# Help!! Complex Fractions

• Nov 13th 2006, 03:21 AM
whoa
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 :D
• Nov 13th 2006, 10:56 AM
OReilly
Quote:

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 :D

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

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

3) $\displaystyle \frac{{\frac{{a - b}}{{a + b}} + \frac{b}{a}}}{{\frac{a}{b} - \frac{{a - b}}{{a + b}}}} = \frac{{\frac{{a(a - b) + b(a + b)}}{{a(a + b)}}}}{{\frac{{a(a + b) - b(a - b)}}{{b(a + b)}}}} = \frac{{\frac{{a^2 - ab + ab + b^2 }}{{a(a + b)}}}}{{\frac{{a^2 + ab - ab + b^2 }}{{b(a + b)}}}} = \frac{{\frac{{a^2 + b^2 }}{{a(a + b)}}}}{{\frac{{a^2 + b^2 }}{{b(a + b)}}}} = \frac{{(a^2 + b^2 )b(a + b)}}{{(a^2 + b^2 )a(a + b)}} = \frac{b}{a}$
• Nov 13th 2006, 01:02 PM
whoa
thanks! by the way how did you get the final answer?
• Nov 13th 2006, 01:07 PM
Soroban
Hello, whoa!

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

Quote:

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

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

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

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

Quote:

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

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

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

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

Quote:

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

The LCD is $\displaystyle ab(a+b)$

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

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

• Nov 14th 2006, 01:14 AM
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}

:confused:
• Nov 14th 2006, 02:49 AM
OReilly
Quote:

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}

:confused:

General rule when having number raised by negative number is:

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

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

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

• Nov 14th 2006, 03:23 AM
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;)
• Nov 14th 2006, 03:56 AM
OReilly
Quote:

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:
$\displaystyle \frac{{\frac{1}{{3a}} - \frac{1}{{2b}}}}{{\frac{1}{{2b}} + \frac{1}{{3a}}}} = \frac{{\frac{{2b - 3a}}{{6ab}}}}{{\frac{{3a + 2b}}{{6ab}}}} = \frac{{6ab(2b - 3a)}}{{6ab(3a + 2b)}} = \frac{{2b - 3a}}{{3a + 2b}}$

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

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

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

Now, try to do 3) again.
• Nov 14th 2006, 04:07 AM
whoa
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
• Nov 14th 2006, 04:46 AM
Soroban
Hello, whoa!

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

Quote:

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

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

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

Quote:

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

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

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

Quote:

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

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

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

• Nov 14th 2006, 05:22 AM
whoa
thanks for lending your time to me you guys:) I'd really appreciate it! I learned a lot!

@ soroban

Quote:

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

thanks
• Nov 15th 2006, 02:04 AM
whoa
ok know it already! thanks anyways:)