Results 1 to 10 of 10

Math Help - algebraic expressions

  1. #1
    Junior Member
    Joined
    Jan 2009
    From
    in space
    Posts
    60

    algebraic expressions

    What is the proper way to work this probelm...4cd =14...c=1/2d=7
    What is the right way to wok this problem ?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie
    Joined
    Jan 2009
    Posts
    4
    Quote Originally Posted by Leona_Marie View Post
    What is the proper way to work this probelm...4cd =14...c=1/2d=7
    What is the right way to wok this problem ?
    the way i would solve it is by just replacing what we all ready know.

    4(c)(d)=14 c=1 d=7/2=3.5

    4(1)(3.5)=14

    hope that helps.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member ursa's Avatar
    Joined
    Jan 2009
    From
    Delhi
    Posts
    60
    What is the proper way to work this probelm...4cd =14...c=1/2d=7
    What is the right way to wok this problem ?
    yup darkness9375 is right
    there are 1 equation and 2 unknown variable
    we have to assume 1 variable to get the value of another variable
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Jan 2009
    From
    in space
    Posts
    60

    puzzled

    Quote Originally Posted by darkness9375 View Post
    the way i would solve it is by just replacing what we all ready know.

    4(c)(d)=14 c=1 d=7/2=3.5

    4(1)(3.5)=14

    hope that helps.
    could you break it down to me like...
    4(c)(d)
    4(1)(d)
    4(1)(7/2)
    How did you get 1/2 to =1 and why are you x 7/2 and afterwards how are you x3.5 to get =14 I thought 4x7 is 28 -1/2=14
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Jan 2009
    From
    in space
    Posts
    60

    4cd=14

    Quote Originally Posted by ursa View Post
    yup darkness9375 is right
    there are 1 equation and 2 unknown variable
    we have to assume 1 variable to get the value of another variable
    Ursa when using frac, with algebra...1st equation is 4 sec is 1/2 which is(c)third is7 or(d). So 1/2 x7/1=1x7and2x1=7/2. 2divided by 7=3 1/2=3.5...3.5x4=14 yes???
    Follow Math Help Forum on Facebook and Google+

  6. #6
    A riddle wrapped in an enigma
    masters's Avatar
    Joined
    Jan 2008
    From
    Big Stone Gap, Virginia
    Posts
    2,551
    Thanks
    11
    Awards
    1
    Quote Originally Posted by Leona_Marie View Post
    What is the proper way to work this probelm...4cd =14...c=1/2d=7
    What is the right way to wok this problem ?
    Hi Leona,

    I'm not sure that you have stated your question properly. Try and be a little clearer. Use parentheses, separate lines, anything to set apart what it is you are trying to convey.

    Looks like your original equation is 4cd=14

    Then, the second part is ambiguous.

    Dose c=\frac{1}{2}d and does \frac{1}{2}d=7?

    If that's the case, then substitute 7 for c and solve for d. But I'm not sure that's the case.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Jan 2009
    From
    in space
    Posts
    60
    Quote Originally Posted by masters View Post
    Hi Leona,

    I'm not sure that you have stated your question properly. Try and be a little clearer. Use parentheses, separate lines, anything to set apart what it is you are trying to convey.

    Looks like your original equation is 4cd=14

    Then, the second part is ambiguous.

    Dose c=\frac{1}{2}d and does \frac{1}{2}d=7?

    If that's the case, then substitute 7 for c and solve for d. But I'm not sure that's the case.
    4cd is the equation Im not sure how to use (c) which =1/2 to get the answer =14 Im guessing you x (c) which=1/2 x (d) which =7 . I just thought that you would use multiplying fractions ...1/2 x7/1 =3.5. 3.5 x 4 =14 4cd =14 please varify or change leona
    Follow Math Help Forum on Facebook and Google+

  8. #8
    A riddle wrapped in an enigma
    masters's Avatar
    Joined
    Jan 2008
    From
    Big Stone Gap, Virginia
    Posts
    2,551
    Thanks
    11
    Awards
    1
    Quote Originally Posted by Leona_Marie View Post
    4cd=14 is the equation Im not sure how to use (c) which =1/2 to get the answer =14 Im guessing you x (c) which=1/2 x (d) which =7 . I just thought that you would use multiplying fractions ...1/2 x7/1 =3.5. 3.5 x 4 =14 4cd =14 please varify or change leona
    Ok, Leona, I think you intend to state the problem this way.

    4cd=14

    c=\frac{1}{2}

    Substituting, we get

    4\left(\frac{1}{2}\right)d=14

    2d=14

    \frac{2d}{2}=\frac{14}{2}

    d=7

    Checking, we substitute the values we found for c and d back into the original equation and get:

    4\left(\frac{1}{2}\right)(7)=14

    4(3.5)=14

    14=14
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Junior Member
    Joined
    Jan 2009
    From
    in space
    Posts
    60
    Quote Originally Posted by masters View Post
    Ok, Leona, I think you intend to state the problem this way.

    4cd=14

    c=\frac{1}{2}

    Substituting, we get

    4\left(\frac{1}{2}\right)d=14

    2d=14

    \frac{2d}{2}=\frac{14}{2}

    d=7

    Checking, we substitute the values we found for c and d back into the original equation and get:

    4\left(\frac{1}{2}\right)(7)=14

    4(3.5)=14

    14=14
    Thanks p.s how do you give thanks
    Follow Math Help Forum on Facebook and Google+

  10. #10
    A riddle wrapped in an enigma
    masters's Avatar
    Joined
    Jan 2008
    From
    Big Stone Gap, Virginia
    Posts
    2,551
    Thanks
    11
    Awards
    1
    Quote Originally Posted by Leona_Marie View Post
    Thanks p.s how do you give thanks
    Click on the "Thanks" button below the post.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Algebraic expressions
    Posted in the Algebra Forum
    Replies: 7
    Last Post: March 20th 2011, 09:27 AM
  2. Dividing algebraic expressions
    Posted in the Algebra Forum
    Replies: 1
    Last Post: October 24th 2009, 12:20 PM
  3. algebraic expressions
    Posted in the Algebra Forum
    Replies: 2
    Last Post: September 3rd 2009, 05:52 PM
  4. Simplifying Algebraic Expressions
    Posted in the Algebra Forum
    Replies: 3
    Last Post: May 6th 2009, 09:28 AM
  5. algebraic expressions
    Posted in the Algebra Forum
    Replies: 4
    Last Post: January 22nd 2009, 07:04 AM

Search Tags


/mathhelpforum @mathhelpforum