Results 1 to 4 of 4

Math Help - Am I doing This correctly?

  1. #1
    Newbie
    Joined
    Jun 2008
    Posts
    6

    Am I doing This correctly?

    5x + 6y = 18 solve for y

    5x - 5x + 6y = 18 + 5x
    6y = 18 + 5x
    6y/6 = 18 + 5x/6
    y = 3 + 5x

    This has 2 variables, do I need to solve for x or will y=3+5x work out?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Feb 2008
    From
    Westwood, Los Angeles, CA
    Posts
    176
    Thanks
    1
    When the problem says "solve for y," the intent is simply to get y by itself.

    You're on the right track, but:

    5x + 6y = 18

    5x - 5x + 6y = 18 - 5x (here, you added 5x to the right hand side)

    6y = 18 - 5x

    6y/6 = (18-5x)/6

    y = 3 - 5x/6 (When you divide by 6, you have to divide EACH TERM by 6 - this means you need to do 18/6 AND 5x/6)

    So, you were on the right track, just a couple small things.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jun 2008
    Posts
    6
    Thank you, a simple + or - can break me I now know this.

    So when I plug in everything should it look like this?

    5x + 6y = 18 solve for y
    5x - 5x + 6y = 18 - 5x
    6y = 18 - 5x
    6y/6 = 18 + 5x/6
    y = 3 - 5x/6

    5x + 6(3 - 5x/6) = 18
    5x + 18 - 30x/36 = 18
    5x + 18 - 18 - 30x/36 = 18 -18
    5x - 30x/6 = 0
    30x/6 - 30x/6 = 0
    x = 0

    5(0) + 6y = 18
    0 + 6y = 18
    6y = 18
    6y/6 = 18/6
    y = 3

    (0,3)

    I will be so happy if I'm on the right trail.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Feb 2008
    From
    Westwood, Los Angeles, CA
    Posts
    176
    Thanks
    1
    Not quite. When you plug it back in, you get:

    5x + 6(3 - 5x/6) = 18

    5x + 18 - 5x = 18 (Note that [tex]6 \times \frac{5x}{6} = \frac{30x}{6} = 5x[\math].

    18 = 18.

    This means you've done it right - if you ended up with something like 12 = 18, you'd know you made a mistake.

    The problem is NOT asking you to find numbers for x and y. The only thing you have to do when it says "solve for y" is to get y by itself and everything else on the other side. Your final answer is:

    y = 3 - \frac{5x}{6}.

    That's it!

    Note that this does describe a relationship between x and y, and (0,3) does satisfy this relationship. However, note that there are an infinite number of combinations of x and y that work. If you put in, say, 6 for x, you get:

    y = 3 - \frac{30}{6} = 3 - 5 = -2

    so you get the point (6, -2).

    If you were to graph ALL the points (there are an infinite number of them) that satisfy your equation, you'd get a straight line.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Did I do this correctly?
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: January 30th 2010, 10:13 PM
  2. did i do this correctly?
    Posted in the Calculus Forum
    Replies: 2
    Last Post: September 20th 2009, 04:17 PM
  3. did i do this correctly?
    Posted in the Calculus Forum
    Replies: 2
    Last Post: August 27th 2009, 05:26 PM
  4. Is this done correctly?
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 16th 2007, 04:03 PM
  5. Is this done correctly?
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 2nd 2007, 06:38 PM

Search Tags


/mathhelpforum @mathhelpforum