Linear equations addition

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

Question: Find the value of c and d.

$\displaystyle 3d=13-2c$

$\displaystyle \frac{3c+d}{2}=8$

Rearranged, simplified and multiply each equation by 2:

$\displaystyle 6d+4c=26$

$\displaystyle d+3c=16$

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

$\displaystyle 18d+12c=78$

$\displaystyle -4d-12c=-64$

Then add them and find what d is worth:

$\displaystyle 14d=14$

$\displaystyle d=1$

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

$\displaystyle 3(1)+2c=13$

$\displaystyle 3+2(c)=13$

$\displaystyle c=5$

$\displaystyle d=1, c=5$

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

Re: Linear equations addition

You aren't doing anything wrong, that solution is correct. Perhaps you entered the fraction incorrectly into Wolfram Alpha

Re: Linear equations addition

You could have avoided one step and that is multiplying the first equation by 2.

Re arrange the equations and we have

2c + 3d = 13 ---- (1)

3c + d = 16 ------(2)

Multiply the second equation by -3 and re write the equations

2c + 3d = 13 ---- (1)

-9c -3 d = -48 ------(3)

Add the equations and you have:

-7c = -35 i.e., c = 5. Now get d from equation (2) we get d = 1