Linear equations using addition

[uperkurk]

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.

 

[jbunniii]

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

 

[uperkurk]

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}

 

[jbunniii]

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.

 

[AlephZero]

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

 

[uperkurk]

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?

 

[Mark44]

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.

 

[micromass]

Yes, looking back that is what is shows under "Input Result" How would I input the correct format?

Use correct brackets.