Results 1 to 7 of 7

Math Help - Basic Linear Algebra Help

  1. #1
    Junior Member
    Joined
    Oct 2012
    From
    Australia
    Posts
    26

    Basic Linear Algebra Help

    5(x+3)/6=4+3(x-1)/5 ANSWER: x = 27/7. Could someone please show me the steps as to how this final answer was derived? Thank you.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member PaulAdrienMauriceDirac's Avatar
    Joined
    Oct 2012
    From
    T'bilisi
    Posts
    33
    Thanks
    1

    Re: Basic Linear Algebra Help

    \frac{5(x+3)}{6}=4+\frac{3(x-1)}{5}

    \frac{5(x+3)}{6}-\frac{3(x-1)}{5}=4

    \frac{25(x+3)-18(x-1)}{30}=4

    25x+75-18x+18=120

    7x+93=120

    7x=27

    x=\frac{27}{7}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Oct 2012
    From
    Australia
    Posts
    26

    Re: Basic Linear Algebra Help

    Quote Originally Posted by PaulAdrienMauriceDirac View Post
    \frac{5(x+3)}{6}=4+\frac{3(x-1)}{5}

    \frac{5(x+3)}{6}-\frac{3(x-1)}{5}=4

    \frac{25(x+3)-18(x-1)}{30}=4

    25x+75-18x+18=120

    7x+93=120

    7x=27

    x=\frac{27}{7}
    How did you get the number '75'? That's the part I'm missing. Is it just 25*18/6? Also, how did you get the figure "120" and where did the '7x' come from in the final simplification? Thanks!
    Last edited by MathClown; November 2nd 2012 at 06:48 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member PaulAdrienMauriceDirac's Avatar
    Joined
    Oct 2012
    From
    T'bilisi
    Posts
    33
    Thanks
    1

    Re: Basic Linear Algebra Help

    Quote Originally Posted by MathClown View Post
    How did you get the number '75'? That's the part I'm missing. Is it just 25*18/6? Also, how did you get the figure "120" and where did the '7x' come from in the final simplification? Thanks!
    Sorry, I kind of skipped some parts.

    \frac{5(x+3)}{6}=4+\frac{3(x-1)}{5}

    Now, we have to movie the terms that have variable x in them, to try to isolate it, so we subtract both sides by \frac{3(x-1)}{5}:

    \frac{5(x+3)}{6}-\frac{3(x-1)}{5}=4

    Next we find common denominator of these two fractions, which is 30:

    \frac{25(x+3)-18(x-1)}{30}=4

    Next we distribute 25(x+3)=25x+25*3=25x+75 and -18(x-1)=-18x-18(-1)=-18x+18, and we multiply both sides by 30 to get rid of the denominator on the left side and we get 120 on the right side by multiplying 30 to 4.

    25x+75-18x+18=120

    Next we subtract 93 to both sides.

    7x+93=120

    7x=27

    ...and finally, divide by 7 to both sides.

    x=\frac{27}{7}

    I hope this is more clear.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,393
    Thanks
    1327

    Re: Basic Linear Algebra Help

    Quote Originally Posted by MathClown View Post
    5(x+3)/6=4+3(x-1)/5 ANSWER: x = 27/7. Could someone please show me the steps as to how this final answer was derived? Thank you.
    \frac{5(x+ 3)}{6}= 4+ \frac{3(x- 1)}{5}
    Athough it is really the same thing, instead of getting a common denominator I would multiply through by 30:
    (30)(5)(x+ 3)/6= 4(30)+ (30)(3)(x- 1)/5 and, since 30/6= 5 and 30/5= 6,

    (5)(5)(x+ 3)= 4(30)+ 6(3)(x- 1)

    25(x+ 3)= 120+ 18(x- 1).

    Now we apply the "distributive law": a(b+ c)= ab+ ac. That is, 25(x+ 3)= 25x+ 25(3) and 18(x- 1)= 18x- 18(1). Of course, 25(3)= 75 so this becoes 25x+ 75= 120+ 18x- 18= (120- 18)+ 18x= 102+ 18x. Now, subtract 18 from both sides and subtract 75 from both sides to get
    25x- 18x= 7x= 102- 75= 27. The last step is to divide both sides by 7: (7x)/7= x= 27/7.
    Last edited by HallsofIvy; November 2nd 2012 at 07:12 AM.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,317
    Thanks
    697

    Re: Basic Linear Algebra Help

    here is how i would do it:

    \frac{5(x+3)}{6} = 4 + \frac{3(x-1)}{5}

    first i would get the 2nd side to "be one fraction", which means changing 4 to (1)(4) = (5/5)(4) = 20/5:

    \frac{5(x+3)}{6} = \frac{20}{5} + \frac{3(x-1)}{5}

    now we can add the numerators together, and we have:

    \frac{5(x+3)}{6} = \frac{3(x-1) + 20}{5}

    since 3(x-1) = 3x - 3, 3(x-1) + 20 = 3x - 3 + 20 = 3x + 17 (i just took the 3 from the 20: the fewer terms, the better)

    \frac{5(x+3)}{6} = \frac{3x - 17}{5}

    now, "cross-multiply": if a/b = c/d, then ad = bc. why does this work?

    b(a/b) = b(c/d) (because a/b = c/d, so multplying them both by b still preserves the equality)

    ab/b = bc/d...but ab/b = (a/1)(b/b) = (a/1)(1) = a/1 = a. so:

    a = bc/d, and multplying both sides by d:

    ad = (bc/d)d = (bc/d)(d/1) = (bc)(d/d) = (bc)(1) = bc, which is what i said was true up there a few lines ago.

    so when we cross-multiply, we get:

    5(5(x+3)) = 6(3x - 17) look, no fractions! (i HATE fractions, every sane person should).

    we can simplify the left side a bit, before we continue:

    5(5(x+3)) = 5(5x+15) = 25x + 75

    and we can also simplify the right side, as well:

    6(3x + 17) = 18x + 102 (i used a calculator to calculate 6*17...so sue me)

    so now we have:

    25x + 75 = 18x + 102 this is looking better already.

    subtracting 18x from both sides gives us:

    7x + 75 = 102

    subtracting 75 from both sides gives us:

    7x = 27

    and finally dividing both sides by 7 gives us:

    x = \frac{27}{7}

    it should comfort you that 3 different people using 3 different methods all give the same answer. we're all probably right. how do you make sure?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Oct 2012
    From
    Australia
    Posts
    26

    Re: Basic Linear Algebra Help

    Thanks for all your help, guys! I will no doubt stumble across a similar question and once again ask for help. :S I've repped everyone in this thread for his useful help.
    Last edited by MathClown; November 2nd 2012 at 10:43 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Basic Linear Algebra Problem
    Posted in the Advanced Algebra Forum
    Replies: 6
    Last Post: January 14th 2011, 11:20 AM
  2. Basic Linear Algebra - Linear Transformations Help
    Posted in the Advanced Algebra Forum
    Replies: 6
    Last Post: December 7th 2010, 03:59 PM
  3. Basic Linear Algebra question
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: May 22nd 2009, 05:09 PM
  4. basic linear algebra questions
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: April 18th 2009, 06:17 AM
  5. Basic Linear Algebra Proof
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: January 21st 2008, 01:01 PM

Search Tags


/mathhelpforum @mathhelpforum