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Math Help - Fractions in multi-step equations

  1. #1
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    Fractions in multi-step equations

    Having a bit of trouble wrapping my head around the following, as I am studying on my own, after years away from school. Got the basic concept down, but was hoping someone could explain how and WHY this works in different words so I can hopefully get a firmer grasp on it.

    The first equation I have figured out pretty well, but I will explain it the way I understand it, and see if this is correct.

    3/4x-6=10

    Add six to both sides and you have:

    3/4x=16

    Now, when you multiply both sides by the reciprocal, the faction and the reciprocal will cancel each other out on the left side? Is that correct? Meaning, if 3/4 and 4/3 cancel each other out, you are left with

    x=4/3*16/1?

    If that is correct up until there, then I think I have a firm grasp on it. Someone explaining it in other words would be a help however.


    Now, the second equation I am having a bit more trouble with. It involves two factions, where the variable is the numerator in one of the fractions.

    x-8 = x/3+1/6

    Now, apparently I want to get rid of the fractions. So I would multiply both sides by the least common multiple of the denominators, coming up with the following?

    6(x-8) = 6(x/3+1/6)

    Is that correct up until there? If it is, can someone explain to me why this works? Assuming this is correct, it would leave me with:

    6x-48 = 2x+1

    Then, to finish up, isolate 4x on the left side, isolate 49 on the left side, and end up with an answer of x = 49/4

    I'm pretty sure this second equation is figured out correctly. I'm just having trouble grasping the concept of WHY multiplying by the least common multiple of all the denominators works.

    Any help you guys can provide would be awesome. It'd be even awesome-er if someone could provide me with a few problems to work out, seeing as I'm studying on my own, and don't have a ready source of problems like the second one to work with.
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  2. #2
    A Plied Mathematician
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    You are correct in your thinking about the first problem:

    \frac{4}{3}\times\frac{3}{4}=1, and

    1\times x=x for any x.

    So it does cancel out.

    For the second problem, you have

    x-8=\frac{x}{3}+\frac{1}{6}.

    The first step here is to get all the x terms over to one side, and all the other terms over to the other side thus:

    x-\frac{x}{3}=\frac{1}{6}+8.

    The next step is to add the fractions on each side. Repeat after me: adding fractions is a matter of getting a common denominator. That's your mantra. Can you continue from here?
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  3. #3
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    Geez. All this stuff that seemed so easy in middle school is now suddenly perplexing now that I'm 26 . I appreciate your answer, am rather impressed by the list of qualifications you list on your signature, and am now finally realizing what an undertaking this is going to be to compile a similar list of my own. Good thing I've finally found the dedication. Anyways, after staring blankly at my screen for about 5 minutes trying to recall everything I learned 12+ years ago, here's what I came up with. (I'll figure out how to enter the equations you did later for my future posts, which I'm sure will be numerous).

    First, in keeping with the new found mantra of "adding decimals is a matter of finding the least common denominator," we find that 6 is the LCD of 1, 3, and 6. Working out the following:

    6(x-x/3) = 6(1/6+8)

    Gives us something like:

    6x-2x = 1+48

    Simplified to:

    4x = 49

    And finally dividing both sides by 4 gives us:

    x = 49/4 or 12 1/4 or 12.25, whatever form the problem had originally asked for the answer to be written in.


    The same answer as I came up with the first time. I have a bit more insight now as to why this works. My question now becomes, is one way preferential to the other in terms of reaching this answer?
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by MDS1005 View Post
    The same answer as I came up with the first time. I have a bit more insight now as to why this works. My question now becomes, is one way preferential to the other in terms of reaching this answer?
    Either method is acceptable, but it's better if you know how to do both just in case.

    -Dan
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  5. #5
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    Hello, MDS1005!

    The technique you used on the second problem
    . . works for all equations with fractions.

