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Solving Linear Equations (fractions)

  1. Feb 17, 2012 #1
    I'm taking a step back as I've found a concept I'm struggling with. If I have an equation:

    2(x-1)/3 = (x/4)+1

    I've been told I need to multiply by 12 to give me:

    8(x-1) = 3x+12

    However I'm struggling with the multiplication of fractions in his instance. Can anyone advise me how that calculation works? I understand the remainder of the working. Thanks :)
  2. jcsd
  3. Feb 17, 2012 #2


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    Do you remember the rules of fractions?

    If you have some fraction, say, [itex]\frac{2}{3}[/itex] and you multiply it by 9, there are various ways to represent the same number. You could have:

    [tex]=9\cdot \frac{2}{3}[/tex] (note, the dot just means multiply, as x could be confused with a variable)

    [tex]=\frac{9\cdot 2}{3}[/tex]


    [tex]=\frac{9}{3}\cdot 2[/tex]

    [tex]=3\cdot 2[/tex]


    Now notice the expression [itex]\frac{9}{3}\cdot 2=3\cdot 2[/itex]. We essentially cancel out the denominator this way, which is what you're trying to do in your problem.

    So why do we multiply your question by 12 and not just any other number? Well, what we're looking for is something called the Lowest Common Denominator (LCD) of 3 and 4. What this means is that we want the first number that both 3 and 4 multiply into.

    Multiples of 3:
    3, 6, 9, 12, 15, 18, 21, 24, ...

    Multiples of 4:
    4, 8, 12, 16, 20, 24, ....

    Notice the common multiples are 12 and 24. But we want the LCD which would be 12.

    So now when you multiply [itex]\frac{2(x-1)}{3}[/itex] by 12, you get [tex]12\cdot \frac{2(x-1)}{3}=\frac{12}{3}\cdot 2(x-1)=4\cdot 2(x-1)=8(x-1)[/tex]

    and if you multiply the other side by 12, you get [tex]12\left(\frac{x}{4}+1\right)=\frac{12x}{4}+12=3x+12[/tex]

    Just remember the common rules of fractions:

    [tex]a\cdot\frac{b}{c}=\frac{ab}{c}=\frac{a}{c}\cdot b=a\cdot b\cdot\frac{1}{c}[/tex]

  4. Feb 18, 2012 #3
    That's great thank you :) Following on from that I know that Y/x1 = y/x2 cancels down to give x1 = x2 but can you explain why?
  5. Feb 18, 2012 #4

    I like Serena

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    Can you multiply the left side of the equation with x1 and simultaneously also multiply the right side with x1?

    After that multiply left and right with x2.

    And finally divide left and right by y.
  6. Feb 18, 2012 #5


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    What I like Serena said.

    Also, you can think of it logically. If [tex]\frac{2}{x}=\frac{2}{y}[/tex] then for these to be equal, don't x and y need to be the same?
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