Results 1 to 5 of 5

Math Help - Simplifying expressions

  1. #1
    Newbie
    Joined
    Nov 2006
    Posts
    9

    Simplifying expressions

    Ok this is probably REALLY easy, but i have this as part of a few tests I have to do for college work, i dont take maths and havent done since school (about 5 years ago) so im really rusty on this sort of stuff so if anyone could help me out with the answers and if possible how they did it, it would be of great help!

    Ok here it goes!

    (------ means divide and (sq) means squared!)

    3x - 6
    --------
    4x - 8

    Question 2

    2a - b
    ----------
    4a(sq) - b(sq)

    Question 3

    4a - 2 x 5a(squared) - 25
    ---------- ------------------
    3a(sq) - 15 8a - 4


    Ok! any help with these is much appreciated!!!!!!!

    sorry one more!

    x - 3 - x-5
    -------- ------------
    4 3


    Thank you!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member OReilly's Avatar
    Joined
    Mar 2006
    Posts
    340
    Quote Originally Posted by jimmybob View Post
    Ok this is probably REALLY easy, but i have this as part of a few tests I have to do for college work, i dont take maths and havent done since school (about 5 years ago) so im really rusty on this sort of stuff so if anyone could help me out with the answers and if possible how they did it, it would be of great help!

    Ok here it goes!

    (------ means divide and (sq) means squared!)

    3x - 6
    --------
    4x - 8

    Question 2

    2a - b
    ----------
    4a(sq) - b(sq)

    Question 3

    4a - 2 x 5a(squared) - 25
    ---------- ------------------
    3a(sq) - 15 8a - 4


    Ok! any help with these is much appreciated!!!!!!!

    sorry one more!

    x - 3 - x-5
    -------- ------------
    4 3


    Thank you!

    1) \frac{{3x - 6}}{{4x - 8}} = \frac{{3(x - 2)}}{{4(x - 2)}} = \frac{3}{4}

    2) \frac{{2a - b}}{{4a^2  - b^2 }} = \frac{{2a - b}}{{(2a - b)(2a + b)}} = \frac{1}{{2a + b}}

    3) \frac{{4a - 2}}{{3a^2  - 15}} \cdot \frac{{5a^2  - 25}}{{8a - 4}} = \frac{{2(2a - 1)}}{{3(a^2  - 5)}} \cdot \frac{{5(a^2  - 5)}}{{4(2a - 1)}} = \frac{2}{3} \cdot \frac{5}{4} = \frac{{10}}{{12}} = \frac{5}{6}

    4) \frac{{x - 3}}{4} - \frac{{x - 5}}{3} = \frac{{3(x - 3) - 4(x - 5)}}{{12}} = \frac{{3x - 9 - 4x + 20}}{{12}} = \frac{{11 - x}}{{12}}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Nov 2006
    Posts
    9
    Wow thats great, thank you!

    Just a few more i need to complete i have

    Q1. 8p - 3q +5p - 2q + 4p

    Q2. 5x + 5y divide 4x + 4y
    ---------- --------------
    2x(sq) +6x 6x


    Q3. x + x + x
    -- -- --
    4 + 5 + 6

    Q4. 5x - 2 - 2x - 5
    -------- ---------
    10 15


    Sorry for asking so many questions

    Really appreciate the help!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,830
    Thanks
    123
    Quote Originally Posted by jimmybob View Post
    ...
    Just a few more...

    Q1. 8p - 3q +5p - 2q + 4p

    Q2. 5x + 5y divide 4x + 4y
    ---------- --------------
    2x(sq) +6x 6x


    Q3. x + x + x
    -- -- --
    4 + 5 + 6

    Q4. 5x - 2 - 2x - 5
    -------- ---------
    10 15
    Hello, Jimmybob,

    to Q1: you can only add the same variables:

    8p - 3q +5p - 2q + 4p = 8p + 5p + 4p - 3q - 2q = 17p - 5q

    to Q2: I assume that you know that division by a fraction could be rewritten as a product by the inverse fraction:
    \frac{5x+5y}{2x^2+6x} \div \frac{4x+4y}{6x} = \frac{5x+5y}{2x^2+6x} \cdot \frac{6x}{4x+4y}=.
    Now factorize and cancel out equal factors:

