by uperkurk
 P: 159 I'm probably making a silly mistake or Wolfram Alpha is lying to me. Question: Find the value of c and d. $3d=13-2c$ $\frac{3c+d}{2}=8$ Rearranged, simplified and multiply each equation by 2: $$6d+4c=26$$ $$d+3c=16$$ Now find the common multiple which in my case I will use 12: $$18d+12c=78$$ $$-4d-12c=-64$$ Then add them and find what d is worth: $$14d=14$$ $$d=1$$ Now when I plug this back into the equation, I will use the first one: $$3(1)+2c=13$$ $$3+2(c)=13$$ $$c=5$$ $$d=1, c=5$$ What am I doing wrong? Sorry if this is the long winded way to do it.
 Sci Advisor HW Helper PF Gold P: 2,890 Your answer is correct, as you can verify by plugging ##d = 1## and ##c = 5## into the two given equations.
P: 159
 Quote by jbunniii Your answer is correct, as you can verify by plugging ##d = 1## and ##c = 5## into the two given equations.
So Wolfram is lying to me it seems?

Wolfram says the answer is $$c=\frac{35}{16}, d=\frac{23}{8}$$

HW Helper
PF Gold
P: 2,890

 Quote by uperkurk So Wolfram is lying to me it seems? Wolfram says the answer is $$c=\frac{35}{16}, d=\frac{23}{8}$$
It seems more likely that you didn't enter the problem correctly into Wolfram Alpha.
 Engineering Sci Advisor HW Helper Thanks P: 6,339 I think you told Wolfram the second equation was $$3c + \frac d 2 = 8$$
P: 159
 Quote by AlephZero I think you told Wolfram the second equation was $$3c + \frac d 2 = 8$$
Yes, looking back that is what is shows under "Input Result" How would I input the correct format?
Mentor
P: 20,933
 Quote by uperkurk Yes, looking back that is what is shows under "Input Result" How would I input the correct format?
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.
Mentor
P: 16,512
 Quote by uperkurk Yes, looking back that is what is shows under "Input Result" How would I input the correct format?
Use correct brackets.

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