Linear equations using addition

  1. I'm probably making a silly mistake or Wolfram Alpha is lying to me.

    Question: Find the value of c and d.

    [itex]3d=13-2c[/itex]

    [itex]\frac{3c+d}{2}=8[/itex]

    Rearranged, simplified and multiply each equation by 2:

    [tex]6d+4c=26[/tex]
    [tex]d+3c=16[/tex]

    Now find the common multiple which in my case I will use 12:

    [tex]18d+12c=78[/tex]
    [tex]-4d-12c=-64[/tex]

    Then add them and find what d is worth:

    [tex]14d=14[/tex]

    [tex]d=1[/tex]

    Now when I plug this back into the equation, I will use the first one:

    [tex]3(1)+2c=13[/tex]
    [tex]3+2(c)=13[/tex]
    [tex]c=5[/tex]

    [tex]d=1, c=5[/tex]

    What am I doing wrong? Sorry if this is the long winded way to do it.
     
  2. jcsd
  3. jbunniii

    jbunniii 3,377
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    Your answer is correct, as you can verify by plugging ##d = 1## and ##c = 5## into the two given equations.
     
  4. So Wolfram is lying to me it seems?

    Wolfram says the answer is [tex]c=\frac{35}{16}, d=\frac{23}{8}[/tex]
     
  5. jbunniii

    jbunniii 3,377
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    It seems more likely that you didn't enter the problem correctly into Wolfram Alpha.
     
  6. AlephZero

    AlephZero 7,300
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    I think you told Wolfram the second equation was
    $$3c + \frac d 2 = 8$$
     
  7. Yes, looking back that is what is shows under "Input Result" How would I input the correct format?
     
  8. Mark44

    Staff: Mentor

    If you meant this:
    $$ \frac{3c + d}{2}$$

    you should have written it as (3c + d)/2.

    Also, there was some wasted effort when you multiplied the first equation by 2. You didn't need to do that.
     
  9. micromass

    micromass 18,844
    Staff Emeritus
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    Use correct brackets.
     
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