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Linear equations using addition 
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#1
Mar2013, 08:49 PM

P: 159

I'm probably making a silly mistake or Wolfram Alpha is lying to me.
Question: Find the value of c and d. [itex]3d=132c[/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]4d12c=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
Mar2013, 08:58 PM

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Your answer is correct, as you can verify by plugging ##d = 1## and ##c = 5## into the two given equations.



#3
Mar2013, 09:02 PM

P: 159

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


#4
Mar2013, 09:05 PM

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Linear equations using addition



#5
Mar2013, 09:35 PM

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P: 7,279

I think you told Wolfram the second equation was
$$3c + \frac d 2 = 8$$ 


#6
Mar2013, 09:51 PM

P: 159




#7
Mar2013, 10:31 PM

Mentor
P: 21,397

$$ \frac{3c + d}{2}$$ 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. 


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