    \frac{5(x+y)}{2x(x+3)} \cdot \frac{6x}{4(x+y)} =\frac{15}{4(x+3)}

    to Q3: The LCM of 4, 5 and 6 is 60. That's the new denominator:
    \frac{x}{4}+\frac{x}{5}+\frac{x}{6}=x\left(\frac{1  5}{60}+\frac{12}{60}+\frac{10}{60}\right) = \frac{37}{60}x

    to Q4: The LCM of 10 and 15 is 30. This is the new denominator:

    \frac{5x-2}{10}-\frac{2x-5}{15}=\frac{3(5x-2)}{3\cdot 10}-\frac{2(2x-5)}{2 \cdot 15}= \frac{11x+4}{30}

    EB
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,860
    Thanks
    742
    Hello, jimmybob!

    Q1)\;\;8p - 3q +5p - 2q + 4p

    We can combine only 'like' terms ("apples and oranges").

    We have: . (8p + 5p + 4p) + (-3q - 2q) \;= \;17p - 5q



    Q2)\;\;\frac{5x + 5y}{2x^2+6x} \div \frac{4x + 4y}{6x}

    To divide fractions: invert (the second fraction) and multiply.

    We have: . \frac{5x + 5y}{2x^2 + 6x}\cdot \frac{6x}{4x+4y}

    Factor: . \frac{5(x+y)}{2x(x+3)} \cdot \frac{6x}{4(x+y)}

    Reduce: . \frac{5(x+y)}{\not{2}\not{x}(x+3)} \cdot \frac{\not{6}^3\not{x}}{4(x+y)} .
    and the (x+y) also cancels out

    And we have: . \frac{15}{4(x+3)}



    Q3)\;\;\frac{x}{4} + \frac{x}{5} + \frac{x}{6}

    To add or subtract fractions, they must have a common denominator.
    The LCD for this problem is 60.
    We must convert all the fractions to have 60 in their denominators.

    We have: . \frac{15}{15}\!\cdot\!\frac{x}{4} + \frac{12}{12}\!\cdot\!\frac{x}{5} + \frac{10}{10}\!\cdot\!\frac{x}{6} \:=\:\frac{15x}{60} + \frac{12x}{60} + \frac{10x}{60}

    Now they can be added: combine the numerators and "keep" the denominator.

    . . \frac{15x + 12x + 10x}{60} \;=\;\frac{37x}{60}



    Q4)\;\;\frac{5x-2}{10} - \frac{2x-5}{15}

    The LCD is 30.

    We have: . \frac{3}{3}\!\cdot\!\frac{5x-2}{10} - \frac{2}{2}\!\cdot\!\frac{2x-5}{15} \;= \;\frac{15x - 6}{30} - \frac{4x - 10}{30}

    Combine: . \frac{(15x - 6) - (4x - 10)}{30} \;= \;\frac{15x - 6 - 4x + 10}{30} \;= \;\frac{11x + 4}{30}


    Thanks for the head-up, EB!
    Last edited by Soroban; November 15th 2006 at 01:32 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Need help simplifying some expressions!
    Posted in the Algebra Forum
    Replies: 1
    Last Post: November 11th 2011, 02:49 PM
  2. Simplifying Expressions
    Posted in the Algebra Forum
    Replies: 31
    Last Post: October 22nd 2011, 07:15 PM
  3. Simplifying Expressions
    Posted in the Algebra Forum
    Replies: 1
    Last Post: September 21st 2011, 06:30 PM
  4. Simplifying expressions
    Posted in the Calculus Forum
    Replies: 10
    Last Post: November 10th 2009, 11:41 AM
  5. Simplifying log. expressions.
    Posted in the Algebra Forum
    Replies: 10
    Last Post: July 7th 2008, 09:20 PM

Search Tags


/mathhelpforum @mathhelpforum