    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

    QUOTE

    . . . . (3/4)x - 6 .= .10

    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~



    Multiply both sides by the LCD, 4:

    . . 4[(3/4)x - 6] .= .4(10)

    . . . . . . 3x - 24 .= .40

    . . . . . . . . . .3x .= .64

    . . . . . . . . . . .x .= .64/3



    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

    QUOTE

    . . x - 8 .= .x/3 + 1/6

    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


    Multiply both sides by the LCD, 6:

    . . 6 [x - 8] .= .6[x/3 + 1/6]

    . . .6x - 48 .= .2x + 1

    . . . . . . 4x .= .49

    . . . . . . . x .= .49/4

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  6. #6
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    Awesome. Thanks guys. I wonder if anyone could give me a problem or two to work out so that I can now see if I really do get this (which I'm pretty sure I do.)
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  7. #7
    A Plied Mathematician
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    Sure. Solve

    \frac{x}{3}+\frac{5}{7}=\frac{2x}{5}-\frac{8}{11} and

    \frac{3x}{2}+\frac{31}{3}-\frac{42}{6}=\frac{7}{8}-\frac{5x}{10}.
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  8. #8
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by MDS1005 View Post
    I wonder if anyone could give me a problem or two to work out so that I can now see if I really do get this (which I'm pretty sure I do.)
    Have a care when you ask for that kind of favor around here. You'll get a work-out.

    -Dan
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  9. #9
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    Well, I decided to take the weekend off from studying, walk away, and make sure I remembered what to do when I re-opened my workbook. Read the first problem, read Dan's comment about getting a workout, and said to myself, "how hard could this first one really be." Well, 20 minutes later, feeling like I was on the wrong track, resorting to a LCD calculator when I was doing LCD's mentally and skipped over 1155 (I got to 1320 as a factor of 11 before I decided I probably missed something), forgetting how to work the problem the way it was explained at the beginning of this thread, then going back to my original way of eliminating the fractions first, thinking the answer I got was wrong, checking my answer and seeing that it works.... Well, you get the idea, quite the workout. And I'm still not sure my answer is correct......

    X = 21.62 (rounded to the nearest hundredth)

    I wasnt going to write out all how I came to this conclusion, but I may as well so that hopefully it can be discussed, and improvements in my method can be suggested.

    First step, eliminate the fractions. Do this by multiplying both sides by the LCD, which I found to be 1155. (Here's where I will have my first question on my method.)

    1155(x/3+5/7) The way I am looking at this is as follows. How many times does the LCD divide by each denominator? Take the number of times it divides, and multiply the numerator by that number. In the case of x/3, 1155/3=385, x*385=385x. Repeated with 5/7 yields 1155/7=165, 165*5=825. So this side of the problem works out to 385x+825.

    Is my idea of dividing the LCD by the denominator, then multiplying by the numerator the correct way of thinking of this? Any suggestions on other ways to view it?

    1155(2x/5 - 8/11)...... 1155/5=231, 231*2x=462x..... 1155/11=105, 105*8=840. This side equals 462x-840

    The problem now looks like:

    385x+825 = 462x-840

    At this point, I decided I was going the the wrong direction. I decided to continue on and see what happened. Isolating x's on one side, I came up with:

    1665=77x

    1665/77 = 21.62 (rounded to nearest hundredth)

    I looked at this number, and positively decided I was wrong. For the hell of it though, I substituted 21.62 for all the x's in the original problem, rounding everything to the nearest hundredth, and I came to the result:

    7.91 = 7.92

    Close enough that I figured I'd write this long post and see what comments may come as far as correcting my methods or thinking.....
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  10. #10
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    Really hoping someone can take a look at my post one up from this one and tell me if I am correct.
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  11. #11
    Super Member Quacky's Avatar
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    Quote Originally Posted by MDS1005 View Post
    Really hoping someone can take a look at my post one up from this one and tell me if I am correct.
    You are correct; your method is fine as far as I can tell. Good example, Ackbeet.
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  12. #12
    A Plied Mathematician
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    As quacky said, good on the first one. Did you do the second? I got (only look AFTER you've done the problem - it will do you no good to look at the spoiler in advance. You're not being graded on this, for crying out loud!)

    Spoiler:
    x = - 59/48.
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