=\;\frac{a^2+b^2}{1}\;=\;a^2+b^2$

11. thanks for lending your time to me you guys I'd really appreciate it! I learned a lot!

@ soroban

. . $\frac{6ab\left(\frac{1}{3a} - \frac{1}{2b}\right)}{6ab\left(\frac{1}{2b} + \frac{1}{3a}\right)} \;= \;\frac{\not{6 }^2\!\!\not{a}b\cdot\frac{1}{\not{3}\not{a}} - \not{6 }^3a\!\!\not{b}\cdot\frac{1}{\not{2}\not{b}}}{\not {6}^3a\!\!\not{b}\cdot\frac{1}{\not{2}\not{b}} - \not{6}^2\!\!\not{a}b\cdot\frac{1}{\not{3}\not{a}} } \;=\;\frac{2b -3a}{3a+2b}$ . . . [/size]cannot be reduced
why cant it be reduced?

thanks

12. ok know it already! thanks anyways